text
stringlengths 60
1.05M
| metadata
stringlengths 1.02k
2.67k
|
|---|---|
# MATH 433 Applied Algebra Lecture 26: Order of an element in a group.
```MATH 433
Applied Algebra
Lecture 26:
Order of an element in a group.
Subgroups.
Groups
Definition. A group is a set G , together with a binary
operation ∗, that satisfies the following axioms:
(G1: closure)
for all elements g and h of G , g ∗ h is an element of G ;
(G2: associativity)
(g ∗ h) ∗ k = g ∗ (h ∗ k) for all g , h, k ∈ G ;
(G3: existence of identity)
there exists an element e ∈ G , called the identity (or unit)
of G , such that e ∗ g = g ∗ e = g for all g ∈ G ;
(G4: existence of inverse)
for every g ∈ G there exists an element h ∈ G , called the
inverse of g , such that g ∗ h = h ∗ g = e.
The group (G , ∗) is said to be commutative (or Abelian) if
(G5: commutativity) g ∗ h = h ∗ g for all g , h ∈ G .
Basic properties of groups
• The identity element is unique.
• The inverse element is unique.
• (g −1)−1 = g . In other words, h = g −1 if and
only if g = h−1.
• (gh)−1 = h−1g −1.
• (g1g2 . . . gn )−1 = gn−1 . . . g2−1g1−1.
• Cancellation properties: gh1 = gh2 =⇒
h1 = h2 and h1 g = h2 g =⇒ h1 = h2 for all
g , h1, h2 ∈ G .
Indeed, gh1 = gh2 =⇒ g −1 (gh1 ) = g −1 (gh2 )
=⇒ (g −1 g )h1 = (g −1 g )h2 =⇒ eh1 = eh2 =⇒ h1 = h2 .
Similarly, h1 g = h2 g =⇒ h1 = h2 .
Equations in groups
Theorem Let G be a group. For any a, b, c ∈ G ,
• the equation ax = b has a unique solution
x = a−1b;
• the equation ya = b has a unique solution
y = ba−1;
• the equation azc = b has a unique solution
z = a−1bc −1 .
Problem. Solve an equation in the group S(5):
(1 2 4)(3 5)π(2 3 4 5) = (1 5).
−1
Solution: π = (1 2 4)(3 5) (1 5)(2 3 4 5)−1
= (3 5)−1 (1 2 4)−1 (1 5)(2 3 4 5)−1
= (5 3)(4 2 1)(1 5)(5 4 3 2) = (1 3)(2 4 5).
Powers of an element
Let g be an element of a group G . The positive powers of g
are defined inductively:
g 1 = g and g k+1 = g · g k for every integer k ≥ 1.
The negative powers of g are defined as the positive powers of
its inverse: g −k = (g −1)k for every positive integer k.
Finally, we set g 0 = e.
Theorem Let g be an element of a group G and r , s ∈ Z.
Then
(i) g r g s = g r +s ,
(ii) (g r )s = g rs ,
(iii) (g r )−1 = g −r .
Idea of the proof: First one proves the theorem for positive
r , s by induction (induction on r for (i) and (iii), induction on
s for (ii) ). Then the general case is reduced to the case of
positive r , s.
Order of an element
Let g be an element of a group G . We say that g has finite
order if g n = e for some positive integer n.
If this is the case, then the smallest positive integer n with this
property is called the order of g and denoted o(g ).
Otherwise g is said to have the infinite order, o(g ) = ∞.
Theorem If G is a finite group, then every element of G has
finite order.
Proof: Let g ∈ G and consider the list of powers:
g , g 2, g 3 , . . . . Since all elements in this list belong to the
finite set G , there must be repetitions within the list. Assume
that g r = g s for some 0 < r < s. Then g r e = g r g s−r
=⇒ g s−r = e due to the cancellation property.
Theorem 1 Let G be a group and g ∈ G be an element of
finite order n. Then g r = g s if and only if r ≡ s mod n.
In particular, g r = e if and only if the order n divides r .
Theorem 2 Let G be a group and g ∈ G be an element of
infinite order. Then g r 6= g s whenever r 6= s.
Theorem 3 Let g and h be two commuting elements of a
group G : gh = hg . Then
(i) the powers g r and hs commute for all r , s ∈ Z,
(ii) (gh)r = g r hr for all r ∈ Z.
Theorem 4 Let G be a group and g , h ∈ G be two
commuting elements of finite order. Then gh also has a
finite order. Moreover, o(gh) divides lcm o(g ), o(h) .
Examples
• G = S(10), g = (1 2 3 4 5 6), h = (7 8 9 10).
g and h are disjoint cycles, in particular, gh = hg .
We have o(g ) = 6, o(h) = 4, and
o(gh) = lcm(o(g ), o(h)) = 12.
• G = S(6), g = (1 2 3 4 5 6),
h = (1 3 5)(2 4 6).
Notice that h = g 2 . Hence gh = hg = g 3 = (1 4)(2 5)(3 6).
We have o(g ) = 6, o(h) = 3, and
o(gh) = 2 < lcm(o(g ), o(h)).
• G = S(5), g = (1 2 3), h = (3 4 5).
gh = (1 2 3 4 5), hg = (1 2 4 5 3) 6= gh.
We have o(g ) = o(h) = 3 and o(gh) = o(hg ) = 5.
Subgroups
Definition. A group H is a called a subgroup of a group G if
H is a subset of G and the group operation on H is obtained
by restricting the group operation on G .
Theorem Let H be a nonempty subset of a group G and
define an operation on H by restricting the group operation of
G . Then the following are equivalent:
(i) H is a subgroup of G ;
(ii) H is closed under the operation and under taking the
inverse, that is, g , h ∈ H =⇒ gh ∈ H and
g ∈ H =⇒ g −1 ∈ H;
(iii) g , h ∈ H =⇒ gh−1 ∈ H.
Corollary If H is a subgroup of G then (i) the identity
element in H is the same as the identity element in G ; (ii) for
any g ∈ H the inverse g −1 taken in H is the same as the
inverse taken in G .
Examples of subgroups: • (Z, +) is a subgroup of (R, +).
• (Q \ {0}, ×) is a subgroup of (R \ {0}, ×).
• The alternating group A(n) is a subgroup of the symmetric
group S(n).
• The special linear group SL(n, R) is a subgroup of the
general linear group GL(n, R).
• Any group G is a subgroup of itself.
• If e is the identity element of a group G , then {e} is the
trivial subgroup of G .
Counterexamples: • (R \ {0}, ×) is not a subgroup of
(R, +) since the operations do not agree.
• (Zn , +) is not a subgroup of (Z, +) since Zn is not a
subset of Z (although every element of Zn is a subset of Z).
• (Z \ {0}, ×) is not a subgroup of (R \ {0}, ×) since
(Z \ {0}, ×) is not a group.
```
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://studylib.net/doc/10503835/math-433-applied-algebra-lecture-26--order-of-an-element-...", "fetch_time": 1619016400000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618039544239.84/warc/CC-MAIN-20210421130234-20210421160234-00017.warc.gz", "warc_record_offset": 641100470, "warc_record_length": 11630, "token_count": 2006, "char_count": 5533, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2021-17", "snapshot_type": "latest", "language": "en", "language_score": 0.8440600037574768, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00020-of-00064.parquet"}
|
# BAB III LIMIT FUNGSI DAN KEKONTINUAN - PowerPoint PPT Presentation
1 / 17
BAB III LIMIT FUNGSI DAN KEKONTINUAN. 4.4 Limit fungsi trigonometri. Bukti Perhatikan Gambar 4.4 berikut !. y. T. Q. r. . x. 0. P. Gambar 4.4. Luas OPQ < Sektor OPQ < OPT (*). (**). (***). (****).
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
BAB III LIMIT FUNGSI DAN KEKONTINUAN
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
#### Presentation Transcript
BAB III
LIMIT FUNGSI DAN KEKONTINUAN
4.4 Limit fungsitrigonometri
Bukti
PerhatikanGambar 4.4 berikut!
y
T
Q
r
x
0
P
Gambar 4.4
LuasOPQ < Sektor OPQ < OPT (*)
(**)
(***)
(****)
Substitusipersamaan (**) s/d (****) kepersamaan (*) didapat,
Gunakanteoremaapit!
(4.16)
(4.17)
(4.18)
Bukti
(terbukti)
(4.19)
Bukti
Bukti
Bukti
Bukti
Bukti
3.5 Limit fungsitrigonometriinvers
(4.22)
Bukti
(4.22)
Bukti
(4.22)
Bukti
(4.24)
Bukti
(4.25)
Bukti
(4.26)
Bukti
(4.27)
Bukti
3.6 Limit takhingga
mungkinakandidapatbahwa f(x) membesarataumengecil
berikut.
y
x
0
2
Gambar 4.5
titik 2 dariarahkananmaka f(x) membesartanpabatas
• Sedangkanpadasaat x mendekati 2 dariarahkirimaka f(x)
• bahwa limit f(x) untuk x mendekati 2 dariarahkanan
Sedangkan limit f(x) untuk x mendekati 2 dariarahkiriadalah –
Karena limit kiri limit kanan, makatidakada
(lihatpersamaan 4.14)
Untukmemecahkan limit takhinggaperhatikanteoremaberikut!
Bukti
Jikasemuasukudibagidenganxmmaka,
Jika m < n, maka
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.slideserve.com/charla/bab-iii-limit-fungsi-dan-kekontinuan", "fetch_time": 1493336042000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-17/segments/1492917122629.72/warc/CC-MAIN-20170423031202-00626-ip-10-145-167-34.ec2.internal.warc.gz", "warc_record_offset": 666129547, "warc_record_length": 16791, "token_count": 719, "char_count": 1931, "score": 4.0, "int_score": 4, "crawl": "CC-MAIN-2017-17", "snapshot_type": "longest", "language": "en", "language_score": 0.25456658005714417, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00052-of-00064.parquet"}
|
# Maximum Likelihood Estimate for Linear Gaussian Conditional Probability Distributions in Python
by Elias Hernandis • Published Feb. 6, 2023 • Tagged maths, probability, statistics
A random variable $X$ has a linear Gaussian conditional probability distribution on its set of parents $\{U_1, \ldots, U_k\}$ if
$P(X \mid U_1, \ldots, U_k ) = \mathcal N (w_0 + w_1 U_1 + \ldots + w_k U_k, \sigma^2)$
where $w_0, \ldots, w_k \in \mathbb R$ are generally known as weights.
Suppose we have some data where each tuple is an observation of $X$ and $U_1, \ldots, U_k$. Then we can estimate the parameters $w_0, \ldots, w_k, \sigma^2$ using maximum likelihood relatively easily.
The detailed mathematical explanation of how this is done can be found in Section 17.2.4 of Koller and Friedman. Here we just provide which works by using the sufficient statistics outlined in that method (and hence is quite fast!).
from typing import Any, Sequence
import numpy as np
import pandas as pd
def mle(self, node: Any, parents: Sequence[Any], data: pd.DataFrame):
'''
Find maximum likelihood estimate for mean, variance and weights given
the data and the dependencies on the parents.
Parameters
----------
data : pandas.DataFrame
A DataFrame with one row per observation and one column per
variable.
Returns
-------
A two-tuple where the first argument is the vector of weights and the second
is the variance.
Notes
-----
Maximum likelihood estimation of parameters is computed using the
*sufficient statistics* approach described in section 17.2.4 of [1]_.
References
----------
.. [1] D. Koller and N. Friedman, Probabilistic graphical models:
principles and techniques. Cambridge, MA: MIT Press, 2009.
'''
M = len(data)
k = len(parents)
x_sum = data[node].sum()
u_sums = data[list(parents)].sum().to_numpy()
xu_sums = [(data[node] * data[p]).sum() for p in parents]
uu_sums = [[(data[ui] * data[uj]).sum() for uj in parents]
for ui in parents]
# solve A*beta = b
A = np.block([[np.reshape([M], (1, 1)),
np.reshape(u_sums, (1, k))],
[np.reshape(u_sums, (k, 1)),
np.reshape(uu_sums, (k, k))]])
b = [x_sum] + xu_sums
beta = np.linalg.solve(A, b)
# extract parameters
mean, weights = beta[0], beta[1:]
x_var = data[node].var()
cov_d = data[list(parents)].cov()
var = x_var - sum([
sum([
weights[i] * weights[j] * cov_d[pi][pj]
for j, pj in enumerate(parents)
]) for i, pi in enumerate(parents)
])
return beta, var
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://hernandis.me/2023/02/06/mle-for-linear-gaussian-cpds-in-python.html?ref=previous-post", "fetch_time": 1714034437000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712297290384.96/warc/CC-MAIN-20240425063334-20240425093334-00803.warc.gz", "warc_record_offset": 270564580, "warc_record_length": 6705, "token_count": 675, "char_count": 2412, "score": 3.546875, "int_score": 4, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.7721028923988342, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00054-of-00064.parquet"}
|
Sunteți pe pagina 1din 3
# Lesson Plan in Mathematics 9 August 8, 2017
LC: transforms the quadratic function defined by y = ax2 + bx + c into the form y = a(x – h)2 + k.
I. Objectives:
At the end of 60 minutes, with 75% proficiency, the students will be able to:
a. transform the quadratic function defined by y=ax2+bx+c into the form
y=a(x-h)2+k
b. identify the values of h and k.
## II. Subject Matter:
a. Topic: Transforming the quadratic function defined by y=ax2+bx+c into the form
y=a(x-h)2+k
b. Reference: Mathematics Learner’s Material 9, pp. 129-132 ; CG: (M9AL-Ih-1)
c. Instructional Materials: Cartolina, Manila Paper, Chalk
d. Value Focus: Accuracy and Cooperation
III. Procedure:
Teachers’ Activity Students Activity
A. PRE-ACTIVITY
1. Review
Determine the number that must be added to make each
a perfect square trinomial.
a. x2 - 4x +__ c. x2 + 6x +__ a. x2 - 4x + 4 c. x2 + 6x + 9
b. x2 + 3x +__ d. x2 +10x +__ b. x2 + 3x + 9/4 d. x2 +10x + 25
2. Motivation
Write each in factored form (x-h)2
a. x2 - 4x + 4 c. x2 + 6x + 9 a. ( x – 2 )2 c. ( x + 3)2
b. x2 + 3x + 9/4 d. x2 +10x +25 b. ( x + 3/2 )2 d. (x + 5)2
B. ACTIVITY
Setting of Standards:
1. Find a Partner.
2. Get ½ crosswise sheet of padpaper.
4. 5 minutes are allotted time to finish the activity.
Activity Proper:
Transform the given quadratic functions into the
form y=a(x-h)2+k by following the steps below. Fill in the
blanks. Work in pairs.
1. y = x2 - 4x + 10 Rewrite the function in this
2
2. y = x + 10x + 9 manner:
1. y= (x2 - 4x + ______) + 10 - ___
Steps Task y = (x2 – 4x + 4) +10 - 4
1. Group the terms containing = (x-2)2 + 6
‘x’. h=2;k=6
2. Complete the expression in 2. y= (x2 + 10x + ______) + 9 - ___
parenthesis to make it a perfect y = (x2 + 10x + 25) + 9-25
square trinomial. = (x+5)2 -16
3. Express the perfect square h = -5 ; k = -16
trinomial as the square of
binomial.
4. Give the value of h and k.
C. POST ACTIVITY
1. Analysis
a. How did you determine the number that must be
b. Why do we need to transform the general form
of quadratic function to its standard form or vertex form?
c. What are the values of h and k? How do we 1. Group the terms containing ‘x’.
identify the value of h and k? 2. Complete the expression in
parenthesis to make it a perfect
2. Abstraction square trinomial.
a. How do we transform the quadratic function 3. Express the perfect square
2 2
defined by y=ax +bx+c into the form y=a(x-h) +k? What trinomial as the square of
are the steps? binomial.
4. Give the value of h and k.
3. Application: 1. y=(x2 + 6x + 9) +3 – 9
Transform the given quadratic function into its y=(x+3)2 - 6
vertex form. Identify the value of h and k. h = -3 ; k = -6
1. y = x2 + 6x + 3 2. y = x2 + 2x + 2 2. y=(x2 + 2x + 1) + 2– 1
y=(x-1)2 +1
h=1;k=1
IV. Assessment:
Direction: Transform the given quadratic function into its vertex form. Identify the value of h
and k.
1. y = x2 + 10x + 4 2. y = x2 + 4x + 2
V. Assignment
Direction: Identify the values of h and k of the ff. quadratic functions. Transform into Standard
form if needed.
1. y = x2
2. y = 2x2 + 4x – 3
3. y = ( x + 2 )2 + 3
4. f(x) = 2 (x - 2)2
Prepared by:
Salazar F. Española
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://ro.scribd.com/document/447701880/Lesson-Plan-in-Mathematics-9-August-8-docx", "fetch_time": 1603228242000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-45/segments/1603107874135.2/warc/CC-MAIN-20201020192039-20201020222039-00109.warc.gz", "warc_record_offset": 523825939, "warc_record_length": 77594, "token_count": 1162, "char_count": 3158, "score": 4.375, "int_score": 4, "crawl": "CC-MAIN-2020-45", "snapshot_type": "latest", "language": "en", "language_score": 0.6739565134048462, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00000-of-00064.parquet"}
|
# Thread: Volume of solid of revolution.
1. ## Volume of solid of revolution.
1. Find the volume of the solid obtained by rotating the region bounded by the curve
y^2 - y^3 - x = 0, x = 0
about the line x = 0
2. Integral of (cos x + xsinx)/x(x+cosx) dx
2. Originally Posted by devil10078
1. Find the volume of the solid obtained by rotating the region bounded by the curve y^2 - y^3 - x = 0, x = 0 about the line x = 0
The intersection points are $(0,0)$ and $(0,1)$ . Then,
$V=\pi\displaystyle\int_0^1(y^2-y^3)^2\;dy=\ldots$
Fernando Revilla
3. $\displaystyle \frac{\cos{x} + x\sin{x}}{x(x+\cos{x})}=-\left(\frac{-\cos{x} - x\sin{x}}{x^2 + x\cos{x}}\right)$
$\displaystyle = -\left(\frac{2x + \cos{x} - x\sin{x} - 2x- 2\cos{x} }{x^2 + x\cos{x}}\right)$
$\displaystyle = -\left(\frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}\right) - \left(\frac{-2x - 2\cos{x}}{x^2 + x\cos{x}}\right)$
$\displaystyle = -\left(\frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}\right) + \frac{2(x + \cos{x})}{x(x + \cos{x})}$
$\displaystyle = \frac{2}{x} - \frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}$.
Therefore $\displaystyle \int{\frac{\cos{x} + x\sin{x}}{x(x + \cos{x})}\,dx} = \int{\frac{2}{x} - \frac{2x + \cos{x} - x\sin{x}}{x^2 + x\cos{x}}\,dx}$.
To integrate the second term, make the substitution $\displaystyle u = x^2 + x\cos{x}$.
4. good call, but it's actually easier than that, since
\begin{aligned}
\int{\frac{\cos x+x\sin x}{x(x+\cos x)}\,dx}&=\int{\frac{x+\cos x+x\sin x-x}{x(x+\cos x)}\,dx} \\
& =\ln \left| x \right|-\int{\frac{1-\sin x}{x+\cos x}\,dx} \\
& =\ln \left| x \right|-\ln \left| x+\cos x \right|+k \\
& =\ln \left| \frac{x}{x+\cos x} \right|+k.
\end{aligned}
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/calculus/166384-volume-solid-revolution.html", "fetch_time": 1505959434000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-39/segments/1505818687592.20/warc/CC-MAIN-20170921011035-20170921031035-00026.warc.gz", "warc_record_offset": 212858970, "warc_record_length": 10756, "token_count": 697, "char_count": 1689, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2017-39", "snapshot_type": "longest", "language": "en", "language_score": 0.476386696100235, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00003-of-00064.parquet"}
|
# estimate value of $\sqrt[30]{0.05}$
Yesterday I got an exam in which there was a problem and its solution results in $$\sqrt[30]{0.05}$$
I didn't go further calculation. Still I can't.
My lecturer said, even I'm still not sure if he made ironic humor, that if there is a math operation requiring higher than mid-school level math you to do, not do that, let it leave as it is. As a note, my department is not math.
Did he really make a humor?
Is there way/methods to figure out/estimate its result without calculator w.r.t thinking in exam(time limit) and not in exam?
• I don't think that there's a nice way to evaluate $$\sqrt[30]{0.05}$$ Have a look here Apr 9, 2019 at 9:31
• Are you allowed log tables or slide rules in the exam? Apr 9, 2019 at 9:32
• @Henry we are never allowed to make use of any additional sources sir. Apr 9, 2019 at 9:33
• @Dr.Mathva I tried it to see maybe wolfram gives hint to solve that, but don't. Apr 9, 2019 at 9:34
• There are methods to calculate this such as Taylor Series expansion, or using the n-th root formula or using Newton's approximation or using Log function (or its expansion), but non results-in a simple calculation. I don't think in high school you know about any of this with the exception of Log function which may be what you should use. Some reference is here: en.wikipedia.org/wiki/Nth_root Apr 9, 2019 at 11:15
We know that $$\sqrt[30]{0.05}$$ is a number a little smaller than $$1$$ because $$\sqrt[n]{0.05}$$ converges to $$1$$ for $$n$$ to infinity. So set $$\sqrt[30]{0.05}=1-a$$ and then try to estimate $$a$$. $$a$$ satisfies the equation $$(1-a)^{30}=\frac{1}{20}$$ Writing out the first few terms gives $$1-30a+\frac{30\cdot 29}{2}a^2 + ...= \frac{1}{20}$$ Note that because $$a$$ is small, the further coefficients are decreasing quickly.
Using just $$1-30a\sim \frac{1}{20}$$ yields $$\sqrt[30]{0.05} \sim 0.93$$ without calculator. If you use more terms you should get better approximations.
Edit: It turns out this doesn't work quite as well as I thought. While it is true that the later coefficients with higher powers of $$a$$ are decreasing quickly, the biggest coefficient in the series is at $$a^3$$. So in order to get something that is actually an approximation of $$a$$ one would have to compute at least until $$a^4$$ or $$a^5$$. This leads to a polynomial which is not really easy to solve by hand. Computing further terms would increase accuracy but I'm not sure whether this is helpful in a no calculator scenario.
• However, if you try to use the quadratic term, we do not get a solution because the discriminant $\Delta < 0$. So, you should use more terms which entails very complicated computations compared to the original one. Conclusion: your method "works" only for a linear approximation. Apr 9, 2019 at 9:58
• I get $a \approx 0.03$ and so $\sqrt[30]{0.05} \approx 0.97$.
– lhf
Apr 9, 2019 at 10:01
• @lhf, since the the real value is around $0.9049661$, even the linear approximation does not work very well . So, I do not understand the massive upvotes. Apr 9, 2019 at 10:17
You can take it step-by-step: $$\sqrt[30]{0.05}=\sqrt[5]{\sqrt[3]{\sqrt{\frac5{100}}}}\approx\sqrt[5]{\sqrt[3]{\frac{2.2}{10}}}=\sqrt[5]{\sqrt[3]{\frac{220}{1000}}}\approx \sqrt[5]{\frac{6}{10}}=\sqrt[5]{\frac{60000}{100000}}\approx\frac9{10}.$$
This is how the ancients might have attempted this, using the method of false position.
Solve $$x^2=0.05$$, starting with an initial guess of $$x=0.2$$.
$$0.05=(0.2+h)^2\approx0.04+0.4h\implies h=0.02$$ $$0.05=(0.22+h)^2\approx0.0484+0.44h\implies h=0.0036$$ So $$x=0.2236$$.
Now solve $$y^3=0.2236$$, starting with $$0.6$$.
$$0.2236=(0.6+h)^3\approx0.216+1.08h\implies h=0.007$$ $$0.2236=(0.607+h)^3\approx0.2236+1.1h\implies h=0$$
Finally solve $$z^5=0.607$$, starting with $$0.9$$.
$$0.607=(0.9+h)^5\approx0.5905+3.28h\implies h=0.005$$ so $$z\approx0.905$$.
Disclaimer: I used a calculator!
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 33, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/3180773/estimate-value-of-sqrt300-05", "fetch_time": 1709338861000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-10/segments/1707947475711.57/warc/CC-MAIN-20240301225031-20240302015031-00655.warc.gz", "warc_record_offset": 377876260, "warc_record_length": 36815, "token_count": 1243, "char_count": 3928, "score": 3.625, "int_score": 4, "crawl": "CC-MAIN-2024-10", "snapshot_type": "latest", "language": "en", "language_score": 0.936633825302124, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00047-of-00064.parquet"}
|
### Angle
An angle is the figure formed by two rays sharing a common endpoint, called the vertex of the angle.
The size of an angle
Imagine that the ray OB is rotated about the point O until it lies along OA. The amount of turning is called the size of the angle AOB.
A revolution is the amount of turning required to rotate a ray about its endpoint until it falls back onto itself. The size of 1 revolution is 360o.
A straight angle is the angle formed by taking a ray and its opposite ray. A straight angle is half of a revolution, and so has size equal to
180o.
Right angle
Let AOB be a line, and let OX be a ray making equal angles with the ray OA and the ray OB. Then the equal angles ∠AOX and ∠BOX are called right angles.
A right angle is half of a straight angle, and so is equal to 90o.
Classification of angles
Angles are classified according to their size.
We say that
• An angle with size α is acute if 0o < α < 90o,
• An angle with size α is obtuse if 900 < α < 180o,
• An angle with size α is reflex if 1800 < α < 360o
Two angles at a point are called adjacent if they share a common ray and a common vertex and lie on opposite sides of the common ray.
Hence, in the diagram,
• AOC and ∠BOC are adjacent
Two angles that add to 90o are called complementary. For example, 23o and 67o are complementary angles.
In each diagram the two marked angles are called corresponding angles.
If the lines are parallel, then each pair of corresponding angles are equal.
Conversely, if a pair of corresponding angles are equal, then the lines are parallel.
Two angles that add to 180o are called supplementary angles. For example, 45o and 135o are supplementary angles.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://k10outline.scsa.wa.edu.au/home/teaching/curriculum-browser/mathematics-v8/overview/glossary/angle", "fetch_time": 1669673355000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-49/segments/1669446710662.60/warc/CC-MAIN-20221128203656-20221128233656-00112.warc.gz", "warc_record_offset": 386865298, "warc_record_length": 11171, "token_count": 413, "char_count": 1688, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2022-49", "snapshot_type": "latest", "language": "en", "language_score": 0.9299619793891907, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00023-of-00064.parquet"}
|
# Understanding Ordered Factors in a Linear Model
Consider the following data from the text Design and Analysis of Experiments, 7 ed (Montgomery, Table 3.1). It has two variables: power and rate. Power is a discrete setting on a tool used to etch circuits into a silicon wafer. There are four levels to choose from. Rate is the distance etched measured in Angstroms per minute. (An Angstrom is one ten-billionth of a meter.) Of interest is how (or if) the power setting affects the etch rate. Below we enter the small data set by hand into R and create a quick plot. The plot suggests higher power leads to higher etch rates.
# Table 3.1
power <- rep(c(160, 180, 200, 220), 5)
rate <- c(575, 565, 600, 725,
542, 593, 651, 700,
530, 590, 610, 715,
539, 579, 637, 685,
570, 610, 629, 710)
etch <- data.frame(rate, power)
stripchart(rate ~ power, data = etch, pch = 1,
vertical = TRUE, xlab = "power")
A formal statistical analysis of this data requires a linear model. To perform the analysis in R we need to define the power variable as a factor. This tells R that power is a categorical variable and to treat it as such in a modeling procedure. But in addition, we may want to further specify that the levels of power are ordered. Power level 160 is less than 180, and 180 is less than 200, and so on. This is additional information we may want to account for. We can define power as an ordered factor in R using the ordered function. We do that below and save the ordered factor version as “powerF”. Notice that calling head to view the first 6 values of powerF shows us the ordering of the levels: 160 < 180 < 200 < 220
etch$powerF <- ordered(etch$power)
head(etch$powerF) ## [1] 160 180 200 220 160 180 ## Levels: 160 < 180 < 200 < 220 It’s important to note that R is storing powerF internally as integers, which we can see by calling unclass or as.integer on the variable. (We’ll make use of this fact in the next plot we create.) unclass(etch$powerF)
## [1] 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
## attr(,"levels")
## [1] "160" "180" "200" "220"
Now let’s formally analyze the data using the lm function in R. This is equivalent to performing a one-way ANOVA. Below we model rate as a function of the ordered factor “powerF” and pipe the result in the summary function using base R’s native forward pipe operator, |>, introduced in version 4.1.0.
lm(rate ~ powerF, data = etch) |> summary()
##
## Call:
## lm(formula = rate ~ powerF, data = etch)
##
## Residuals:
## Min 1Q Median 3Q Max
## -25.4 -13.0 2.8 13.2 25.6
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 617.750 4.085 151.234 < 2e-16 ***
## powerF.L 113.011 8.169 13.833 2.56e-10 ***
## powerF.Q 22.700 8.169 2.779 0.0134 *
## powerF.C 9.347 8.169 1.144 0.2694
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 18.27 on 16 degrees of freedom
## Multiple R-squared: 0.9261, Adjusted R-squared: 0.9122
## F-statistic: 66.8 on 3 and 16 DF, p-value: 2.883e-09
At the bottom of the output is the familiar ANOVA test result. The F-statistic is large and the p-value is tiny which provides evidence that the mean rates between the four power levels are different. (We have statistically sanctified what we saw in the plot.) In addition the adjusted R-squared suggests power explains about 91% of the variability we’re seeing in the etch rates. But what about the rest? What are these mysterious “power.L”, “power.Q”, and “power.C” labels? What do their estimates mean? What are the hypothesis tests associated with them?
The quick answer to these questions is that the L, Q and C stand for Linear, Quadratic, and Cubic; that the coefficients don’t have a ready interpretation; and that each hypothesis is a test for trend.
The “power.L” entry is a test for linear trend. The null is no linear trend (ie, a flat line). The large t value provides evidence against this. Again, this is not a surprise. Looking at the plot above we can visualize a straight line over the points.
The “power.Q” entry is a test for quadratic trend. The null is no quadratic trend (ie, a straight line). The t value of 2.8 provides some evidence against this. This is less obvious in the plot, though we can perhaps imagine a slight curve superimposed over the data.
The “power.C” entry is a test for cubic trend. The null is no cubic trend (ie, a straight or quadratic line). The t value of 1.1 is small and the p-value is quite large. There doesn’t appear to be any cubic trend in the data.
We can visualize these trends using ggplot. Below we call geom_smooth three times to plot linear, quadratic and cubic lines using lm. Notice we have to unclass the powerF variable to make this work. Defining the colors as “1”, “2” and “3” is an easy way to force ggplot to create a legend and place the labels in that order. We notice there isn’t much difference between the fitted quadratic and cubic lines, visually demonstrating why the cubic trend test was not “significant”.
library(ggplot2)
ggplot(etch) +
aes(x = powerF, y = rate) +
geom_point(shape = 1) +
geom_smooth(aes(x = unclass(powerF), color = "1"),
formula = y ~ x,
method = lm, se = FALSE) +
geom_smooth(aes(x = unclass(powerF), color = "2"),
formula = y ~ poly(x, 2),
method = lm, se = FALSE) +
geom_smooth(aes(x = unclass(powerF), color = "3"),
formula = y ~ poly(x, 3),
method = lm, se = FALSE) +
scale_color_discrete("Trend", labels = c("linear", "quadratic", "cubic")) +
theme_classic()
If you’re thinking these tests for trend look a lot like a polynomial model, you’re right. That’s basically what it is. In fact we get identical test results (ie, same test statistics and p-values) if we simply fit a 3-degree polynomial model to the original numeric power variable.
lm(rate ~ poly(power, 3), data = etch) |> summary()
##
## Call:
## lm(formula = rate ~ poly(power, 3), data = etch)
##
## Residuals:
## Min 1Q Median 3Q Max
## -25.4 -13.0 2.8 13.2 25.6
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 617.750 4.085 151.234 < 2e-16 ***
## poly(power, 3)1 252.700 18.267 13.833 2.56e-10 ***
## poly(power, 3)2 50.759 18.267 2.779 0.0134 *
## poly(power, 3)3 20.900 18.267 1.144 0.2694
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 18.27 on 16 degrees of freedom
## Multiple R-squared: 0.9261, Adjusted R-squared: 0.9122
## F-statistic: 66.8 on 3 and 16 DF, p-value: 2.883e-09
We also get the same test results if we unclass the ordered factor version of power and fit a 3-degree polynomial model.
lm(rate ~ poly(unclass(powerF), 3), data = etch) |> summary()
##
## Call:
## lm(formula = rate ~ poly(unclass(powerF), 3), data = etch)
##
## Residuals:
## Min 1Q Median 3Q Max
## -25.4 -13.0 2.8 13.2 25.6
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 617.750 4.085 151.234 < 2e-16 ***
## poly(unclass(powerF), 3)1 252.700 18.267 13.833 2.56e-10 ***
## poly(unclass(powerF), 3)2 50.759 18.267 2.779 0.0134 *
## poly(unclass(powerF), 3)3 20.900 18.267 1.144 0.2694
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 18.27 on 16 degrees of freedom
## Multiple R-squared: 0.9261, Adjusted R-squared: 0.9122
## F-statistic: 66.8 on 3 and 16 DF, p-value: 2.883e-09
Did you notice the coefficients and standard errors of these two models differ from the first model with the ordered factor? That’s because the ordered factor model uses a contrast. A contrast is a matrix that transforms a series of 0/1 dummy variables into columns that can be estimated in a modeling routine. The default contrast for ordered factors in R is the polynomial contrast. We can see the contrast R uses by calling the contr.poly function. Simply tell it how many levels your categorical variable has. In our case we have 4.
contr.poly(4)
## .L .Q .C
## [1,] -0.6708204 0.5 -0.2236068
## [2,] -0.2236068 -0.5 0.6708204
## [3,] 0.2236068 -0.5 -0.6708204
## [4,] 0.6708204 0.5 0.2236068
That’s the contrast used to convert dummy variables of an ordered factor into a model matrix. Let’s demonstrate. Below we generate a 4-column matrix of dummy variables for the power variable. There is one row per observation and one column per level of power. A 1 in a row identifies membership in one of the four power categories.
Xpower <- sapply(c(160, 180, 200, 220),
function(x)ifelse(etch$power==x, 1, 0)) colnames(Xpower) <- c("160", "180", "200", "220") head(Xpower) ## 160 180 200 220 ## [1,] 1 0 0 0 ## [2,] 0 1 0 0 ## [3,] 0 0 1 0 ## [4,] 0 0 0 1 ## [5,] 1 0 0 0 ## [6,] 0 1 0 0 Now we multiply the dummy variable matrix by the polynomial contrast to obtain a model matrix that is estimable. Xpower %*% contr.poly(4) ## .L .Q .C ## [1,] -0.6708204 0.5 -0.2236068 ## [2,] -0.2236068 -0.5 0.6708204 ## [3,] 0.2236068 -0.5 -0.6708204 ## [4,] 0.6708204 0.5 0.2236068 ## [5,] -0.6708204 0.5 -0.2236068 ## [6,] -0.2236068 -0.5 0.6708204 ## [7,] 0.2236068 -0.5 -0.6708204 ## [8,] 0.6708204 0.5 0.2236068 ## [9,] -0.6708204 0.5 -0.2236068 ## [10,] -0.2236068 -0.5 0.6708204 ## [11,] 0.2236068 -0.5 -0.6708204 ## [12,] 0.6708204 0.5 0.2236068 ## [13,] -0.6708204 0.5 -0.2236068 ## [14,] -0.2236068 -0.5 0.6708204 ## [15,] 0.2236068 -0.5 -0.6708204 ## [16,] 0.6708204 0.5 0.2236068 ## [17,] -0.6708204 0.5 -0.2236068 ## [18,] -0.2236068 -0.5 0.6708204 ## [19,] 0.2236068 -0.5 -0.6708204 ## [20,] 0.6708204 0.5 0.2236068 Notice this is the same model.matrix that is used when we model the ordered factor with lm. (The column of ones is for the intercept.) lm(rate ~ powerF, data = etch) |> model.matrix() ## (Intercept) powerF.L powerF.Q powerF.C ## 1 1 -0.6708204 0.5 -0.2236068 ## 2 1 -0.2236068 -0.5 0.6708204 ## 3 1 0.2236068 -0.5 -0.6708204 ## 4 1 0.6708204 0.5 0.2236068 ## 5 1 -0.6708204 0.5 -0.2236068 ## 6 1 -0.2236068 -0.5 0.6708204 ## 7 1 0.2236068 -0.5 -0.6708204 ## 8 1 0.6708204 0.5 0.2236068 ## 9 1 -0.6708204 0.5 -0.2236068 ## 10 1 -0.2236068 -0.5 0.6708204 ## 11 1 0.2236068 -0.5 -0.6708204 ## 12 1 0.6708204 0.5 0.2236068 ## 13 1 -0.6708204 0.5 -0.2236068 ## 14 1 -0.2236068 -0.5 0.6708204 ## 15 1 0.2236068 -0.5 -0.6708204 ## 16 1 0.6708204 0.5 0.2236068 ## 17 1 -0.6708204 0.5 -0.2236068 ## 18 1 -0.2236068 -0.5 0.6708204 ## 19 1 0.2236068 -0.5 -0.6708204 ## 20 1 0.6708204 0.5 0.2236068 ## attr(,"assign") ## [1] 0 1 1 1 ## attr(,"contrasts") ## attr(,"contrasts")$powerF
## [1] "contr.poly"
The latter two models that explicitly modeled a 3-degree polynomial did not use a contrast. Instead the actual power values were squared and cubed.
Is there an advantage to using one over the other? Maybe we should have just used polynomial regression instead of creating an ordered factor? Fox and Weisberg (2019) tell us contr.poly is appropriate when the levels of your ordered factor are equally spaced and balanced. That is the case for the power variable. Each level is separated by 20 units and we have five observations for each power level. If we have an ordered factor that does not have equal spacing and/or is not balanced, we’re better off performing polynomial regression via the poly function.
It’s worth noting that even though we have power saved as an ordered factor and modeled with a polynomial contrast, we can still perform follow-up analyses usually associated with standard unordered categorical variables. For example, using the emmeans package we can make pairwise comparisons.
library(emmeans)
lm(rate ~ powerF, data = etch) |>
emmeans(specs = "powerF") |>
pairs()
## contrast estimate SE df t.ratio p.value
## 160 - 180 -36.2 11.6 16 -3.133 0.0294
## 160 - 200 -74.2 11.6 16 -6.422 <.0001
## 160 - 220 -155.8 11.6 16 -13.485 <.0001
## 180 - 200 -38.0 11.6 16 -3.289 0.0216
## 180 - 220 -119.6 11.6 16 -10.352 <.0001
## 200 - 220 -81.6 11.6 16 -7.063 <.0001
##
## P value adjustment: tukey method for comparing a family of 4 estimates
Hopefully you now have a better understanding of what’s happening in R when you set a factor as ordered and use it in a linear model.
References
For questions or clarifications regarding this article, contact the UVA Library StatLab: [email protected]
View the entire collection of UVA Library StatLab articles.
Clay Ford
Statistical Research Consultant
University of Virginia Library
July 1, 2021
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://data.library.virginia.edu/understanding-ordered-factors-in-a-linear-model/", "fetch_time": 1638107663000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00327.warc.gz", "warc_record_offset": 280958901, "warc_record_length": 10534, "token_count": 4297, "char_count": 12657, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2021-49", "snapshot_type": "latest", "language": "en", "language_score": 0.8456212282180786, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00014-of-00064.parquet"}
|
# Why von Neumann entropy requires diagonalization and linear entropy doesn't?
The linear entropy for a state $$\rho$$ is defined as $$S_L = 1 - Tr[\rho^2]$$, while as von Neumann entropy as $$S_{N} = -Tr[\rho \ln \rho]$$. According to quantiki, the computation of $$S_{N}$$ requires diagonalization but $$S_L$$. But this is not clear to me why.
• Well how would you compute $\rho \log \rho$ and $\rho^2$? <Fun fact: technically you wouldn't actually need diagonalization to compute the logarithm of full rank density matrices, you could also do it via a power series>. Commented Feb 1, 2022 at 12:50
The linear entropy $$S_L = 1 - {\rm Tr}[\rho^2]$$ certainly doesn't need diagonalization because $$\rho^2$$ is straightforward to calculate, as is its trace.
The von Neumann entropy $$S_N=-{\rm Tr}[\rho \ln\rho]$$ seems to require diagonalization because of the factor $$\ln \rho$$. Taking logarithms is not a "natural" operation for a matrix, and it's not at all clear at first glance what it should even mean.
But, physicists find it useful to say that any scalar function $$f(x)$$ can be applied to any matrix $$A$$ in the following way:
1. Factor (or "diagonalize") the matrix: $$A = U\Lambda U^\dagger$$, so that $$U$$ is a unitary transformation and $$\Lambda$$ is a diagonal matrix.
2. Apply the function $$f(x)$$ on each element of $$\Lambda$$. Call the resulting matrix $$f(\Lambda)$$.
3. Define the matrix $$f(A) \equiv U f(\Lambda) U^\dagger$$.
Essentially we have applied the scalar function to each element of $$A$$ when represented in its eigenbasis.
So, the very definition of $$\ln \rho$$, and therefore $$S_N$$, implicitly includes a diagonalization.
But as @Rammus points out in a comment, the joys of calculus do let us sometimes use alternate means of calculation such as Taylor series. Actually, it seems the function $$f(A)$$ is often (maybe usually!) defined in terms of the Taylor series. See for example:
I was a little embarrassed to find that even a mathematical physics textbook defines it that way. My physics teachers are perhaps a little iconoclastic. ;)
• That looks cool! Could you suggest some reference(s) regarding the above 3 steps? Commented Feb 1, 2022 at 21:27
• That is an excellent question. In fact I've just checked three different sources and they all contradict me, defining $f(A)$ in terms of the Taylor series! ^_^ I'll edit the end of my answer to acknowledge this and give a couple links. Commented Feb 1, 2022 at 22:27
• @jecado don't be so quick to dismiss your physics teachers: you can have operator functions that are not analytic! They are then defined by their action on eigenstates. When $A|\psi_i\rangle=\lambda_i|\psi_i\rangle$, we define operators $f(A)$ by $f(A)|\psi_i\rangle=f(\lambda_i)|\psi_i\rangle$, plain and simple. Then if $f$ happens to have a Taylor series expansion we can do other things too Commented Feb 2, 2022 at 16:00
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://quantumcomputing.stackexchange.com/questions/23897/why-von-neumann-entropy-requires-diagonalization-and-linear-entropy-doesnt", "fetch_time": 1719063829000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198862396.85/warc/CC-MAIN-20240622112740-20240622142740-00774.warc.gz", "warc_record_offset": 415742264, "warc_record_length": 40413, "token_count": 761, "char_count": 2909, "score": 3.53125, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.9161264896392822, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00062-of-00064.parquet"}
|
+0
# Writing an Exponential Function
0
161
2
what is the annual growth rate of the painting's value? Write the answer as a percentage rounded to the nearest tenth.
The function V=250(1.28)^d models the value of a rare painting, in thousands of dollars, d decades after it sold at an auction for \$250,000.
Mar 13, 2019
#1
+111437
0
V=250(1.28)^d
Since one year = d / 10.........we have
V = 250 (1.28)^(d/10) =
250 [(1.28)^(1/10)]^d
250 [≈ 1.025]^d
250 [ 1 + .025 ]^d
So....the annual growth rate is .025 ≈ 2.5%
Mar 13, 2019
#2
0
Thank you so much!!!!
Guest Mar 15, 2019
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://web2.0calc.com/questions/writing-an-exponential-function", "fetch_time": 1596823002000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-34/segments/1596439737206.16/warc/CC-MAIN-20200807172851-20200807202851-00590.warc.gz", "warc_record_offset": 550841811, "warc_record_length": 6679, "token_count": 219, "char_count": 594, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2020-34", "snapshot_type": "latest", "language": "en", "language_score": 0.8124027252197266, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00060-of-00064.parquet"}
|
Physics Forums (http://www.physicsforums.com/index.php)
- Science Fiction & Fantasy (http://www.physicsforums.com/forumdisplay.php?f=219)
- - Can you help me understand g's and acceleration for a fictional Star Wars spaceship? (http://www.physicsforums.com/showthread.php?t=158997)
ledopmi Mar3-07 05:12 AM
Can you help me understand g's and acceleration for a fictional Star Wars spaceship?
I read that a space ship can reach 5,100g in space. Is there a way to determine how many kilometers per hour this would be? As I understand it, the only way to feel g force is to be accelerating. The acceleration should slow down as speed increases and eventually stop so there would be 0g at that point.
I saw somewhere that 5,100g would be equal to 51,000kph but there was no explanation as to how the anwer was calculated.
cristo Mar3-07 05:53 AM
The g is an acceleration (the acceleration due to gravity felt on the surface of the earth). Therefore, 5100g is an acceleration and so, no, it cannot be converted into kilometres per hour, since this is a unit of velocity.
russ_watters Mar3-07 08:44 AM
Since 1g is 9.8m/s/s, 5100g is 50,000m/s/s or 50km/s/s.
ledopmi Mar3-07 08:27 PM
OK thanks. Can you explain the meters per second per second to me. I searched the web and found formulas but no examples.
Meters divided by time divided by time again. Can you give me an example of this formula with actual numbers?
russ_watters Mar4-07 10:29 AM
That's the units of acceleration. If you accelerate at 1m/s/s for 10 seconds, you'll be going 10m/s. speed = acceleration * time.
ledopmi Mar5-07 10:01 PM
Thank you.:smile:
Taremos Mar6-07 12:21 AM
Now that you know what a g is, also know that 5,100g would crush pretty much anything experiencing that magnitude of acceleration.
Check out this article on the effects of acceleration on the human body:
http://quest.nasa.gov/saturn/qa/new/...n_the_body.txt
Even prolonged exposure to 3g's can cause serious health problems.
Prolonged exposure to 10gs would kill a person, even in a pressures suit.
Even if inertial dampeners, a device that would somehow less the effects of acceleration and currently only science fiction, were used, I doubt they would be able to compensate for 5,100 times earth gravity.
All times are GMT -5. The time now is 07:14 AM.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.physicsforums.com/printthread.php?t=158997", "fetch_time": 1409660086000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2014-35/segments/1409535921957.9/warc/CC-MAIN-20140901014521-00415-ip-10-180-136-8.ec2.internal.warc.gz", "warc_record_offset": 1362492579, "warc_record_length": 2823, "token_count": 603, "char_count": 2326, "score": 3.859375, "int_score": 4, "crawl": "CC-MAIN-2014-35", "snapshot_type": "latest", "language": "en", "language_score": 0.9314188957214355, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00015-of-00064.parquet"}
|
# math
Benny is traveling at a speed of 15 mi/h. What is his speed in feet per second?
1. 👍
2. 👎
3. 👁
1. A few facts:
1 mile = 5,280 feet
15 mi = 15 * 5280 = ? total feet
1 hour = 60 * 60 = ? total seconds
Divide the total number of feet by the total number of seconds.
1. 👍
2. 👎
👤
Ms. Sue
## Similar Questions
1. ### Calculus
A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as –32 feet per second squared
2. ### Science
#1. Kilometers per hour describes the speed of an object. ture false **** #2. Which of the following reference points would allow you to observe the speed of a car that you are traveling in? a) an airplane flying across the sky b)
3. ### Math
The distance require to stop a car varies directly as the square of its speed. If 250 feet are required to stop a car traveling 60 miles per hour, how many feet are required to stop a car traveling 96 miles per hour!
4. ### physics
Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts his horn, which has frequency of 544 Hz. The other cyclist hears the frequency as 522 Hz. If the speed of sound in air is 344
1. ### Math
the stopping distance of an automobile is directly proportional to the square of its speed v. a car required 90 feet to stop when its speed was 70 miles per hour. find a mathematical model that gives the stopping distance d in
2. ### Physics
At a busy intersection a 1460-kg car traveling west with a speed of 12 m/s collides head-on with a minivan traveling east with a speed of 9.4 m/s. The cars stick together and move with an initial velocity of 1.5 m/s to the east
3. ### physics
A car with a mass of 980 kg is initially traveling east toward an intersection with a speed of vc = 18.9 m/s and a 1500 kg pickup is traveling north toward the same intersection. The car and truck collide at the intersection and
4. ### Precalculus
1.A plane takes off at an angle of 6 degrees traveling at the rate of 200 feet/second. If it continues on this flight apth at the same speed, how many minutes will it take to reach a n altitude of 8000 feet? 2. A plane takes off
1. ### Physics
When you approach a blinking source of light, the frequency of its blinking is seen to: decrease, increase, or remain the same always. I THINK IT IS INCREASING. If you are traveling at close to the speed of light and shine a
2. ### Math
A cyclist rides his bike at a speed of 18 miles per hour. What is this speed in feet per second? How many feet will the cyclist travel in 10 seconds? In your computations, use the fact that mile is equal to feet. Do not round your
3. ### Science
2. Which of the following reference points would allow you to observe the speed of a car that you are traveling in? A. An airplane flying acroos the sky. B. A car traveling next to you at the same speed you are traveling. C. A car
4. ### Physics Conservation of Energy
A car traveling at 20 miles/hour stops in a distance of 15 feet when its brakes are locked. Assuming that the frictional force doesn't depend on speed, that same car traveling at 60 miles/hour will stop in a distance closest to a.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/questions/73678/benny-is-traveling-at-a-speed-of-15-mi-h-what-is-his-speed-in-feet-per-second", "fetch_time": 1623907415000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-25/segments/1623487629209.28/warc/CC-MAIN-20210617041347-20210617071347-00355.warc.gz", "warc_record_offset": 740886194, "warc_record_length": 5012, "token_count": 831, "char_count": 3275, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2021-25", "snapshot_type": "latest", "language": "en", "language_score": 0.9194059371948242, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00022-of-00064.parquet"}
|
# The line parallel to the x-axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0, where (a, b) ? (0, 0) is ?
SHAIK AASIF AHAMED
9 years ago
Hello student,
The line passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0 is
ax + 2by + 3b+$\dpi{80} \lambda$(bx - 2ay - 3a)=0...............(1)
(a+b$\dpi{80} \lambda$)x+(2b-2a$\dpi{80} \lambda$)y+3b-3$\dpi{80} \lambda$a=0
As the line is parallel to x axis
a+b$\dpi{80} \lambda$=0 so $\dpi{80} \lambda$=(-b/a)
Putting $\dpi{80} \lambda$=(-b/a) in equation (1) we get
ax + 2by + 3b+(-a/b)(bx - 2ay - 3a)=0
y(2b+2a2/b)+3b+3a2/b=0
on simplifying we get y=-3/2
so it is 3/2 units below x-axis
Thanks and Regards
Shaik Aasif
Latika Leekha
9 years ago
The given lines are ax + 2by + 3b = 0 and bx – 2ay – 3a = 0.
Then the required equation is of the form
ax + 2by + 3b + l(bx – 2ay – 3a) = 0.
(a + bl)x + (2b – 2al)y + (3b-3a) = 0
Since, the lline is given to be parallel to x-axis, hence we must have the coefficient of x to be zero.
Hence, a+bl = 0
l = -a/b
So, the given equation reduces to (2b + 2a2/b)y + (3b + 3a2/b) = 0
This yields 2y + 3 = 0, which is a line parallel to x-axis and since, y = -3/2, so it lies below the x-axis at a distnce of 3/2 from the origin.
Thansk & Regards
Latika Leekha
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.askiitians.com/forums/Algebra/the-line-parallel-to-the-x-axis-and-passing-throug_75862.htm", "fetch_time": 1713561828000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712296817455.17/warc/CC-MAIN-20240419203449-20240419233449-00488.warc.gz", "warc_record_offset": 582801617, "warc_record_length": 43970, "token_count": 556, "char_count": 1316, "score": 4.46875, "int_score": 4, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.7950319051742554, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00052-of-00064.parquet"}
|
# Median of two sorted arrays
Posted by chunyang on April 26, 2019
### 题目
There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
You may assume nums1 and nums2 cannot be both empty.
Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example 2:
nums1 = [1, 2] nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
### 解法
#### 思路
• 每次都将需要查找的 target 降低一半
• 当其中一个为空时，我们直接返回另外一个数组里面对应的元素即可
#### 代码
class Solution {
double findMedianSortedArraysHelper(
vector<int>& nums1, int s1, int e1,
vector<int>& nums2, int s2, int e2, int target) {
if (e1 - s1 > e2 - s2) {
return findMedianSortedArraysHelper(nums2, s2, e2, nums1, s1, e1, target);
}
if (s1 >= e1) return nums2[s2 + target];
if (s2 >= e2) return nums1[s1 + target];
if (target == 0) return nums1[s1] < nums2[s2] ? nums1[s1] : nums2[s2];
int mid1 = s1 + (target - 1) / 2;
if (mid1 >= e1) mid1 = e1-1;
int mid2 = s2 + (target - 1) - (mid1 - s1);
cout << "num1: " << s1 << " " << e1 << " " << mid1 << endl;
cout << "num2: " << s2 << " " << e2 << " " << mid2 << endl;
cout << "target: " << target << " index: " << (mid1 - s1 + mid2 - s2 + 2) << endl;
// assert(target >= 0);
// assert(target == (mid1 - s1 + mid2 - s2 + 2));
if (nums1[mid1] == nums2[mid2]) return nums1[mid1];
else if (nums1[mid1] > nums2[mid2]) {
auto diff = mid2 - s2 + 1;
return findMedianSortedArraysHelper(nums1, s1, e1, nums2, mid2+1, e2, target - diff);
} else {
auto diff = mid1 - s1 + 1;
return findMedianSortedArraysHelper(nums1, mid1+1, e1, nums2, s2, e2, target - diff);
}
}
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int m = nums1.size();
int n = nums2.size();
int sum = m + n;
if ((sum & 0x1) == 0) {
// even
auto a = findMedianSortedArraysHelper(nums1, 0, m, nums2, 0, n, sum / 2);
// cout << "A: " << a << endl;
auto b = findMedianSortedArraysHelper(nums1, 0, m, nums2, 0, n, sum / 2 - 1);
// cout << "B: " << b << endl;
return (a + b) / 2.0;
} else {
// odd
return findMedianSortedArraysHelper(nums1, 0, m, nums2, 0, n, sum / 2);
}
}
};
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.chunyangwen.com/algorithm/leetcode-median-of-array.html", "fetch_time": 1563860949000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-30/segments/1563195528869.90/warc/CC-MAIN-20190723043719-20190723065719-00153.warc.gz", "warc_record_offset": 666477421, "warc_record_length": 6231, "token_count": 819, "char_count": 2144, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2019-30", "snapshot_type": "latest", "language": "en", "language_score": 0.3181088864803314, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00013-of-00064.parquet"}
|
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## Integral Calculus
### Course: Integral Calculus>Unit 5
Lesson 7: Alternating series test
# Alternating series test
When a series alternates (plus, minus, plus, minus,...) there's a fairly simple way to determine whether it converges or diverges: see if the terms of the series approach 0.
## Want to join the conversation?
• From to - what is the second condition given for?
Isn't it clear, that if lim(b_n) = 0 when n -> infinity, then {b_n} should be decreasing sequence only? How this sequence could be not decreasing if the limit of b_n (when n -> infinity) is equal zero?
• I can't really explain how any of this works, but I can give a counter-example: sin(pi*n/4)/n is not a decreasing series, but still tends to zero as n approaches infinity.
• The restrictions on this test seem redundant.
from to , Sal gives three restrictions on the series:
1) Bn ≥ 0 for all relevant n (namely positive integers n).
2) lim as Bn→∞ = 0
3) {Bn} is a decreasing sequence.
Don't the first two rules imply the third?
• Consider the function f(n) = x*e^(-n). This will be our Bn.
What do you get at n= 0, n=1, n=2?
f(0) = 0,
f(1) = e^-1 = 0.37,
f(2) = 2/e^-2 = 0.27
So the first 3 terms of the sequence are: [0, 0.37, 0.27].
Notice the b2> b1 this function is not always decreasing! Yet all terms are greater than zero. And the limit as n→∞ = 0. Some functions like this rise a little bit, then fall back down as they go on. So the third rule is necessary.
• Does this test work for an increasing sequence?
• This test is used to determine if a series is converging. A series is the sum of the terms of a sequence (or perhaps more appropriately the limit of the partial sums).
This test is not applicable to a sequence.
Also, to use this test, the terms of the underlying sequence need to be alternating (moving from positive to negative to positive and so on). Therefore the sequence would not be increasing but rather 'oscillating' between positive and negative terms.
• At -: Why is it necessary that the alternating sign be expressed by either (-1)^n or (-1)^(n+1). Aren't there a whole slew of other ways to produced an alternating sign? What about (-1)^(n+2), (-1)^(n^2), or any polynomial exponent with odd integer coefficients?
• We are only talking about the form the series takes on. We know that it alternates, so the question is, is a negative term first, or a positive term. Given n goes from 1 to infinity, the first term of the (-1)^n series will be negative, and the first term of the (-1)^(n+1) series will be positive. That is all that is meant by the form of the series. Why make it any more complicated? It is an alternating series, either the first term is positive, or the first term is negative.
• What is an example of a series that fails the alternating series test but still converges, as Sal mentions at the end?
• At He says that we can use other techniques like the limit comparison test, I tired using the limit comparison test but I can't figure out another functions that behaves like this function and that I can figure out It converges ... can someone please tell me how :)
• I don't think we can use the limit comparison test in this case, since the limit comparison require terms in both sequences greater that 0
Maybe he's saying the ratio test, the ratio test also involve limit, and can prove the series converges
• While attempting some practice problems, I couldn't get the correct answer, and this came up as a hint. "This series meets all the conditions for the alternating series test and hence it converges. However, since we can show that ∑n=1∞ n+1n2 diverges by using a comparison test with ∑n=1∞1n. Thus the series converges conditionally." I do not understand what this means, What are the "conditions"?
• I had a similar experience, I'm not sure the hints are properly done on some of these later topics..
• Is this test also called the Leibnitz test for alternating series??
• Yes it is the same! Some books/profs call it alternating and some call it the Leibnitz test.
(1 vote)
• Can you use the ratio test on an alternating series?
• So I just tried the first practice problem in the "Alternating Series" challenge, and I found that one of the hints contradicts what Sal says in this video. At , Sal says, "Now once again, if something does not pass the Alternating Series Test, that does not necessarily mean that it diverges; it just means that you couldn't use the Alternating Series Test to prove that it converges."
The problem hint, however, says, "...Since lim n→∞ an = 1 ≠ 0, the series fails the Alternating Series Test and therefore diverges."
Help? Can the Alternating Series Test be used to prove DIVERGENCE, or only conclusively prove CONVERGENCE?
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.khanacademy.org/math/integral-calculus/ic-series/ic-alternating-series/v/alternating-series-test", "fetch_time": 1701830115000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-50/segments/1700679100575.30/warc/CC-MAIN-20231206000253-20231206030253-00428.warc.gz", "warc_record_offset": 927917154, "warc_record_length": 101368, "token_count": 1225, "char_count": 4941, "score": 4.15625, "int_score": 4, "crawl": "CC-MAIN-2023-50", "snapshot_type": "latest", "language": "en", "language_score": 0.9125615954399109, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00037-of-00064.parquet"}
|
Search Memberlist Usergroups
Page 1 of 1 [7 Posts]
Author Message
Mike1170
Joined: 17 Sep 2005
Posts: 74
Posted: Fri Jun 23, 2006 3:05 am Post subject: Re: Buckets and balls... help please.
"Mike" wrote in message
Quote: First, put some tighter restrictions on m and n: 1 < m < N, 1 < t < N and m + t >= N If m or t >= N, then probability is 0, or if m + t < N, then probability is > 1.
Oops, tighter restrictions should be on m and t
and they should each be greater than or equal to 1.
Mike1170
Joined: 17 Sep 2005
Posts: 74
Posted: Fri Jun 23, 2006 2:51 am Post subject: Re: Buckets and balls... help please.
"BEZMan" wrote in message
Quote: ... Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I randomly put all my red balls in different buckets (such that no two red balls are in the same bucket), and then do the same with my blue balls. Buckets will then contain [no ball], [one red ball], [one blue ball] or [one red and one blue ball] at the end. When I am done, what are odds that I end up with an empty bucket. Also, if I repeat this operation S times, what are the final odds that I obtain empty bucket at least once?
First, put some tighter restrictions on m and n:
1 < m < N, 1 < t < N and m + t >= N
(If m or t >= N, then probability is 0, or if m + t < N, then probability is
1.)
After red balls are placed, there are m occupied and N-m unoccupied buckets.
The required probability is the complement of the probability that all of
the
N-m empty buckets get filled when placing the t blue balls, i.e., that N-m
of
the t blue balls are placed in the N-m empty buckets and the rest are placed
in
or among the m occupied buckets.
Probability = P = 1 - C(N-m,N-m) x C(m,t-N+m) / C(N,t) where
C(A,B) = the number of combinations of A things taken B at a time
C(A,B) = A! / [ B! x (A-B)! ], 0! = 1, 1! = 1, 2!=2, 3!=6, 4!=24, 5!=120,
....
P = 1 - m! t! / N! x (m+t-N)!
(Odds [against] = (1/P - 1) to 1. Thus, if P = 0.25, then odds [against]
are 3 to 1.)
The probability of obtaining an empty bucket at least once in S repititions
of the
operation is the complement of not obtaining an empty bucket in S
operations.
P(S) = 1 - (1-P)^S
Mike1170
Joined: 17 Sep 2005
Posts: 74
Posted: Sat Jun 17, 2006 7:19 pm Post subject: Re: Buckets and balls... help please.
"BEZMan" wrote
Quote: Hi all, I am struggling with this problem, and was hoping somebody would help me. Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I randomly put all my red balls in different buckets (such that no two red balls are in the same bucket), and then do the same with my blue balls. Buckets will then contain [no ball], [one red ball], [one blue ball] or [one red and one blue ball] at the end. When I am done, what are odds that I end up with an empty bucket. Also, if I repeat this operation S times, what are the final odds that I obtain empty bucket at least once?
Given that red and blue balls are each restricted to buckets not already
occupied by a ball of the same color, try using a "sampling without
replacement" approach instead of the ball and urn model. Instead of N
buckets, think of a population of N items (say the integers from 1 to N) and
instead of randomly putting balls in buckets (in a way that prevents two
balls of the same color in the same bucket) think of sampling the N integers
without replacement. Randomly putting all of the red balls in different
buckets is equivalent to drawing a sample of size m from a population of
size N. Ditto the blue balls. Randomly putting all of the blue balls in
different buckets is equivalent to drawing a sample of size t from the same
population of size N. With this appreach, I think you'll have better luck
figuring the probabilities of unsampled, singly-sampled and doubly-sampled
items (buckets, integers or whatever).
Pavel314
Joined: 29 Apr 2005
Posts: 78
Posted: Fri Jun 16, 2006 9:57 pm Post subject: Re: Buckets and balls... help please.
"BEZMan" <[email protected]> wrote in message
Quote: Hi all, I am struggling with this problem, and was hoping somebody would help me. Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I randomly put all my red balls in different buckets (such that no two red balls are in the same bucket), and then do the same with my blue balls. Buckets will then contain [no ball], [one red ball], [one blue ball] or [one red and one blue ball] at the end.
After you put all the red balls into the buckets, do you have to put the
blue balls into empty buckets first, filling all empty buckets before you
can but a blue ball in with a red one? Or do you put the blue balls randomly
into the buckets, just avoiding two blue balls in the same bucket? Depending
on the rules governing the blue ball placement, you will get different
Quote: When I am done, what are odds that I end up with an empty bucket.
If you have to fill the empty buckets with blue balls first, it depends on
the relationship of N, m, and t.
1. m + t < N, you get N - m - t empty buckets so the odds are 100%.
2. m + t >= N, no empty buckets.
Quote: Also, if I repeat this operation S times, what are the final odds that I obtain empty bucket at least once?
This seems to indicate that the blue balls don't have to go into empty
buckets, otherwise the results would always be the same.
Paul
Quote: Thanx.
BEZMan
science forum beginner
Joined: 16 Jun 2006
Posts: 3
Posted: Fri Jun 16, 2006 7:06 am Post subject: Re: Buckets and balls... help please.
BEZMan wrote:
Quote: BEZMan wrote: Hi all, I am struggling with this problem, and was hoping somebody would help me. Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I randomly put all my red balls in different buckets (such that no two red balls are in the same bucket), and then do the same with my blue balls. Buckets will then contain [no ball], [one red ball], [one blue ball] or [one red and one blue ball] at the end. When I am done, what are odds that I end up with an empty bucket. Also, if I repeat this operation S times, what are the final odds that I obtain empty bucket at least once? Thanx. What I meant was in fact: "When I am done, what are odds that I end up with AT LEAST one empty bucket." Sorry....
Also, what are the odds of having NO bucket containing 2 balls: that is
all bucket have at most one ball?
Again, thank you very much in advance.
BEZMan
science forum beginner
Joined: 16 Jun 2006
Posts: 3
Posted: Fri Jun 16, 2006 7:00 am Post subject: Re: Buckets and balls... help please.
BEZMan wrote:
Quote: Hi all, I am struggling with this problem, and was hoping somebody would help me. Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I randomly put all my red balls in different buckets (such that no two red balls are in the same bucket), and then do the same with my blue balls. Buckets will then contain [no ball], [one red ball], [one blue ball] or [one red and one blue ball] at the end. When I am done, what are odds that I end up with an empty bucket. Also, if I repeat this operation S times, what are the final odds that I obtain empty bucket at least once? Thanx.
What I meant was in fact:
"When I am done, what are odds that I end up with AT LEAST one empty
bucket."
Sorry....
BEZMan
science forum beginner
Joined: 16 Jun 2006
Posts: 3
Posted: Fri Jun 16, 2006 6:57 am Post subject: Buckets and balls... help please. Hi all, I am struggling with this problem, and was hoping somebody would help me. Let say I have N buckets, m (<=N) red ball and t (<=N) blue balls. I randomly put all my red balls in different buckets (such that no two red balls are in the same bucket), and then do the same with my blue balls. Buckets will then contain [no ball], [one red ball], [one blue ball] or [one red and one blue ball] at the end. When I am done, what are odds that I end up with an empty bucket. Also, if I repeat this operation S times, what are the final odds that I obtain empty bucket at least once? Thanx.
Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
Page 1 of 1 [7 Posts]
The time now is Mon Sep 24, 2018 2:16 am | All times are GMT
Jump to: Select a forum-------------------Forum index|___Science and Technology |___Math | |___Research | |___num-analysis | |___Symbolic | |___Combinatorics | |___Probability | | |___Prediction | | | |___Undergraduate | |___Recreational | |___Physics | |___Research | |___New Theories | |___Acoustics | |___Electromagnetics | |___Strings | |___Particle | |___Fusion | |___Relativity | |___Chem | |___Analytical | |___Electrochem | | |___Battery | | | |___Coatings | |___Engineering |___Control |___Mechanics |___Chemical
Topic Author Forum Replies Last Post Similar Topics Hausdorff dimension and balls in R^n Amanda Math 5 Wed Jul 05, 2006 4:04 pm gold balls Mark Spahn Recreational 3 Sun May 21, 2006 9:21 am red balls, white balls... Gino Martelli Math 3 Fri May 19, 2006 2:44 pm balls and metric eugene Math 14 Sat Apr 29, 2006 12:28 pm picking balls Andersen Math 17 Thu Apr 27, 2006 7:51 pm
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://sci4um.com/about32054-Buckets-and-balls...-help-please..html", "fetch_time": 1537755402000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-39/segments/1537267160085.77/warc/CC-MAIN-20180924011731-20180924032131-00234.warc.gz", "warc_record_offset": 212845299, "warc_record_length": 8065, "token_count": 2668, "char_count": 9280, "score": 3.8125, "int_score": 4, "crawl": "CC-MAIN-2018-39", "snapshot_type": "longest", "language": "en", "language_score": 0.883402407169342, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00011-of-00064.parquet"}
|
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A240908 The sequency numbers of the 8 rows of a version of the Hadamard-Walsh matrix of order 8. 4
0, 7, 3, 4, 1, 6, 2, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS The Hadamard (Hadamard-Walsh) matrix is widely used in telecommunications and signal analysis. It has 3 well-known forms which vary according to the sequency ordering of its rows: "natural" ordering, "dyadic" or Payley ordering, and sequency ordering. In a mathematical context the sequency is the number of zero crossings or transitions in a matrix row (although in a physical signal context, it is half the number of zero crossings per time period). The matrix row sequencies are a permutation of the set [0,1,2,...n-1], where n is the order of the matrix. For spectral analysis of signals the sequency-ordered form is needed. Unlike the dyadic ordering (given by A153141), the natural ordering requires a separate list for each matrix order. This sequence is the natural sequency ordering for an order 8 matrix. LINKS N. J. A. Sloane, A Library of Hadamard Matrices Eric Weisstein's World of Mathematics, Hadamard Matrix Wikipedia, Walsh matrix FORMULA Recursion: H(2)=[1 1; 1 -1]; H(n) = H(n-1)*H(2), where * is Kronecker matrix product. EXAMPLE This is a fixed length sequence of only 8 values, as given. CROSSREFS Cf. A240909 "natural order" sequencies for Hadamard-Walsh matrix, order 16. Cf. A240910 "natural order" sequencies for Hadamard-Walsh matrix, order 32. Cf. A153141 "dyadic order" sequencies for Hadamard-Walsh matrix, all orders. Cf. A000975(n) is sequency of last row of H(n). - William P. Orrick, Jun 28 2015 Sequence in context: A097517 A127559 A066747 * A117043 A013664 A154173 Adjacent sequences: A240905 A240906 A240907 * A240909 A240910 A240911 KEYWORD nonn,fini,full AUTHOR Ross Drewe, Apr 14 2014 EXTENSIONS Definition of H(n) corrected by William P. Orrick, Jun 28 2015 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified August 5 22:41 EDT 2021. Contains 346488 sequences. (Running on oeis4.)
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://oeis.org/A240908", "fetch_time": 1628221092000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-31/segments/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00155.warc.gz", "warc_record_offset": 435488070, "warc_record_length": 4333, "token_count": 681, "char_count": 2437, "score": 3.515625, "int_score": 4, "crawl": "CC-MAIN-2021-31", "snapshot_type": "latest", "language": "en", "language_score": 0.796556293964386, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00032-of-00064.parquet"}
|
• K-5 AIS Math
with Mrs. Smith
• What is AIS math?
AIS math is a remedial service that is provided to students who are below grade level. My services are provided outside of the regular classroom and, depending on the students' needs, are provided for 30 minutes either every other day or on a daily basis. My groups are organzied by both grade and skill level and consist of only one, two, or three students.The small group dynamics allow me to provide intensive services that consist of research based strategies and activities that are aligned to the Common Core Standards and are geared towards meeting each student's individual needs.
• Classroom Rules
1. Respect each other, our classroom, and out materials
2. Follow directions the first time they are given
• What Does My Child Need To Know?
Here is a list of many, but not all, of the skills
will be learning this year. For a complete
list of the Common Core State Standards
Kindergarten
- Counting to 100 by ones and tens
- Identifying and writing numbers 0 to 20
- Comparing two numbers between 0 and 10
- Counting objects using one to one correspondence
-Identifying 2D shapes and describing their attributes
(ex. number of sides, corners)
-Finding the missing number in a sequence of numbers
-Identifying 1 more/1 less
-Counting backwards from 10
-Adding and subtracting using pictures and manipulatives
- Solving basic addition facts using preferred strategies
(ex. doubles, doubles +1, ten facts)
-Solveing basic subtraction facts using a variety of strategies
(ex. number line, counting backwards, counting on)
-Relating addition to subtraction and vice versa
(ex. 3 + 2 = 5, 5 - 3 = 2, 5 - 2 = 3)
- Understanding place value and value up to the tens place
- Skip counting
-Comparing 2-digit numbers
(ex. 1 + __ = 6)
-Telling time to the whole hour
-Using math vocabulary words
(ex. addend, sum, digit, difference, fact families)
- Building fluency with the addition facts up to 10+10
- Adding/subtracting up to 3-digit numbers with regrouping
-Identifying numbers as even or odd
-Understanding place value up to the hundreds place
(ex. expanded, written, standard forms)
-Building fluency with the basic subtraction facts
-Comparing up to 3-digit numbers using >, < = symbols
-Adding/subtracting 10 mentally from a 2-digit number
-Adding/subtracting 100 mentally from a 3-digit number
- Rounding up to 4-digit numbers to any place value
- Understanding place value up to the ten thousands place
(ex. expanded, standard written form)
- Adding/subtracting multi-digit numbers with regrouping
-Understanding what multiplication is
- Memorizing the basic multiplication facts up to 10 x 10
- Place value up to 1,000,000
(ex. expanded, standard, written form)
- Rounding multi-digit numbers to any place value
-Finding factors and multiples
-Solving multi-step word problems
-Multiplying up to 4-digit numbers by a 1-digit number
-Multiplying 2-digit by 2-digit numbers
-Dividing up to a 4-digit dividend by a 1-digit divisor
- Multiplying up to 4-digit numbers by 3-digit numbers
(including both whole numbers and decimals)
- Solving expressions using the Order of Operations
-Identifying prime/composite numbers
-Prime factorization
-Place value
(whole numbers up to billions, decimals to thousandths)
-Rounding whole numbers and decimals
Division with up to a 4-digit dividend and 2-digit divisor
(including both whole numbers and decimals)
-Long Division with interpreting remainders
-Solving multi-step word problems
• Family Game Night
Games not only enhance math skills, but are a great way to spend time with your loved ones. Pick one set night during the week to play games with your family...and HAVE FUN!!
Here are some classic games that use math skills:
Yahtzee, Monopoly (non-electronic version), Go Fish, Cribbage, Skip Bo (card game), Phase 10 (card game), and any games that require you to keep score (Rummy, Scrabble)
• DON'T FORGET...
You can practice your math skills online as well. Check out all of the websites under the "Useful Links" tab.
• MATH-A-THON
Math-A-Thon, America's largest education-based fundraiser, benefits St. Jude Children’s Research Hospital. To participate, kids ask family members and friends for pledges to solve math problems in the St. Jude Math-A-Thon Funbook, developed by Scholastic.
Hartford has proudly participated in this program
for over 25 years!
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "www.hartfordcsd.org", "fetch_time": 1716962784000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-22/segments/1715971059206.29/warc/CC-MAIN-20240529041915-20240529071915-00521.warc.gz", "warc_record_offset": 691483955, "warc_record_length": 82036, "token_count": 1063, "char_count": 4434, "score": 3.828125, "int_score": 4, "crawl": "CC-MAIN-2024-22", "snapshot_type": "longest", "language": "en", "language_score": 0.8572558164596558, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00059-of-00064.parquet"}
|
Home / Transformer / Determination of Transformer Equivalent Circuit Parameters
# Determination of Transformer Equivalent Circuit Parameters
Want create site? Find Free WordPress Themes and plugins.
In order for a model to be useful, there must be a way to determine the values of the model parameters. Two simple tests are used to determine the values for the parameters of the transformer equivalent circuit of Fig. 1 (a) and (b). The two tests are the short-circuit and open-circuit tests.
Fig.1: Approximate Equivalent Circuit (a) and (b)
If it is desired to find the parameters of the exact equivalent circuit of Fig. 2, it is customary to assume R1=a2R2 and X1 =a2X2. This assumption allows the decomposition of the equivalent resistance and reactance into the primary and secondary components.
Fig.2: Transformer Equivalent Circuit in Phasor Form
Let the primary (winding 1) be the high-voltage side and the secondary (winding 2) the low-voltage side for the transformer of Fig.3.
Fig.3: An Actual Transformer
## Transformer Open-Circuit Test
In the open-circuit test, the transformer rated voltage is applied to the low-voltage side of the transformer with the high-voltage side left open. Measurement of power, current, and voltage are made on the low-voltage side as shown in Fig.4.
Fig.4: Connections for Open Circuit Test
Since the high-voltage side is open, the input current Ioc is equal to the exciting current through the shunt excitation branch as shown in the equivalent circuit of Fig 5. Because this current is very small, about 5% of rated value, the voltage drop across the low-voltage winding and the winding copper losses are neglected.
Fig.5: Equivalent Circuit for Open Circuit Test
The magnitude of the admittance of the shunt excitation branch of the equivalent circuit referred to the low-voltage side is calculated as follows:
$\begin{matrix} \left| {{Y}_{o2}} \right|=\frac{{{I}_{oc}}}{{{V}_{oc}}} & {} & \left( 1 \right) \\\end{matrix}$
The phase angle of the admittance is found as
$\begin{matrix} -{{\theta }_{o2}}=-{{\cos }^{-1}}\left( \frac{{{P}_{oc}}}{{{V}_{oc}}{{I}_{oc}}} \right) & {} & \left( 2 \right) \\\end{matrix}$
Thus, the complex admittance may be expressed as
$\begin{matrix} {{Y}_{o2}}=\left| {{Y}_{o2}} \right|\angle -{{\theta }_{o2}}={{G}_{c2}}-j{{B}_{m2}} & {} & \left( 3 \right) \\\end{matrix}$
The corresponding resistance and reactance parameter of Fig. 1 (b) are derived from the conductance and susceptance, respectively:
$\begin{matrix} {{R}_{c2}}=\frac{1}{{{G}_{c2}}} & {} & \left( 4 \right) \\\end{matrix}$
$\begin{matrix} j{{X}_{m2}}=\frac{1}{-j{{B}_{m2}}} & {} & \left( 5 \right) \\\end{matrix}$
These parameters may be referred to the high-voltage side to give the parameters of the equivalent circuit of Fig. 1 (a).
$\begin{matrix} {{R}_{c1}}={{a}^{2}}{{R}_{c2}} & {} & \left( 6 \right) \\\end{matrix}$
$\begin{matrix} {{X}_{m1}}={{a}^{2}}{{X}_{m2}} & {} & \left( 7 \right) \\\end{matrix}$
Note that when a rated voltage at the rated frequency is applied during the open-circuit test, the power input Poc is practically equal to the rated core loss. It most applications, this value of core loss is typically assumed to remain constant for different load levels.
## Transformer Short-Circuit Test
In the short-circuit test, the low-voltage side is short-circuited and the high-voltage side is connected to a variable, low-voltage source. Measurements of power, current, and voltage are made on the high-voltage side as shown in Fig. 6.
Fig.6: Connections for Short Circuit Test
The applied voltage is adjusted until rated short-circuit current flows in the windings. This voltage is generally much smaller than the rated voltage, in the range of 0.05 to 0.10 per unit. Therefore, the current through the magnetizing branch is negligible, and the applied voltage may be assumed to occur wholly as a voltage drop across the transformer series impedance as shown in the equivalent circuit of Fig.7.
Fig.7: Equivalent Circuit for Short Circuit Test
The magnitude of the series impedance of referred to the high-voltage (primary) side may be calculated as follows:
$\begin{matrix} \left| {{Z}_{e1}} \right|=\frac{{{V}_{sc}}}{{{I}_{sc}}} & {} & \left( 8 \right) \\\end{matrix}$
The equivalent series resistance referred to the high-voltage side is
$\begin{matrix} {{R}_{e1}}=\frac{{{P}_{sc}}}{I_{sc}^{2}}={{R}_{1}}+{{a}^{2}}{{R}_{2}} & {} & \left( 9 \right) \\\end{matrix}$
Corresponding equivalent series reactance referred to the high-voltage side is
$\begin{matrix} {{X}_{e1}}=\sqrt{{{\left| {{Z}_{e1}} \right|}^{2}}-R_{e1}^{2}}={{X}_{1}}+{{a}^{2}}{{X}_{2}} & {} & \left( 10 \right) \\\end{matrix}$
The values of these parameters derived from the short-circuit test are used in conjunction with Fig. 1(a). These parameters may be referred to the low-voltage (secondary) side to derive the corresponding values for use with Fig. 1 (b) as follows:
$\begin{matrix} {{R}_{e2}}=\left( \frac{1}{{{a}^{2}}} \right){{R}_{e1}} & {} & \left( 11 \right) \\\end{matrix}$
$\begin{matrix} {{X}_{e2}}=\left( \frac{1}{{{a}^{2}}} \right){{X}_{e1}} & {} & \left( 12 \right) \\\end{matrix}$
Note that when rated current flows through the windings during the short-circuit test, the power input Psc is equal to the rated copper loss.
## Tests on Three-phase Transformers
When a three-phase transformer under-goes open-circuit and short-circuit tests, it must be remembered that the power being measured is total three-phase power, the measured voltage is line-to-line voltage, and the measured current is line current. The impedance parameters of interest are to be calculated on a per-phase basis. Therefore, before using the formulas derived above to calculate the resistances and reactances, the measured values must also be converted to per-phase values.
Did you find apk for android? You can find new Free Android Games and apps.
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://electricalacademia.com/transformer/determination-transformer-equivalent-circuit-parameters/", "fetch_time": 1544725370000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-51/segments/1544376825029.40/warc/CC-MAIN-20181213171808-20181213193308-00208.warc.gz", "warc_record_offset": 91103303, "warc_record_length": 15126, "token_count": 1588, "char_count": 5919, "score": 3.78125, "int_score": 4, "crawl": "CC-MAIN-2018-51", "snapshot_type": "longest", "language": "en", "language_score": 0.8345917463302612, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00028-of-00064.parquet"}
|
# Binary Calculator
## Binary Calculation—Add, Subtract, Multiply, or Divide
+ - × ÷ = ?
## Convert Binary Value to Decimal Value
Binary Value: = ?
## Result
Binary value: 01011111101
Decimal Value: = ?
RelatedHex Calculator | IP Subnet Calculator
The binary system is a numerical system that functions virtually identically to the decimal number system that people are likely more familiar with. While the decimal number system uses the number 10 as its base, the binary system uses 2. Furthermore, although the decimal system uses the digits 0 through 9, the binary system uses only 0 and 1, and each digit is referred to as a bit. Apart from these differences, operations such as addition, subtraction, multiplication, and division are all computed following the same rules as the decimal system.
Almost all modern technology and computers use the binary system due to its ease of implementation in digital circuitry using logic gates. It is much simpler to design hardware that only needs to detect two states, on and off (or true/false, present/absent, etc.). Using a decimal system would require hardware that can detect 10 states for the digits 0 through 9, and is more complicated.
Below are some typical conversions between binary and decimal values:
Binary/Decimal Conversion
Decimal Binary 0 0 1 1 2 10 3 11 4 100 7 111 8 1000 10 1010 16 10000 20 10100
While working with binary may initially seem confusing, understanding that each binary place value represents 2n, just as each decimal place represents 10n, should help clarify. Take the number 8 for example. In the decimal number system, 8 is positioned in the first decimal place left of the decimal point, signifying the 100 place. Essentially this means:
8 × 100 = 8 × 1 = 8
Using the number 18 for comparison:
(1 × 101) + (8 × 100) = 10 + 8 = 18
In binary, 8 is represented as 1000. Reading from right to left, the first 0 represents 20, the second 21, the third 22, and the fourth 23; just like the decimal system, except with a base of 2 rather than 10. Since 23 = 8, a 1 is entered in its position yielding 1000. Using 18, or 10010 as an example:
18 = 16 + 2 = 24 + 21
10010 = (1 × 24) + (0 × 23) + (0 × 22) + (1 × 21) + (0 × 20) = 18
The step by step process to convert from the decimal to the binary system is:
1. Find the largest power of 2 that lies within the given number
2. Subtract that value from the given number
3. Find the largest power of 2 within the remainder found in step 2
4. Repeat until there is no remainder
5. Enter a 1 for each binary place value that was found, and a 0 for the rest
Using the target of 18 again as an example, below is another way to visualize this:
2n 24 23 22 21 20 Instances within 18 1 0 0 1 0 Target: 18 18 - 16 = 2 → 2 - 2 = 0
Converting from the binary to the decimal system is simpler. Determine all of the place values where 1 occurs, and find the sum of the values.
EX: 10111 = (1 × 24) + (0 × 23) + (1 × 22) + (1 × 21) + (1 × 20) = 23
24 23 22 21 20 1 0 1 1 1 16 0 4 2 1
Hence: 16 + 4 + 2 + 1 = 23.
Binary addition follows the same rules as addition in the decimal system except that rather than carrying a 1 over when the values added equal 10, carry over occurs when the result of addition equals 2. Refer to the example below for clarification.
Note that in the binary system:
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0, carry over the 1, i.e. 10
EX:
10 11 11 10 1 + 1 0 1 1 1 = 1 0 0 1 0 0
The only real difference between binary and decimal addition is that the value 2 in the binary system is the equivalent of 10 in the decimal system. Note that the superscripted 1's represent digits that are carried over. A common mistake to watch out for when conducting binary addition is in the case where 1 + 1 = 0 also has a 1 carried over from the previous column to its right. The value at the bottom should then be 1 from the carried over 1 rather than 0. This can be observed in the third column from the right in the above example.
### Binary Subtraction
Similar to binary addition, there is little difference between binary and decimal subtraction except those that arise from using only the digits 0 and 1. Borrowing occurs in any instance where the number that is subtracted is larger than the number it is being subtracted from. In binary subtraction, the only case where borrowing is necessary is when 1 is subtracted from 0. When this occurs, the 0 in the borrowing column essentially becomes "2" (changing the 0-1 into 2-1 = 1) while reducing the 1 in the column being borrowed from by 1. If the following column is also 0, borrowing will have to occur from each subsequent column until a column with a value of 1 can be reduced to 0. Refer to the example below for clarification.
Note that in the binary system:
0 - 0 = 0
0 - 1 = 1, borrow 1, resulting in -1 carried over
1 - 0 = 1
1 - 1 = 0
EX1:
-11 20 1 1 1 – 0 1 1 0 1 = 0 1 0 1 0
EX2:
-11 2-10 0 – 0 1 1 = 0 0 1
Note that the superscripts displayed are the changes that occur to each bit when borrowing. The borrowing column essentially obtains 2 from borrowing, and the column that is borrowed from is reduced by 1.
### Binary Multiplication
Binary multiplication is arguably simpler than its decimal counterpart. Since the only values used are 0 and 1, the results that must be added are either the same as the first term, or 0. Note that in each subsequent row, placeholder 0's need to be added, and the value shifted to the left, just like in decimal multiplication. The complexity in binary multiplication arises from tedious binary addition dependent on how many bits are in each term. Refer to the example below for clarification.
Note that in the binary system:
0 × 0 = 0
0 × 1 = 0
1 × 0 = 0
1 × 1 = 1
EX:
1 0 1 1 1 × 1 1 1 0 1 1 1 + 1 0 1 1 1 0 = 1 0 0 0 1 0 1
As can be seen in the example above, the process of binary multiplication is the same as it is in decimal multiplication. Note that the 0 placeholder is written in the second line. Typically the 0 placeholder is not visually present in decimal multiplication. While the same can be done in this example (with the 0 placeholder being assumed rather than explicit), it is included in this example because the 0 is relevant for any binary addition / subtraction calculator, like the one provided on this page. Without the 0 being shown, it would be possible to make the mistake of excluding the 0 when adding the binary values displayed above. Note again that in the binary system, any 0 to the right of a 1 is relevant, while any 0 to the left of the last 1 in the value is not.
EX:
1 0 1 0 1 1 0 0
= 0 0 1 0 1 0 1 1 0 0
≠ 1 0 1 0 1 1 0 0 0 0
### Binary Division
The process of binary division is similar to long division in the decimal system. The dividend is still divided by the divisor in the same manner, with the only significant difference being the use of binary rather than decimal subtraction. Note that a good understanding of binary subtraction is important for conducting binary division. Refer to the example below, as well as to the binary subtraction section for clarification.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.calculator.net/binary-calculator.html?d2bnumber1=765&calctype=d2b&x=65&y=21", "fetch_time": 1675862872000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764500813.58/warc/CC-MAIN-20230208123621-20230208153621-00113.warc.gz", "warc_record_offset": 699368881, "warc_record_length": 6291, "token_count": 1956, "char_count": 7112, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.9080761671066284, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00011-of-00064.parquet"}
|
# Math Instruction Doesn't Add Up
It's time to tell students the truth, writes James E. Rosenbaum in American Educator: If you do poorly in high school, you'll do poorly in college and on the job. A useful sidebar explains what you need to do in high school to be successful.
The vast majority of high school seniors say they plan to get a college degree. Yet less than 40 percent will earn a two or four-year degree. Success is linked closely to high school performance: While 64 percent of "A" students with college plans earn a two-year or higher degree, only 14 percent of college-bound seniors with averages of "C" or lower earn any sort of degree. Half of the "C" and "D" students will not earn a single college credit. They'll take remedial classes and then give up.
Almost 40 percent of college-bound students “believed that school effort had little relevance for their future careers," writes Rosenbaum, a Northwestern professor. Wrong. “Over half the students who do more than 10 hours of homework a week will get a four-year college degree; only about 16 percent of those doing less than three hours of homework a week will earn a bachelor's degree.”
For students who go directly into the workforce, high school grades predict pay. "B" students earn considerably more than "C" students.
Let Math Be Math
The Connected Math curriculum is controversial in Madison, Wisc., — especially with parents who are mathematicians. The Capital Times compares a problem for seventh graders.
A gas station sells soda in three sizes. A 20-ounce cup costs 80 cents, a 32-ounce cup is 90 cents and a 64-ouncer goes for \$1.25.
Traditional: What size offers the most soda for the money?
Connected: If the gas station were to offer an 84-ounce Mega Swig, what would you expect to pay for it?
A student, for instance, could argue that the 84-ouncer would cost what the 20-ounce and 64-ounce cups cost together. Another student could say that soda gets cheaper with volume, and then choose an answer based on some per-ounce price slightly less than what was given for the 64-ounce drink.
Or a student could skip calculating the per-ounce prices and just pick a number: \$1.50 seems about right.
On one side, those who support Connected Math say that engaging students by presenting problems as real-life scenarios, often with no absolute solution or single path to arrive at an answer, fosters innovation and forces students to explain and defend their reasoning as they discover mathematical concepts.
The other side says the approach trades the clear, fundamental concepts of math, distilled through thousands of years of logical reasoning, for verbiage and vagary that may help students learn to debate but will not give them the foundation they need for more advanced mathematical study.
. . . University of Wisconsin-Madison math and computer science Professor Jin-Yi Cai is a critic.
"It goes around and around and things never really get down to the really crisp, elegant, basic fundamental principles," Cai said of the Connected Math texts.
"It takes away the elegance, it takes away the beauty, it takes away the most basic logic structure. And the students are left with a vague, touchy-feely idea," he said.
Cai and UW-Madison mathematician Melania Alvarez, who is running for school board, also object to all the essay writing required by Connected Math. Writing isn't math, they say.
My algebra teacher, Miss Diedrick, said that math is a language of its own.
X + Y = ?
Starting this year, California students are supposed to pass algebra to earn a high school diploma, but many districts are seeking waivers. That includes districts that made algebra a graduation requirement years ago, yet never really enforced it.
The requirement has forced schools to get serious about teaching algebra. For one thing, flunkers get better teachers, reports the San Jose Mercury News.
At Milpitas High School, Lam Le — who has taught mathematics at the college level — teaches two classes of beginning algebra.
Le inherited several students who were failing but says she has only one F student now.
In her class, students actually start work before the bell rings. And they don't start to pack up until it rings again. “Math takes discipline, and clear direction,'' Le said.
...Several students in her class said they support the requirement linking algebra to graduation, despite their struggles. “If you don't know how to do math, you can't get a good job,'' said Jouit Soliano, who came from the Philippines last year.
California's standards call for students to learn algebra in eighth grade. Yet the graduation exam was postponed because so many students were flunking the math portion of the test, which required only a 55 percent. Only the hardest questions required high school math skills.
David Klein, a math professor at Cal State-Northridge, writes about California's math mess. At his campus, the ethnic studies departments run remedial math classes in which everyone passes. Klein teaches an arithmetic class for future elementary teachers, who are allowed to use calculators on the arithmetic exam.
Engineering Is Expendable
San Francisco State's president, forced to cut the budget, is threatening to close the School of Engineering, writes Debra Saunders in the Chronicle. Raza Studies, Recreational and Leisure Studies and Women Studies would remain, preparing students for...Well, leisure studies will come in handy for the permanently unemployed. The Institute on Sexuality, Social Inequality and Health is not threatened. Saunders writes:
It makes you wonder if the guys in Engineering should rename their discipline. You know, call it The School of Engineering, Structural Inequality and Disparity Dynamics. Even better: The School of Social Engineering.
Some 700 engineering students would be out of luck if SF State dumps the department.
Letters
Stephen Chastain of Maine says:
I just graduated with a mechanical engineering degree and agree that the U.S. math and science skills are virtually worthless. The mechanical engineering department is scrambling every year to find qualified students. As a result, at least 70 percent of them are foreign nationals with the majority being Indian. They are there because they are the best in the world, not because of the government dumbing down the program. In contrast, U.S. students are suing to get into law school. What does that tell you about national priorities? Is there any wonder why jobs are going overseas?
Willard Smith of Aurora, Il., says:
In the midst of a four-year technical recession, where tens of thousands of experienced engineers have been out of work for more than two years, the notion that high-tech companies cannot find extremely competent people in the United States is laughable.
Greg Rupper writes:
Reform K12 claims that if Indian programmers demanded U.S. wages they would be out of work. I used to work for a company that recruited programmers from India. They were paid U.S. wages. They were all good programmers. Our office stopped recruiting from India only because our company opened an engineering office in India. U.S. companies are perfectly willing to pay Indian programmers U.S. wages if necessary.
As to poor math and science education, most U.S. high schools only require two years of math. The electrical engineering degree that I got assumed that the students would take calculus in their first year of college. There is no way that a student can be ready to take calculus with only two years of high school math.
Engineering is based on physics. Physics is taught in the language of calculus. All of the high school physical science classes that try to teach science without even using algebra are a joke. I believe you can teach more science with your mouth taped shut then you can without writing any equations.
The main reason why we fail at math and science is not because we don't know how to teach math and science. The problem is that we don't try. If we required high school students to take the math necessary to understand science, then they would learn the material, and be ready to pursue high-tech careers. But, given the choice of taking math or another elective class, many students chose other classes. (Even I have to admit that math is boring. You need a lot of math before it really becomes useful.)
If we want to improve math and science, we need to teach math. It’s that simple.
Chuck Mobley of Douglasville, Ga., writes:
I have a BS in engineering, an MBA, graduate-level statistics coursework, and experience as a U.S. Navy officer. I've been out of work since December, 2001. When my son dropped out of college to work at Best Buy, I just smiled and wished him luck. I certainly couldn't admonish him the way my father would have.
By the way, in my last job, a software development position, most of my co-workers were Indians here on work visas. They weren't that smart, and were very hard to collaborate with, because they saved collaboration for their own kind. If project managers think they can work collaboratively with these people across eleventy-seven time zones, they're going to be very disappointed.
I believed in the ideal of education bringing success in life. I am very disappointed.
Satvik Patel of Grayson, Ga., says:
I have experienced math and science education in both India and in America. When I studied in India, we were not allowed to use calculators. Every math problem had to be done by hand. And once I came to America, I found that my fellow students even for the simplest problems would whip out their calculators. We have become so dependent on these gadgets that we find it difficult to function without one. The students and professionals in India were trained to think with their brains, not their gadgets. This I believe gives them an edge when it comes to jobs and doing them efficiently.
Mark Loper of Jax, Fla., says:
The argument that high tech companies outsource because of poor math and science skills on the part of American workers is just another excuse. I have an aeronautical engineering degree and I have worked in software development the last 15 years. Never during that time was I required to know or perform math or science anywhere near the level that I learned in school. I had about seven semesters of calculus and beyond, yet I never needed anything more than algebra and a little trig.
Companies outsource for one reason — cheap labor. And any other excuse is a lie.
Rich Lewis writes:
I taught eighth grade Language Arts almost eight years ago. I left after two years, because I couldn’t see the light at the end of the tunnel. Most of my students began eighth grade at a sixth grade reading level. There was only so much I could do to prepare them for the Connecticut Mastery Tests each October. The problem was multiplied by consistent behavior problems and a lack of support from parents and the administration. I worked at least 75 hours each week; I made \$29,000 per year.
I now make well over \$100,000 each year. I don’t work nearly as hard as I did when I was a teacher. We are losing an entire generation of teachers because the compensation does not match the requirements for the job. George Bush’s “No Child Left Behind” is a nice slogan, but it can’t erase 15 years of decline. There are too many good teachers leaving the profession for myriad reasons including low wages, violence, unattainable standards, lack of support from parents, lack of support from administration, and just a complete lack of respect.
The real problem is that there are too many kids per classroom, and the children are more unruly then ever. Parents do not teach their children how to behave in a classroom setting. Parents do not teach their children to respect adults. Parents do not spend enough time helping their children with homework. Parents do not spend enough time with their children in general.
Unfortunately, this problem can not be solved with money. Dramatically increasing teachers’ salaries will help to attract and retain top talent, which will lead to creative teaching and higher test scores. But, the real problem is a complete lack of parental involvement. We need someone to teach the parents how to raise their children in an environment that’s more conducive to academic success.
Joanne Jacobs writes about education and other issues at JoanneJacobs.com. She’s writing a book, Ride the Carrot Salad, about a start-up charter high school in San Jose.
Respond to the Writer
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.foxnews.com/story/2004/04/09/math-instruction-doesnt-add-up.html", "fetch_time": 1496007257000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-22/segments/1495463611560.41/warc/CC-MAIN-20170528200854-20170528220854-00220.warc.gz", "warc_record_offset": 627146294, "warc_record_length": 12150, "token_count": 2573, "char_count": 12533, "score": 3.578125, "int_score": 4, "crawl": "CC-MAIN-2017-22", "snapshot_type": "longest", "language": "en", "language_score": 0.966510534286499, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00020-of-00064.parquet"}
|
Friends Pairing Problem (Solution)
Problem Statement
Given a total of n friends, each friend can either remain single or can be paired up with some other friend. The pairing for each friend can be done only once. Find out the total number of ways in which the friends can be single or get paired up.
Sample Test Cases
Input 1: n = 2
Output 1: 2
Explanation 1: There are only 2 ways for the pairings to be made, both friends get paired into 1 group or both friends remain single.
Input 2: n = 3
Output 2: 4
Explanation 2: The pairings are listed below:
{1}, {2}, {3}
{1, 2}, {3}
{1, 3}, {2}
{2, 3}, {1}
Recursive Brute Force Method
The problem can be solved with recursion by considering the choices we have for the nth person. For the nth person,
• The nth person stays single or unpaired, and we recurse for the remaining n – 1 people.
• Nth person pairs up with any for the remaining n – 1 people. The number of ways of doing this (n – 1) * Rec(n – 2).
The recurrence can then be written as:
• Rec(n) = Rec(n – 1) + (n – 1) * Rec(n – 2)
Which can be easily simulated with brute force.
Base Case: Note that if there are 2 friends, there are 2 ways to arrange them, and for a single friend, there is only 1 way to arrange them. So for n <= 2, the answer will be n.
C++ Code
```int countPairings(int n) {
return n <= 2 ? n : countPairings(n - 1) + (n - 1) * countPairings(n - 2);
}```
Java Code
```public static int countPairings(int n) {
return n <= 2 ? n : countPairings(n - 1) + (n - 1) * countPairings(n - 2);
}```
Python Code
```def countPairings(n):
return n if n <= 2 else countPairings(n - 1) + (n - 1) *
countPairings(n - 2)```
Time Complexity: Exponential
Space Complexity: O(1) // If recursion stack space is ignored.
Approach 2: Memoization
Observe from the above recursion tree, that there are many subproblems in the brute force method which are being called again and again. We can memoize these subproblems so as to avoid them from being recalculated again and again with a dynamic programming approach.
Here, the states of our dp will be dp[n]: Number of ways to arrange a total of n friends. The transitions will be the same as in the brute force approach.
C++ Implementation
```int countPairings(int n, vector < int > & dp) {
if (dp[n] != -1) {
return dp[n];
}
dp[n] = n <= 2 ? n : countPairings(n - 1, dp) + (n - 1) * countPairings(n - 2, dp);
return dp[n];
}```
Java Implementation
```public static int countPairings(int n, int[] dp) {
if (dp[n] != -1) {
return dp[n];
}
dp[n] = n <= 2 ? n : countPairings(n - 1, dp) + (n - 1) * countPairings(n - 2, dp);
return dp[n];
}```
Python Implementation
```def countPairings(n, dp):
if dp[n] != -1:
return dp[n]
dp[n] = n if n <= 2 else countPairings(n - 1, dp) + (n - 1) * countPairings(n - 2, dp)
return dp[n]```
Time Complexity: O(n)
Space Complexity: O(n)
Approach 3: Iterative Dynamic Programming
Similar to the recursive memoization-based dynamic programming approach, we can also solve the problem with an iterative dynamic programming-based approach, i.e with tabulation. The states and the transitions for this approach will be the same as the ones for the recursive approach just implemented in an iterative manner.
C++ Implementation
```int countPairings(int n) {
vector < int > dp(n + 1);
for (int i = 0; i <= n; i++) {
if (i <= 2) {
dp[i] = i;
} else {
dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
}
}
return dp[n];
}```
Java Implementation
```public static int countPairings(int n) {
int dp[] = new int[n + 1];
for (int i = 0; i <= n; i++) {
if (i <= 2) {
dp[i] = i;
} else {
dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
}
}
return dp[n];
}```
Python Implementation
```def countPairings(n):
dp = [0] * (n + 1)
for i in range(n + 1):
if i <= 2:
dp[i] = i
else:
dp[i] = dp[i - 1] + (i - 1) * dp[i - 2]
return dp[n]```
Time Complexity: O(n)
Space Complexity: O(n)
Approach 4: Combinatorics Based Approach
The problem given is basically equivalent to the combinatorial problem of “In how many ways can we divide a total of n items into multiple groups (maximum group size here being 2).” This problem is solved by the following formula:
Referring to the above formula, we can precompute the factorials in our problem, and using them calculate the result from the given formula.
C++ Code
```int countPairings(int n) {
vector < int > fact(n + 1);
fact[0] = 1;
for (int i = 1; i <= n; i++) {
fact[i] = fact[i - 1] * i;
}
int groupsOfOnes = n, groupsOfTwos = 1, ans = 0;
while (groupsOfOnes >= 0) {
ans += fact[n] / (groupsOfTwos * fact[groupsOfOnes] * fact[(n - groupsOfOnes) / 2]);
groupsOfOnes -= 2;
groupsOfTwos <<= 1;
}
return ans;
}```
Java Code
```public static int countPairings(int n) {
int fact[] = new int[n + 1];
fact[0] = 1;
for (int i = 1; i <= n; i++) {
fact[i] = fact[i - 1] * i;
}
int groupsOfOnes = n, groupsOfTwos = 1, ans = 0;
while (groupsOfOnes >= 0) {
ans += fact[n] / (groupsOfTwos * fact[groupsOfOnes] * fact[(n - groupsOfOnes) / 2]);
groupsOfOnes -= 2;
groupsOfTwos <<= 1;
}
return ans;
}```
Python Code
```def countPairings(n):
fac = [1] * (n + 1)
for i in range(1, n + 1):
fac[i] = fac[i - 1] * i
groupsOfOnes = n
groupsOfTwos = 1
ans = 0
while groupsOfOnes >= 0:
ans += fac[n] // (
fac[groupsOfOnes] * fac[(n - groupsOfOnes) // 2] * groupsOfTwos
)
groupsOfOnes -= 2
groupsOfTwos <<= 1
return ans```
Time Complexity: O(n)
Space Complexity: O(n)
FAQs
Q. How does the brute force approach have exponential time complexity?
A. From the recursion tree, we can observe that the tree divides into two branches at each level. So the number of states to be visited becomes exponential in number when measured as a function of input n.
Q. How many states does the recursion depend upon in this problem?
A. The recursion is solely dependent on the number of people (n) as the state for this problem.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.interviewbit.com/blog/friends-pairing-problem/", "fetch_time": 1716770287000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00697.warc.gz", "warc_record_offset": 696091600, "warc_record_length": 22048, "token_count": 1814, "char_count": 5828, "score": 4.09375, "int_score": 4, "crawl": "CC-MAIN-2024-22", "snapshot_type": "latest", "language": "en", "language_score": 0.8361843824386597, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00013-of-00064.parquet"}
|
Probabilities are often stated in terms of cumulative
This preview shows pages 21–23. Sign up to view the full content.
Probabilities are often stated in terms of cumulative distribution functions (CDFs). The CDF F x ( a ) of a random variable x is the probability that the random variable x takes on a value less than or equal to a , viz. F x ( a ) = P ( x a ) F x ( a ) is a nondecreasing function of its argument that takes on values between 0 and 1. Another important func- tion is the probability density function (PDF) f x ( x ) of the random variable x , which is related to the CDF through integration/differentiation: F x ( a ) = P ( x a ) = integraldisplay a −∞ f x ( x ) dx dF x ( x ) dx = f x ( x ) The PDF is the probability of observing the variable x in the interval ( x,x + dx ) . Since a random variable always has some value, we must have integraldisplay −∞ f x ( x ) dx = 1 Note that, for continuous random variables, a PDF of 0 (1) for some outcome does not mean that that outcome is prohibited (guaranteed). Either the PDF or the CDF completely determine the behavior of the random variable and are therefore equivalent. Random variables that are normal or Gaussian occur frequently in many contexts and have the PDF: f N ( x ) = 1 σ 2 π e 1 2 ( x μ ) 2 2 which is parametrized by μ and σ . Normal distributions are often denoted by the shorthand N ( μ,σ ) . Another common distribution is the Rayleigh distribution, given by f r ( r ) = braceleftbigg r σ 2 e r 2 / 2 σ 2 ,r 0 0 ,r< 0 which is parametrized just by σ . It can be shown that a random variable defined as the length of a 2d Cartesian vector with component lengths given by Gaussian random variables has this distribution. In terms of radar signals, the envelope of quadrature Gaussian noise is Rayleigh distributed. Still another important random variable called χ 2 has the PDF f χ 2 = 1 2 ν/ 2 Γ( ν/ 2) x 1 2 ν 1 e x/ 2 where Γ is the Gamma function, Γ( x ) = integraltext 0 ξ x 1 e xi , and where the parameter ν is referred to as the number of degrees of freedom. It can be shown that if the random variable x i is N (0 , 1) , then the random variable z = n summationdisplay i =1 x 2 i is χ 2 with n degrees of freedom. If x i refers to voltage samples, for example, then z refers to power estimates based on n voltage samples or realizations. 20
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Mean and variance It is often not the PDF that is of interest but rather just its moments. The expectation or mean of a random variable x , denoted E ( x ) or μ x , is given by E ( x ) = integraldisplay −∞ xf x ( x ) dx For example, it can readily be shown that the expected value for a normal distribution N ( μ,σ ) is E ( x ) = μ . The variance of a random variable, denoted Var ( x ) or σ 2 x , is given by Var ( x ) = E [( x μ x ) 2 ] = E ( x 2 ) μ 2 x = integraldisplay −∞ ( x μ x ) 2 f x ( x ) dx The standard deviation of x meanwhile, which is denoted σ x , is the square root of the variance. In the case of the same normal distribution N ( μ,σ ) , the variance is given by σ 2 , and the standard deviation by σ .
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.coursehero.com/file/p36jj/Probabilities-are-often-stated-in-terms-of-cumulative-distribution-functions/", "fetch_time": 1542173133000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-47/segments/1542039741628.8/warc/CC-MAIN-20181114041344-20181114063344-00556.warc.gz", "warc_record_offset": 854349975, "warc_record_length": 111015, "token_count": 1063, "char_count": 4196, "score": 3.796875, "int_score": 4, "crawl": "CC-MAIN-2018-47", "snapshot_type": "latest", "language": "en", "language_score": 0.9001952409744263, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"}
|
graphing-linear-equations-kristakingmath
# Interactive video lesson plan for: graphing linear equations (KristaKingMath)
#### Activity overview:
► My Algebra 1 course: https://www.kristakingmath.com/algebra-1-course
In this video we learn how to graph linear equations on a cartesian coordinate system. We'll start by ensuring that the equation is in slope-intercept form as y=mx+b, and then mark the y-intercept. From there, we'll add points around the y-intercept using the slope of the line.
● ● ● GET EXTRA HELP ● ● ●
If you could use some extra help with your math class, then check out Krista’s website // http://www.kristakingmath.com
● ● ● CONNECT WITH KRISTA ● ● ●
Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)
Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”
So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, I’ve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student—from basic middle school classes to advanced college calculus—figure out what’s going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here: http://www.kristakingmath.com
INSTAGRAM // https://www.instagram.com/kristakingmath/
PINTEREST // https://www.pinterest.com/KristaKingMath/
QUORA // https://www.quora.com/profile/Krista-King
Tagged under: graphing lines,expert,=mx+, ,graphing,-intercept,slope,graph line,graphing linear equations,Krista King,educational,algebra,slope-intercept form,calculus
Clip makes it super easy to turn any public video into a formative assessment activity in your classroom.
Add multiple choice quizzes, questions and browse hundreds of approved, video lesson ideas for Clip
Make YouTube one of your teaching aids - Works perfectly with lesson micro-teaching plans
Play this activity
1. Students enter a simple code
2. You play the video
3. The students comment
4. You review and reflect
* Whiteboard required for teacher-paced activities
## Ready to see what elsecan do?
With four apps, each designed around existing classroom activities, Spiral gives you the power to do formative assessment with anything you teach.
Quickfire
Carry out a quickfire formative assessment to see what the whole class is thinking
Discuss
Create interactive presentations to spark creativity in class
Team Up
Student teams can create and share collaborative presentations from linked devices
Clip
Turn any public video into a live chat with questions and quizzes
### Spiral Reviews by Teachers and Digital Learning Coaches
@kklaster
Tried out the canvas response option on @SpiralEducation & it's so awesome! Add text or drawings AND annotate an image! #R10tech
Using @SpiralEducation in class for math review. Student approved! Thumbs up! Thanks.
@ordmiss
Absolutely amazing collaboration from year 10 today. 100% engagement and constant smiles from all #lovetsla #spiral
@strykerstennis
Students show better Interpersonal Writing skills than Speaking via @SpiralEducation Great #data #langchat folks!
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://spiral.ac/sharing/amjwqmh/graphing-linear-equations-kristakingmath", "fetch_time": 1638773017000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964363290.39/warc/CC-MAIN-20211206042636-20211206072636-00513.warc.gz", "warc_record_offset": 578992556, "warc_record_length": 10292, "token_count": 768, "char_count": 3397, "score": 3.921875, "int_score": 4, "crawl": "CC-MAIN-2021-49", "snapshot_type": "latest", "language": "en", "language_score": 0.8807855248451233, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00038-of-00064.parquet"}
|
# What is the molar mass of NaHCO_3
## How much sodium hydrogen carbonate is there in a vitamin C tablet?
1st calculation step:
Work orders:
1.
Outρ= m / V [g / L] becomes: m =ρ * V;
2. Calculation of the substance portion m (CO2):
m (CO2) = 1.83 g / L * 0.085 L (average value from 10 student experiments)
m (CO2) = 0.1555 g CO2; This mass of CO2 corresponds to the average weight of 10 tablet pieces of the vitamin C tablet: 0.88 g
1. Intermediate result: 0.88 g of vitamin C tablets produce 0.1555 g of CO2 ;
The Molar mass of citric acid: 192 g / mol;
The molar mass of CO2 : 44 g / mol;
The molar mass of NaHCO3: 84 g / mol;
2nd step: Establishing the reaction equation:
C.6H8O7 3 NaHCO3 —> 3 CO2 + 3 H.2O + C6H5O7N / A3(aq)
Molar masses: 192 g / mol + 252 g / mol ----> 132 g / mol + 54 g / mol + 258 g / mol
Control: sum on the left: 444 g / mol Sum on the right: 444 g / mol
After Stoichiometric equation so arise from 1 Mole citric acid and 3 Moles of sodium hydrogen carbonate 3 Mole CO2 = 132 g CO2 .
3rd step: Calculation of the amount of substance n = m / M
The released portion of substance of m (CO2) = 0.1555 g corresponds to n (CO2) = 0.003535 mol = 3.535 mmol
If n (CO2) = 0.003535 mol, then according to the reaction equation n (NaHCO3) = 0.003535 mol.
Conversion of n (NaHCO3) above m = n * M : m (NaHCO3) = 0.29696 g
4th step: Determination of the mass fraction ω (NaHCO3) on the tablet:
The bulk of the Tablet weight was: 0.88 g = 100%. Then 0.2969 g NaHCO correspond3 = 33,74 %≈34 %
Overall result: mass fraction
ω (NaHCO3) = 34 %
Work orders:
1. Calculate the Portion of fabric m and theMass fractionω (citric acid) in the tablet.
The amount of substance released in the reaction n (CO2) was: n (CO2) = 3.535 mmol
According to the reaction equation, there are 3 moles of CO2 one (1) mole of citric acid;
so the amount of substance n (citric acid) = 1/3 mol CO2 = 1.178 mmol
To m = n * M then the substance portion is citric acid: m (citric acid) = 1.178 mmol * 192 g / mol / 1000 = 0.226 g
The mass fraction of citric acid ω (citric acid) is thus: 25.71% with a tablet mass of 0.88 g = 100%.
2. How high is it Mass fraction of the other substances together? See the table of contents above!
Mass fraction ω (NaHCO3): 34 %
Mass fraction ω (citric acid): 25.71%
Mass fraction ω (vitamin C): 4.5% according to packaging information
Then that is Mass fraction of oligofructose (sugar), starch, sweeteners, flavorings and colorings together:
100% - 64,21 =35,79%
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://satanwarez.xyz/?post=1969", "fetch_time": 1631887222000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-39/segments/1631780055645.75/warc/CC-MAIN-20210917120628-20210917150628-00163.warc.gz", "warc_record_offset": 523632162, "warc_record_length": 4549, "token_count": 848, "char_count": 2517, "score": 4.0625, "int_score": 4, "crawl": "CC-MAIN-2021-39", "snapshot_type": "latest", "language": "en", "language_score": 0.7160194516181946, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00024-of-00064.parquet"}
|
# Search by Topic
#### Resources tagged with Rotations similar to Central Distance:
Filter by: Content type:
Stage:
Challenge level:
### There are 38 results
Broad Topics > Transformations and their Properties > Rotations
### Get Cross
##### Stage: 4 Challenge Level:
A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?
### Weighty Problem
##### Stage: 3 Challenge Level:
The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . .
### Illusion
##### Stage: 3 and 4 Challenge Level:
A security camera, taking pictures each half a second, films a cyclist going by. In the film, the cyclist appears to go forward while the wheels appear to go backwards. Why?
### Turning Triangles
##### Stage: 3 Challenge Level:
A triangle ABC resting on a horizontal line is "rolled" along the line. Describe the paths of each of the vertices and the relationships between them and the original triangle.
### ...on the Wall
##### Stage: 3 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Matching Frieze Patterns
##### Stage: 3 Challenge Level:
Sort the frieze patterns into seven pairs according to the way in which the motif is repeated.
### The Frieze Tree
##### Stage: 3 and 4
Patterns that repeat in a line are strangely interesting. How many types are there and how do you tell one type from another?
### Rollin' Rollin' Rollin'
##### Stage: 3 Challenge Level:
Two circles of equal radius touch at P. One circle is fixed whilst the other moves, rolling without slipping, all the way round. How many times does the moving coin revolve before returning to P?
### Coke Machine
##### Stage: 4 Challenge Level:
The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design...
### Simplifying Transformations
##### Stage: 3 Challenge Level:
How many different transformations can you find made up from combinations of R, S and their inverses? Can you be sure that you have found them all?
### Combining Transformations
##### Stage: 3 Challenge Level:
Does changing the order of transformations always/sometimes/never produce the same transformation?
### Decoding Transformations
##### Stage: 3 Challenge Level:
See the effects of some combined transformations on a shape. Can you describe what the individual transformations do?
### Turning Tangles
##### Stage: 3 Challenge Level:
Look carefully at the video of a tangle and explain what's happening.
### Cubic Rotations
##### Stage: 4 Challenge Level:
There are thirteen axes of rotational symmetry of a unit cube. Describe them all. What is the average length of the parts of the axes of symmetry which lie inside the cube?
### Shaping up with Tessellations
##### Stage: 2 and 3
This article describes the scope for practical exploration of tessellations both in and out of the classroom. It seems a golden opportunity to link art with maths, allowing the creative side of your. . . .
### In a Spin
##### Stage: 4 Challenge Level:
What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse?
### John's Train Is on Time
##### Stage: 3 Challenge Level:
A train leaves on time. After it has gone 8 miles (at 33mph) the driver looks at his watch and sees that the hour hand is exactly over the minute hand. When did the train leave the station?
### Rolling Triangle
##### Stage: 3 Challenge Level:
The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks.
### Symmetric Trace
##### Stage: 4 Challenge Level:
Points off a rolling wheel make traces. What makes those traces have symmetry?
##### Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Frieze Patterns in Cast Iron
##### Stage: 3 and 4
A gallery of beautiful photos of cast ironwork friezes in Australia with a mathematical discussion of the classification of frieze patterns.
### Notes on a Triangle
##### Stage: 3 Challenge Level:
Can you describe what happens in this film?
### Twizzle Arithmetic
##### Stage: 4 Challenge Level:
Arrow arithmetic, but with a twist.
### Transformation Game
##### Stage: 3 Challenge Level:
Why not challenge a friend to play this transformation game?
### Hand Swap
##### Stage: 4 Challenge Level:
My train left London between 6 a.m. and 7 a.m. and arrived in Paris between 9 a.m. and 10 a.m. At the start and end of the journey the hands on my watch were in exactly the same positions but the. . . .
### Napoleon's Theorem
##### Stage: 4 and 5 Challenge Level:
Triangle ABC has equilateral triangles drawn on its edges. Points P, Q and R are the centres of the equilateral triangles. What can you prove about the triangle PQR?
### Overlaid
##### Stage: 2, 3 and 4 Challenge Level:
Overlaying pentominoes can produce some effective patterns. Why not use LOGO to try out some of the ideas suggested here?
### Overlap
##### Stage: 3 Challenge Level:
A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .
### Arrow Arithmetic 3
##### Stage: 4 Challenge Level:
How can you use twizzles to multiply and divide?
### Arrow Arithmetic 1
##### Stage: 4 Challenge Level:
The first part of an investigation into how to represent numbers using geometric transformations that ultimately leads us to discover numbers not on the number line.
### Arrow Arithmetic 2
##### Stage: 4 Challenge Level:
Introduces the idea of a twizzle to represent number and asks how one can use this representation to add and subtract geometrically.
### Attractive Rotations
##### Stage: 3 Challenge Level:
Here is a chance to create some attractive images by rotating shapes through multiples of 90 degrees, or 30 degrees, or 72 degrees or...
### Paint Rollers for Frieze Patterns.
##### Stage: 3 and 4
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach.
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
### Rotations Are Not Single Round Here
##### Stage: 4 Challenge Level:
I noticed this about streamers that have rotation symmetry : if there was one centre of rotation there always seems to be a second centre that also worked. Can you find a design that has only. . . .
### A Roll of Patterned Paper
##### Stage: 4 Challenge Level:
A design is repeated endlessly along a line - rather like a stream of paper coming off a roll. Make a strip that matches itself after rotation, or after reflection
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://nrich.maths.org/public/leg.php?code=132&cl=3&cldcmpid=4729", "fetch_time": 1506107101000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-39/segments/1505818689102.37/warc/CC-MAIN-20170922183303-20170922203303-00326.warc.gz", "warc_record_offset": 256685531, "warc_record_length": 8672, "token_count": 1640, "char_count": 7383, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2017-39", "snapshot_type": "latest", "language": "en", "language_score": 0.8810426592826843, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00040-of-00064.parquet"}
|
An Example on - -
Solved Examples
Attempt following question by selecting a choice to answer. cSolve the linear system.5x - 6y = -10 --- (1)- 6x + 7y = 11 --- (2)BBBB ccccA. c(4 , 5 )B. c(- 4 , 5 )C. c(4 , - 5 )D. c(- 4 , - 5 )cc A Step 1: [Multiply Equation 1 by 6.] 30x - 36y = -60Step 2: [Multiply Equation 2 by 5.]-30x + 35y = 55______________Step 5: [Add the Equations.] - y = -5Step 6: [Solve for y.] y = 5Step 7: [Substitute y = 5 in Equation 1.]5x - 6(5) = -10Step 8: [Multiply.]5x - 30 = -10Step 9: [Add 30 to both sides.]5x = 20Step 10: [Solve for x.]x = 4Step 11: The solution for the linear system is (4, 5).
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.icoachmath.com/solvedexample/sampleworksheet.aspx?process=/__cstlqvxbefxaxbgefgxkhgbmxeajjdxd&.html", "fetch_time": 1412143053000000000, "content_mime_type": "application/xhtml+xml", "warc_filename": "crawl-data/CC-MAIN-2014-41/segments/1412037663365.9/warc/CC-MAIN-20140930004103-00398-ip-10-234-18-248.ec2.internal.warc.gz", "warc_record_offset": 662581019, "warc_record_length": 12002, "token_count": 274, "char_count": 646, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2014-41", "snapshot_type": "longest", "language": "en", "language_score": 0.7298447489738464, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00013-of-00064.parquet"}
|
Parents can use OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Ex 7(C) to provide additional support to their children.
## S Chand Class 9 ICSE Maths Solutions Chapter 7 Logarithms Ex 7(C)
If log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990, log 7 = 0.8451 and log 11 = find the value of the following :
Question 1.
log 6
Solution:
log 6 = log (2 x 3) = log 2 + log 3
= 0.3010 + 0.4771 = 0.7781
Question 2.
log 12
Solution:
log 12 = log (2 x 2 x 3) = log (2² x 3)
= log 2² + log 3 = 2 log 2 + log 3
= 2 (0.3010) + 0.4771 = 0.6020 + 0.4771 = 1.0791
Question 3.
log 15
Solution:
log 15 = log (3 x 5) = log 3 + log 5
= 0.4771 + 0.6990 = 1.1761
Question 4.
log 200
Solution:
log 200 = log (2³ x 5²) = log 2³ + log 5²
= 3 log 2 + 2 log 5
= 3 (0.3010) + 0.6990 x 2
= 0.9030 + 1.3980 = 2.3010
Question 5.
log 36
Solution:
sog 36 – log (4 x 9) = log (2² x 3²)
= log 2² + log 3² = 2 log 2 + 2 log 3
= 2 (0.3010) + 2 (0.4771)
= 0.6020 + 0.9542
= 1.5562
Question 6.
log 80
Solution:
log 80 = log (24 x 5)
= log 24 + log 5
= 4 log 2 + log 5 = 4 (0.3010) + 0.6990
= 1.2040 + 0.6990 = 1.9030
Question 7.
log 2$$\frac { 1 }{ 3 }$$
Solution:
log (2$$\frac { 1 }{ 3 }$$) = log($$\frac { 7 }{ 3 }$$) = log 7 – log 3
= 0.8451 – 0.4771
= 0.3680
Question 8.
log 11³
Solution:
log (11)³ = 3 log 11
= 3 x 1.0414 = 3.1242
Question 9.
log (2$$\frac { 1 }{ 3 }$$)5
Solution:
log (2$$\frac { 1 }{ 3 }$$)5 = log ($$\frac { 7 }{ 3 }$$)5 = 5 log ($$\frac { 7 }{ 3 }$$)
= 5 [log 7 – log 3] = 5 [0.8451 – 0.4771]
= 5 (0.3680) = 1.8400
Question 10.
If log 6 = 0.7782, find the value of log 36.
Solution:
log 6 = 0.7782
log 36 = log (6)² = 2 log 6
= 2 (0.7782) = 1.5564
Question 11.
Given log10 25 = x, log10 75 = y, evaluate without using logarithmic tables, in terms of x and y.
(i) log10 3
(ii) log102
Solution:
Question 12.
Given log 31.87 = x, write down in terms of x.
(i) log (31.87)²
(ii) log10 0.03187
(iii) log10 $$\sqrt{31870}$$
Solution:
log 31.87 = x
(i) log (31.87)² = 2 log (31.87)
= 2x
(ii) log10 0.03187 = log10$$\frac{31.87}{1000}$$
= log10 31.87 – log10 1000
= x – 3 (log 1000 = 3)
(iii) log10 $$\sqrt{31870}$$ = log10 (31870)1/2
= $$\frac { 1 }{ 2 }$$ log10 31870 = $$\frac { 1 }{ 2 }$$ log (31.87 x 1000)
= $$\frac { 1 }{ 2 }$$ [log 31.87 + log 1000]
= $$\frac { 1 }{ 2 }$$ (x + 3) = $$\frac { x+3 }{ 2 }$$
Question 13.
Solve the equation
(i) log10 (x + 1) + log10 (A – 1) = log10 11 + 2 log103
(ii) log (10x + 5) – log (x – 4) = 2
Solution:
(i) log10 (x + 1) + log10 (A – 1) = log10 11 + 2 log103
⇒ log (x + 1) (x – 1) = log (11 x 3²) (∵ 2 log10 3 = log10 3²)
⇒ log (x² – 1) = log (11 x 9) ⇒ log (x² – 1) = log 99
Comparing we get.
x² – 1 = 99 ⇒ x² = 99 + 1 = 100 = (10)²
⇒ x = 10
∴ x = 10
(ii) log (10x + 5) – log (x – 4) = 2
⇒ log $$\frac{10 x+5}{x-4}$$ = log 100 (∵ log 100 = 2)
Comparing both sides,
$$\frac{10 x+5}{x-4}=\frac{100}{1}$$
100x – 400 = 10x + 5
⇒ 100x – 10x = 5 + 400 ⇒ 90x = 405
⇒ x = $$\frac { 405 }{ 90 }$$ = $$\frac { 45 }{ 10 }$$ = 4.5
∴ x = 4.5
Question 14.
(a) Given 2 log10x + $$\frac { 1 }{ 2 }$$ log10y = 1, express y in terms of x.
(b) Express as a single logarithm :
2 log 3 – $$\frac { 1 }{ 2 }$$ log 16 + log 12
Solution:
Question 15.
Given that log10 y + 2log10 x = 2, express y in terms of x.
Solution:
Question 16.
If a = 1 + log10 2 – log10 5, b = 2 log10 3 and c = log10 m – log10 5, find the value of m if a + b = 2c (Do not use log tables).
Solution:
a = 1 + log10 2 – log10 5 = log10 10 + log10 2 – log10 5
= log10 $$\frac { 10×2 }{ 5 }$$– = log10 4
b = 2 log10 3 = log10 3² = log10 9
c = log10 m – log10 5 = log10 $$\frac { m }{ 5 }$$
∵ a + b = 2c
∴ log10 4 + log10 9 = 2 log $$\frac { m }{ 5 }$$
⇒ log (4 × 9) = log ($$\frac { m }{ 5 }$$)² = log$$\frac { m² }{ 25 }$$
Comparing both sides,
4 x 9 = $$\frac { m² }{ 25 }$$ ⇒ m² = 4 x 9 x 25 = 900 = (30)²
∴ m = 30
Question 17.
Express as a single logarithm
2 + $$\frac { 1 }{ 2 }$$ log10 9 – 2 log10 5
Solution:
Question 18.
If a = log 12, b – log 6 and c = 2 log $$\sqrt{2}$$, find
(i) a – b – c
(ii) 9a-b-c
Solution:
a = log 12, b = log 6, c = 2 log $$\sqrt{2}$$ = log
$$(\sqrt{2})^2$$ = log 2
(i) a – b – c = log 12 – log 6 – log 2
= log $$\frac { 12 }{ 6×2 }$$ = log 1 = 0 (∵ log 1 = 0)
(ii) 9a-b-c = 9° [∵ a – b – c = 0 proved in (i)]
= 1 (∵ x° = 1)
Question 19.
If x = log10 12, y = log4 2 x log10 9 and z = log10 (0.4) then find
(i) x – y – z
(ii) prove that 6x – y- z = 6
Solution:
x = log10 12, y = log4 2 x log10 9, z = log10 (0.4)
(i) x = log10 12 = log10 (2² x 3) = log 2² + log 3
= 2 log10 2 + log10 3
y = log4 2 x log109 = log42 x log10
= log4 2 x 2 log10 3 = log4 (4)$$\frac { 1 }{ 2 }$$ x 2 log10 3 (∴ loga = 1)
= log103
z = log10 0.4 = log10 $$\frac { 4 }{ 10 }$$ = log10 4 – log10 10
= log10 2² – 1 = 2 log10 2 – 1
x – y – z = 2 log10 2 + log10 3 – log10 3 – 2 log102 + 1 = 1
(ii) 6x-y-z = 61 = 6 [∵ x – y – z = 1 (proved in (i)]
Hence proved.
Question 20.
If p = log10 20 and q = log10 25, find x such that 2 log10 (x + 1) = 2p – q.
Solution:
p = log10 20 = log10 (22 x 5) = log10 2² + log10 5
= 2 log 2 + log 5
q = log10 25 = log10 (5²) = 2 log10 5
Now,
2p – q = 2 [2 log10 2 + log10 5] – 2 log10 5
= 4 log10 2 + 2 log10 5 – 2 log10 5
= 4 log10 2 = 2 log10
= 2 log10 4
2 log10(x + 1) = 2 log10 4
Comparing, we get
x + 1 = 4 ⇒ x = 4 – 1 = 3
∴ x = 3
Question 21.
Without using logarithm tables, evaluate :
3 + log10 (10-2)
Solution:
3 + log10 (10-2) = 3 + (- 2 log10 10)
= 3 – 2 log10 10 = 3 – 2 x 1
= 3 – 2 = 1 (∵ loga a = 1)
= 1
Question 22.
Given log10 x = a, log10 y = b
(i) Write down 10n-1 in terms of x
(ii) Write down 102b in terms of y
(iii) If log10 P = 2a – b, express P in terms of x and y.
Solution:
(i) log10 x = a, log10y = b
∵ log10 x = a and log10 y = b
∴ 10n = x … (i)
∴ 10b = y … (ii)
Now,
(i) 10n-1 = $$\frac{10^a}{10^1}=\frac{x}{10}$$ [From (i)]
(ii) 102b = (10b)² = y² [From (ii)]
(iii) log10 P = 2a – b
⇒ log10P = 2 log10x – log10y
⇒ log10 P = log x² – log10 y
⇒ log10 P = log $$\frac { x² }{ y }$$
Comparing, we get
p = $$\frac { x² }{ y }$$
Question 23.
Simplify without using tables :
$$2 \log _{10} 5+\log _{10} 8-\frac{1}{2} \log _{10} 4$$
Solution:
Question 24.
Given that log102 = x, log10 3 = y, find
(i) log10 60
(ii) log101.2 in terms of x and y
Solution:
(i) log10 60 = log10 (2 x 3 x 10)
= log10 2 + log10 3 + log10 10
= x + y + 1 (∵ log10 10 = 1)
(ii) log101.2 = log10 $$\frac { 12 }{ 10 }$$ = log10 12 – log10 10
= log (2² x 3) – log10 10
= 2 log 2 + log 3 – log10 10 (∵ log10 10 = 1)
= 2x + y – 1
Question 25.
Given 2 log10 x + 1 = log10 250, find
(i) x
(ii) log102x
Solution:
(i) 2 log10 x + 1 = log10 250
⇒ log10x² + 1 = log10 250
⇒ log10 x² + log10 10 = log10 250 (∵ loga a = 1)
⇒ log (x² x 10) = log10 250
Comparing both side,
10x² = 250 ⇒ x² = 25 = (5)²
∴ x = 5
(ii) log10 2x = log10 2 x 5 = log10 10 = 1
Question 26.
(a) Given that log x = m + n and log y = m – n, express the value of log10 $$\frac { 10x }{ y² }$$ in terms of in and n. (b) Solve for x : log10x = – 2.
Solution:
(a) log x = m + n, log y = m – n
log10 $$\frac{10 x}{y^2}=\log _{10} 10 x-\log _{10} y^2$$
= log10 10 + log10 x – 2 log10 y
= 1 + m + n – 2 (m – n)
= 1 + m + n – 2m + 2n
= 1 – m + 3n
(b) log10 x = – 2 = log10$$\frac { 1 }{ 100 }$$
Comparing we get,
x = $$\frac { 1 }{ 100 }$$
Question 27.
If log $$\left(\frac{p+q}{3}\right)=\frac{1}{2}$$ = (log p + log q) prove that p² + q² = 7pq.
Solution:
Log $$\left(\frac{p+q}{3}\right)=\frac{1}{2}$$ = (log p + log q)
⇒ $$\log \frac{p+q}{3}=\frac{1}{2}(\log p \times q)=\log (p q)^{\frac{1}{2}}$$
Comparing we get,
$$\frac{p+q}{3}=(p q)^{\frac{1}{2}}$$
Squaring both sides,
$$\left(\frac{p+q}{3}\right)^2=p q \Rightarrow \frac{p^2+q^2+2 p q}{9}$$ = pq
⇒ P² + q² + 2 pq
⇒ p² + q² + 2pq = 9pq
⇒ p² + q² = 9pq – 2pq ⇒ 7pq
∴ p² + q² = 2pq
Hence proved.
Question 28.
If x² + y² = 51xy, prove that log $$\frac { 1 }{ 2 }$$ = $$\frac { 1 }{ 2 }$$ (log x + log y).
Solution:
x² + y² = 51xy
⇒ x² + y² – 2xy = 51 xy – 2xy
(Subtracting 2xy)
⇒ (x – y)² = 49xy
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://icsesolutions.com/op-malhotra-class-9-maths-solutions-chapter-7-ex-7c/", "fetch_time": 1702007896000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-50/segments/1700679100710.22/warc/CC-MAIN-20231208013411-20231208043411-00440.warc.gz", "warc_record_offset": 358226337, "warc_record_length": 13926, "token_count": 3970, "char_count": 7956, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2023-50", "snapshot_type": "latest", "language": "en", "language_score": 0.5922195911407471, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00017-of-00064.parquet"}
|
Prime Factorisations
In this worksheet, students select the correct prime factorisation for the given numbers.
Key stage: KS 2
Curriculum topic: Maths and Numerical Reasoning
Curriculum subtopic: Factors and Multiples
Difficulty level:
QUESTION 1 of 10
The list of prime numbers starts as follows:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37........
All other whole numbers are known as composite and can be written as the product of prime factors.
This is known as prime factorisation.
Examples
35 = 5 x 7
14 = 2 x 7
28 = 2 x 2 x 7
24 = 2 x 2 x 2 x 3
Writing 24 = 2 x 3 x 4 is correct but it is not a full prime factorisation because the 4 is not prime.
Select the correct prime factorisation for:
15
1 x 5
3 x 5
3 x 3 x 5
1 x 15
Select the correct prime factorisation for:
30
2 x 15
3 x 10
3 x 3 x 5
2 x 3 x 5
Select the correct prime factorisation for:
24
2 x 12
2 x 2 x 2 x 3
2 x 3 x 4
3 x 8
Select the correct prime factorisation for:
25
25
5 x 5
2 x 5
5 x 20
Select the correct prime factorisation for:
36
2 x 2 x 3 x 3
4 x 9
3 x 3 x 5
5 x 7 + 1
Select the correct prime factorisation for:
20
2 x 2 x 3
4 x 5
2 x 2 x 5
2 x 10
Select the correct prime factorisation for:
16
2 x 2 x 3
4 x 4
2 x 2 x 4
2 x 2 x 2 x 2
Select the correct prime factorisation for:
40
2 x 2 x 10
4 x 2 x 5
2 x 2 x 5
2 x 2 x 2 x 5
Select the correct prime factorisation for:
42
2 x 2 x 7
2 x 3 x 7
2 x 3 x 5
2 x 2 x 2 x 7
Select the correct prime factorisation for:
45
2 x 2 x 5
2 x 3 x 5
3 x 3 x 5
2 x 2 x 2 x 5
• Question 1
Select the correct prime factorisation for:
15
CORRECT ANSWER
3 x 5
• Question 2
Select the correct prime factorisation for:
30
CORRECT ANSWER
2 x 3 x 5
• Question 3
Select the correct prime factorisation for:
24
CORRECT ANSWER
2 x 2 x 2 x 3
• Question 4
Select the correct prime factorisation for:
25
CORRECT ANSWER
5 x 5
• Question 5
Select the correct prime factorisation for:
36
CORRECT ANSWER
2 x 2 x 3 x 3
• Question 6
Select the correct prime factorisation for:
20
CORRECT ANSWER
2 x 2 x 5
• Question 7
Select the correct prime factorisation for:
16
CORRECT ANSWER
2 x 2 x 2 x 2
• Question 8
Select the correct prime factorisation for:
40
CORRECT ANSWER
2 x 2 x 2 x 5
• Question 9
Select the correct prime factorisation for:
42
CORRECT ANSWER
2 x 3 x 7
• Question 10
Select the correct prime factorisation for:
45
CORRECT ANSWER
3 x 3 x 5
---- OR ----
Sign up for a £1 trial so you can track and measure your child's progress on this activity.
What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
Get started
Start your £1 trial today.
Subscribe from £10/month.
• Tuition Partner
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.edplace.com/worksheet_info/11+/keystage2/year5/topic/1161/6267/prime-factorisations", "fetch_time": 1579964004000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00495.warc.gz", "warc_record_offset": 848525407, "warc_record_length": 24991, "token_count": 1005, "char_count": 3036, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2020-05", "snapshot_type": "longest", "language": "en", "language_score": 0.7326207756996155, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00041-of-00064.parquet"}
|
# Chapter 1 - Trigonometric Functions - Section 1.2 Angle Relationships and Similar Triangles - 1.2 Exercises: 39
Equilateral triangle.
#### Work Step by Step
Using the compass to draw a semicircle means that the distance between the compass tip and the pencil is the radius of the circle. The distance from the point marked before to the edge of the circle is the radius. The second length can be used to construct second side. And finally, you can add third side (which is the radius of the circle so congruent). At the end we get triangle with equal sides, therefore, it is equilateral triangle.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.gradesaver.com/textbooks/math/trigonometry/trigonometry-10th-edition/chapter-1-trigonometric-functions-section-1-2-angle-relationships-and-similar-triangles-1-2-exercises-page-17/39", "fetch_time": 1481367927000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-50/segments/1480698543035.87/warc/CC-MAIN-20161202170903-00196-ip-10-31-129-80.ec2.internal.warc.gz", "warc_record_offset": 485483827, "warc_record_length": 37613, "token_count": 168, "char_count": 762, "score": 3.640625, "int_score": 4, "crawl": "CC-MAIN-2016-50", "snapshot_type": "longest", "language": "en", "language_score": 0.8859851956367493, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00054-of-00064.parquet"}
|
# Thread: prove by MI that
1. ## prove by MI that
First, it is given that f(x) is a differentiable function.
and
$\displaystyle \frac{d}{dx}(f(x))^2 = 2 f(x) f '(x)$
Then, I have to prove by MI that
$\displaystyle \frac{d}{dx}(f(x))^n = n (f(x)) ^{n-1}f '(x)$
2. Originally Posted by ling_c_0202
First, it is given that f(x) is a differentiable function.
and
$\displaystyle \frac{d}{dx}(f(x))^2 = 2 f(x) f '(x)$
Then, I have to prove by MI that
$\displaystyle \frac{d}{dx}(f(x))^n = n (f(x)) ^{n-1}f '(x)$
Do this like this.
Proof (by induction) that,
$\displaystyle [f_1(x)\cdot f_2(x)\cdot...\cdot f_n(x)]'$=$\displaystyle f'_1(x)f_2(x)\cdot...\cdot f_n(x)+$$\displaystyle f_1(x)\cdot f'_2(x)\cdot ...\cdot f_n(x)+$$\displaystyle f_1(x)f_2(x)\cdot ...\cdot f'_n(x)$
Then,
$\displaystyle [f(x)]^n=\underbrace{f(x)\cdot f(x)\cdot...\cdot f(x)}_{n}$
Then by the theorem before the derivative is,
$\displaystyle f'(x)f(x)\cdot ...\cdot f(x)+f(x)f'(x)\cdot...\cdot f(x)+$$\displaystyle f(x)\cdot f(x)\cdot...\cdot f'(x) - n times. But each summand can be reduced to, \displaystyle [f(x)]^{n-1}f'(x) n-times. Thus, \displaystyle n[f(x)]^{n-1}f'(x) --- Keep you Calculus question in Calculus section -=USER WARNED=- 3. Originally Posted by ling_c_0202 please help with this question.. First, it is given that f(x) is a differentiable function. and \displaystyle \frac{d}{dx}(f(x))^2 = 2 f(x) f '(x) The above is the base case for the induction with \displaystyle n=2. Then, I have to prove by MI that \displaystyle \frac{d}{dx}(f(x))^n = n (f(x)) ^{n-1}f '(x) Suppose this is true for some \displaystyle k, then \displaystyle \frac{d}{dx}(f(x))^k = k (f(x)) ^{k-1}f '(x) Now (using the product rule): \displaystyle \frac{d}{dx}(f(x))^{k+1} = \frac{d}{dx} f(x) (f(x))^k$$\displaystyle = f(x)\frac{d}{dx}(f(x))^k + \frac{df}{dx} (f(x))^k$
So applying the supposition to the first term:
$\displaystyle \frac{d}{dx}(f(x))^{k+1} $$\displaystyle = k f(x) (f(x))^{k-1}f'(x) + f'(x) (f(x))^k$$\displaystyle =(k+1) (f(x))^k f'(x)$
Which is the required result for $\displaystyle k+1$.
So if we assume the required result for some $\displaystyle k$, it is true for $\displaystyle k+1$, hence
with the base case the result is proven by Mathematical Induction for all $\displaystyle n \ge 2$
(note we could have taken $\displaystyle n=1$ as the base case - which is trivialy true -
then the proof would apply for all $\displaystyle n \ge 1$).
RonL
4. ## thank you
Thank you very much!
I found that i got a mistake when i apply the product rule... stupid me
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/calculus/5040-prove-mi.html", "fetch_time": 1527229372000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-22/segments/1526794867041.69/warc/CC-MAIN-20180525043910-20180525063910-00550.warc.gz", "warc_record_offset": 188363562, "warc_record_length": 10226, "token_count": 881, "char_count": 2553, "score": 4.34375, "int_score": 4, "crawl": "CC-MAIN-2018-22", "snapshot_type": "latest", "language": "en", "language_score": 0.7150132060050964, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00019-of-00064.parquet"}
|
# 5.14: Electric Field as the Gradient of Potential
In Section [m0061_Force_Energy_and_Potential_Difference], it was determined that the electrical potential difference $$V_{21}$$ measured over a path $${\mathcal C}$$ is given by $V_{21} = - \int_{\mathcal C} {\bf E}({\bf r}) \cdot d{\bf l} \label{m0063_eV12}$ where $${\bf E}({\bf r})$$ is the electric field intensity at each point $${\bf r}$$ along $${\mathcal C}$$. In Section [m0064_V_due_to_Point_Charges], we defined the scalar electric potential field $$V({\bf r})$$ as the electric potential difference at $${\bf r}$$ relative to a datum at infinity. In this section, we address the “inverse problem” – namely, how to calculate $${\bf E}({\bf r})$$ given $$V({\bf r})$$. Specifically, we are interested in a direct “point-wise” mathematical transform from one to the other. Since Equation \ref{m0063_eV12} is in the form of an integral, it should not come as a surprise that the desired expression will be in the form of a differential equation.
We begin by identifying the contribution of an infinitesimal length of the integral to the total integral in Equation \ref{m0063_eV12}. At point $${\bf r}$$, this is
$dV = - {\bf E}({\bf r}) \cdot d{\bf l} \label{eVABd}$
Although we can proceed using any coordinate system, the following derivation is particularly simple in Cartesian coordinates. In Cartesian coordinates,
$d{\bf l} = \hat{\bf x}dx + \hat{\bf y}dy + \hat{\bf z}dz$
We also note that for any scalar function of position, including $$V({\bf r})$$, it is true that
$dV = \frac{\partial V}{\partial x} dx + \frac{\partial V}{\partial y} dy + \frac{\partial V}{\partial z} dz$
Note the above relationship is not specific to electromagnetics; it is simply mathematics. Also note that $$dx = d{\bf l} \cdot \hat{\bf x}$$ and so on for $$dy$$ and $$dz$$. Making these substitutions into the above equation, we obtain:
$dV = \frac{\partial V}{\partial x} \left( d{\bf l} \cdot \hat{\bf x} \right) + \frac{\partial V}{\partial y} \left( d{\bf l} \cdot \hat{\bf y} \right) + \frac{\partial V}{\partial z} \left( d{\bf l} \cdot \hat{\bf z} \right)$
This equation may be rearranged as follows:
$dV = \left( \left[ \hat{\bf x} \frac{\partial}{\partial x} + \hat{\bf y} \frac{\partial}{\partial y} + \hat{\bf z}\frac{\partial}{\partial z} \right] V \right) \cdot d{\bf l}$
Comparing the above equation to Equation \ref{eVABd}, we find:
${\bf E}({\bf r}) = - \left[ \hat{\bf x}\frac{\partial}{\partial x} + \hat{\bf y} \frac{\partial}{\partial y} + \hat{\bf z}\frac{\partial}{\partial z} \right] V$
Note that the quantity in square brackets is the gradient operator “$$\nabla$$” (Section [m0098_Gradient]). Thus, we may write
$\boxed{ {\bf E} = - \nabla V } \label{m0063_eEPEDV}$
which is the relationship we seek.
The electric field intensity at a point is the gradient of the electric potential at that point after a change of sign (Equation \ref{m0063_eEPEDV}).
Using Equation \ref{m0063_eEPEDV}, we can immediately find the electric field at any point $${\bf r}$$ if we can describe $$V$$ as a function of $${\bf r}$$. Furthermore, this relationship between $$V$$ and $${\bf E}$$ has a useful physical interpretation. Recall that the gradient of a scalar field is a vector that points in the direction in which that field increases most quickly. Therefore:
The electric field points in the direction in which the electric potential most rapidly decreases.
This result should not come as a complete surprise; for example, the reader should already be aware that the electric field points away from regions of net positive charge and toward regions of net negative charge (Sections [m0002_Electric_Field_Intensity] and/or [m0102_Coulombs_Law]). What is new here is that both the magnitude and direction of the electric field may be determined given only the potential field, without having to consider the charge that is the physical source of the electrostatic field.
Example $$\PageIndex{1}$$: Electric field of a charged particle, beginning with the potential field
In this example, we determine the electric field of a particle bearing charge $$q$$ located at the origin. This may be done in a “direct” fashion using Coulomb’s Law (Section [m0102_Coulombs_Law]). However, here we have the opportunity to find the electric field using a different method. In Section [m0064_V_due_to_Point_Charges] we found the scalar potential for this source was:
$V({\bf r}) = \frac{q}{4\pi\epsilon r}$
So, we may obtain the electric field using Equation \ref{m0063_eEPEDV}:
${\bf E} = - \nabla V = -\nabla \left( \frac{q}{4\pi\epsilon r} \right)$
Here $$V({\bf r})$$ is expressed in spherical coordinates, so we have (Section [m0139_Vector_Operators]):
${\bf E} = -\left[ \hat{\bf r}\frac{\partial}{\partial r} +\hat{\bf \theta}\frac{1}{r}\frac{\partial}{\partial \theta} +\hat{\bf \phi}\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi} \right] \left( \frac{q}{4\pi\epsilon r} \right)$
In this case, $$V({\bf r})$$ does not vary with $$\phi$$ or $$\theta$$, so the second and third terms of the gradient are zero. This leaves
E &= -( )
&= -
&= - ( - )
So we find
E = +
as was determined in Section [m0102_Coulombs_Law].
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Book%3A_Electromagnetics_I_(Ellingson)/5%3A_Electrostatics/5.14%3A_Electric_Field_as_the_Gradient_of_Potential", "fetch_time": 1571508387000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-43/segments/1570986697439.41/warc/CC-MAIN-20191019164943-20191019192443-00556.warc.gz", "warc_record_offset": 457389124, "warc_record_length": 20637, "token_count": 1505, "char_count": 5202, "score": 4.125, "int_score": 4, "crawl": "CC-MAIN-2019-43", "snapshot_type": "latest", "language": "en", "language_score": 0.8364151120185852, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00057-of-00064.parquet"}
|
## Calculus (3rd Edition)
$$y'= (4x-8)e^{(x-1)^2}e^{(x-3)^2}.$$
Taking the $\ln$ on both sides of the equation, we get $$\ln y= \ln \left(e^{(x-1)^2}e^{(x-3)^2}\right)$$ Then using the properties of $\ln$, we can write $$\ln y= \ln e^{(x-1)^2}+\ln e^{(x-3)^2}\\ =(x-1)^2+(x-3)^2.$$ Now taking the derivative for the above equation, we have $$\frac{y'}{y}= 2(x-1)+2(x-3)=4x-8,$$ Hence $y'$ is given by $$y'=y(4x-8)=(4x-8)e^{(x-1)^2}e^{(x-3)^2}$$
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-7-exponential-functions-chapter-review-exercises-page-386/55", "fetch_time": 1685710095000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00154.warc.gz", "warc_record_offset": 853872546, "warc_record_length": 13280, "token_count": 209, "char_count": 445, "score": 4.25, "int_score": 4, "crawl": "CC-MAIN-2023-23", "snapshot_type": "latest", "language": "en", "language_score": 0.6743900179862976, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00004-of-00064.parquet"}
|
# Measurement Worksheets for Ages 5-9
Discover an engaging collection of Measurement Worksheets for Ages 5-9 at Kids Academy. Designed to sharpen young minds, these printable worksheets cover essential measurement concepts such as length, weight, volume, and time. Each activity is thoughtfully crafted to make learning interactive and fun, helping children understand and apply measurement in real-life scenarios. Our worksheets are perfect for both classroom use and at-home practice, ensuring kids build a solid foundation in math. Enrich your child's mathematical journey today with these diverse and educational measurement resources. Visit Kids Academy to explore our complete selection!
Favorites
Interactive
• 5-9
• Measurement
Take your child north to practice data reading with a polar bear friend! This worksheet challenges 3rd graders to interpret data on a graph and solve related math problems.
Worksheet
## Compare Length in Inches and Centimeters Worksheet
This worksheet has your child measuring objects in both centimeters and inches. The first task requires measuring both objects in centimeters, then finding their difference. The second task is the same but in inches.
Compare Length in Inches and Centimeters Worksheet
Worksheet
## Cups, Pints and Quarts With Gallon Man Worksheet
Measurement in cups, pints, quarts and gallons has never been this fun!
Cups, Pints and Quarts With Gallon Man Worksheet
Worksheet
## Flower Math: Measuring Height Worksheet
Let your child learn length with Flower Math! Measure each flower with colorful beads and see how much they measure up to. This worksheet offers a fun and vivid way to learn the concept of length.
Flower Math: Measuring Height Worksheet
Worksheet
## Kids Measurements Worksheet
Help your kids look at the groupings on the bottom right of the worksheet. Compare the heights of the kids shown in the picture to the numbers in each group. Have them check which of the groups of numbers correctly shows the heights.
Kids Measurements Worksheet
Worksheet
## Drawing and Line Plot: Mary's Teddy Bears Worksheet
This worksheet teaches kids to draw and interpret line plots. Mary has 10 teddy bears and has measured their heights. Ask your child to draw a line plot showing the heights of Mary's teddy bears. It's a great way for little ones to learn how to organize information quickly and correctly.
Drawing and Line Plot: Mary's Teddy Bears Worksheet
Worksheet
## Measuring Quest: Inches and Centimeters Worksheet
Your child will measure items in both inches and centimeters and check the correct measurements from the options given. The metric ruler states 2.5 cm equals 1 inch, and 15 cm equals 6 inches. This worksheet helps ensure accurate measurements, despite the different figures.
Measuring Quest: Inches and Centimeters Worksheet
Worksheet
## Tall or Short and Long or Short? Worksheet
This worksheet is fun and helpful for kids to understand the difference between height and length. With pictures of familiar objects, they can compare and choose the box with the right answer. This helps them gain skills and a better foundation for future measuring.
Tall or Short and Long or Short? Worksheet
Worksheet
## Which One Is Longer - Length Worksheet
Let's go to the jungle - kids get excited to practice math with animal-filled worksheets! This one is sure to have your pupil mastering length in no time.
Which One Is Longer - Length Worksheet
Worksheet
## Which Is Taller Worksheet
Help your kindergartner learn about height with this fun worksheet! They'll love comparing zoo animals and deciding which are taller, shorter, or equal. It's a great way to practice math and have fun!
Which Is Taller Worksheet
Worksheet
## Empty and Light or Heavy and Full? Worksheet
Kids can have trouble understanding measurement, like the differences between heavy and light. This worksheet helps them associate full with heavy and empty with light, using familiar images. Plus, it's a fun way to practice fine motor skills, tracing the lines to connect the pictures.
Empty and Light or Heavy and Full? Worksheet
Worksheet
## Years Worksheet
This worksheet is a great way to test your children's knowledge of the calendar. Ask them if they can name the days of the week and months in a year. Read and discuss the questions with them and help them find the right answers. Encourage them to check their answers.
Years Worksheet
Worksheet
## Cups, Pints and Quarts Worksheet
Let your kids explore the world of math with our worksheet! Have them count cups, pints and quarts in pictures and circle the right answer. Get even more free printable math worksheets for kindergarten on our website. Come on, join the adventure!
Cups, Pints and Quarts Worksheet
Worksheet
## Measurement: Compare Volumes Worksheet
Compare volumes with Kids Academy and have fun! (80 words) Practice measuring with Kids Academy and have fun! Compare volumes in pictures and circle which holds more. Keep learning with free math worksheets available on our website. Enjoy the process and have fun!
Measurement: Compare Volumes Worksheet
Worksheet
## Learning about measuring objects in inches, feet and yards Worksheet
Download, cut and let your preschoolers learn and have fun estimating objects like pencils, airplanes and swimming pools in inches, feet and yards!
Learning about measuring objects in inches, feet and yards Worksheet
Worksheet
## Comparing with a Third Object Worksheet
Help your child master measurement skills with this free downloadable math worksheet! Ask them to compare each pair of objects, checking the box for the longer one. Then, read the question and compare the objects across the pairs before ticking the box and continuing to the next group of pictures.
Comparing with a Third Object Worksheet
Worksheet
## Mass Word Problems Worksheet
This worksheet challenges kids with tricky mass word problems. It offers valuable practice with grams. Through division and subtraction, children gain experience handling mass.
Mass Word Problems Worksheet
Worksheet
## Measuring Tools Mix–up Worksheet
Test your child's knowledge of measuring instruments! In this worksheet, have them identify which items are used for measuring length. Talk through the items with them, then ask them to circle the right answer. Use this PDF to challenge your child and learn more about measuring tools.
Measuring Tools Mix–up Worksheet
Worksheet
## Interpret Line Plot Worksheet
Kids can learn to read graphs and analyze data easily with line plots. A fun zoo animal worksheet motivates kids to answer questions by reading the line plot and finding out how many animals a local zoo has.
Interpret Line Plot Worksheet
Worksheet
## Steve's House: Measuring Length Worksheet
Introduce your kids to the various measuring tools and their uses. Explain how a ruler, plastic tape, yard stick, and metal tape measure are used to measure different objects. Guide them as they help Steve measure things in his home with the correct tool.
Steve's House: Measuring Length Worksheet
Worksheet
## Which Has Less? Worksheet
Kids can develop their number sense by comparing and contrasting objects to identify which has less. This worksheet helps them practice: it's bright and colorful, featuring familiar objects and pictures to count, strengthening basic counting skills. It's the perfect warm-up for comparing and contrasting larger numbers.
Which Has Less? Worksheet
Worksheet
## Lets Measure Length Worksheet
It's time to help your child take their measuring skills up a notch! This colorful PDF sheet from Kids Academy introduces kids to measuring with paperclips. Guide them through the page, showing them how to count the clips to measure each image. Then, circle the correct length for each to complete the sheet!
Lets Measure Length Worksheet
Worksheet
## Making a Correct Number Line: Balcony Garden Worksheet
Teach your child to draw and identify a line plot with this worksheet. Have them help Sophia check her plot for the heights of plants in her balcony, ensuring no numbers are skipped. Line plots are a great way to quickly organize information and an essential skill your child needs.
Making a Correct Number Line: Balcony Garden Worksheet
Worksheet
## Taxi Math Worksheet
This 3rd grade worksheet helps kids learn data viz with fun taxis and colors to make the graph come alive. Picture graphs are an engaging way to teach data analysis, and this worksheet is the perfect tool! Kids love picture graphs! This 3rd grade worksheet uses colorful taxis to help them learn to read graphs better. Each image represents a number, teaching data analysis with visuals and fun. An engaging way to learn data viz, this worksheet is perfect for young learners.
Taxi Math Worksheet
Worksheet
Learning Skills
Caring about measurement for ages 5-9 is crucial as it lays the foundation for critical math skills and everyday life competencies. During these formative years, children are naturally curious about the world around them, making it an ideal time to introduce concepts of length, weight, volume, and time. Learning measurement helps children understand and quantify their environment, promoting logical reasoning and problem-solving abilities.
Incorporating measurement into early education enhances cognitive development and encourages practical application of math in real-life contexts. For instance, measuring ingredients for a recipe or determining the time it takes to complete a task helps children see the relevance of math beyond the classroom. Furthermore, grasping measurement principles strengthens their ability in other subjects like science and engineering, fostering a well-rounded academic future.
Parents and teachers who prioritize teaching measurement skills also support the development of fine motor skills through activities like using rulers, scales, and measuring cups. Additionally, it instills a sense of accuracy and attention to detail, encouraging conscientious work habits. Overall, focusing on measurement helps young learners build a strong Math foundation, enhances their analytical thinking, and equips them with practical skills essential for daily life and future academic success.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.kidsacademy.mobi/printable-worksheets/age-5-9/math/measurements/", "fetch_time": 1721631931000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763517833.34/warc/CC-MAIN-20240722064532-20240722094532-00209.warc.gz", "warc_record_offset": 723961193, "warc_record_length": 71032, "token_count": 1965, "char_count": 10212, "score": 3.90625, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.8662489652633667, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00011-of-00064.parquet"}
|
0
# How do you multiply millimeters by millimeters?
Updated: 4/28/2022
Wiki User
10y ago
Exactly the same way you multiply inches by inches, or miles by miles.
Do I need to go through the details ? OK: Multiply the numbers, and
give the product the unit of mm2 (square millimeters). That's an area.
Wiki User
10y ago
Earn +20 pts
Q: How do you multiply millimeters by millimeters?
Submit
Still have questions?
Related questions
### 15 meters to millimeters?
Multiply by 1000 15,000 millimeters.
### How many millimeters is 94 inches?
Multiply by 2540 to convert inches to millimeters
### How would you calculate a cable in millimeters?
Multiply the length in inches by 25.4 to get millimeters.
### What is 18 thousandths in millimeters?
inches to millimeters, multiply inches by 25.4. You do the math.
### why would you multiply to change centimeters to millimeters?
There are 10 millimeters to every 1 centimeter. So, there would be more millimeters in a given length than centimeters. To increase a number you would multiply.
### How can you change meters to millimeters?
Multiply by 1,000
multiply by 1000
### How do you convert from meters to millimeters?
Multiply by 1000.
Multiply by ten.
### How do you convert centimters back to millimeters?
1 centimeter = 10 millimeters Multiply (centimeters) by 10 to get the same length in (millimeters).
### How do you change centimeters in to millimeters?
you multiply the amount of centimeters by 10 to get to millimeters (cm = mm * 10)
### How many centimeters are in 40 millimeters and what do you multiply it by?
400 centimeters. There are 10 millimeters in 1 centimeter
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.answers.com/other-math/How_do_you_multiply_millimeters_by_millimeters", "fetch_time": 1708986674000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-10/segments/1707947474663.47/warc/CC-MAIN-20240226194006-20240226224006-00042.warc.gz", "warc_record_offset": 381293909, "warc_record_length": 45877, "token_count": 414, "char_count": 1653, "score": 3.578125, "int_score": 4, "crawl": "CC-MAIN-2024-10", "snapshot_type": "latest", "language": "en", "language_score": 0.8905039429664612, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00025-of-00064.parquet"}
|
# Alg 1 -- Adding & Subtracting Rational Expressions - Day 2
Subject
Resource Type
Common Core Standards
Product Rating
4.0
3 Ratings
File Type
Compressed Zip File
28 MB
Share
Product Description
I use this lesson to show my students how to add and subtract rational expressions that do not have common denominators. This corresponds to Section 11.4 of the Prentice Hall CA Algebra 1 textbook. I break this lesson into two days -- Day 1: I show them how to add and subtract with like denominators; Day 2: I show them how to add and subtract with unlike denominators. This is the PowerPoint lesson for Day 2.
There is a video in the lesson that describes how to work these problems, including the factoring of quadratic trinomials in order to simplify the answer. That makes this a great lesson for days when you have a substitute teacher, or for home teaching, or you can use that time to take roll while the lesson continues.
This is my fourth lesson in my unit on Rational Expressions & Equations.
Powerpoint Lessons for Unit 14
14.1 Simplifying Rational Expressions
14.2 Multiplying & Dividing Rational Expressions
14.3 Adding & Subtracting Rational Expressions – Day 1
14.4 Adding & Subtracting Rational Expressions – Day 2
14.5 Complex Fractions
14.6 Solving Rational Equations
14.7 Work Problems
Get all 7 lessons & all 7 activities at a 30% discount!
Fun Reviews
Operations with Rational Expressions Relay
Operations with Rational Expressions Scavenger Hunt
Solving Rational Equations Scavenger Hunt
Work Problems Scavenger Hunt
Simplifying Complex Fractions Scavenger Hunt
Solving Rational Equations (easy) Scavenger Hunt
Rational Expressions & Equations Jeopardy
Get all 7 activities at a 30% discount!
For fun ways to review other important topics, try these:
Scavenger Hunts
Fun PowerPoint Review Games
Relays
Tic Tac Toes
Partner Problems
Bingos
Common Core Standards
A-REI 2. Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.
A-APR 7. Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.
Previous California Standards for Algebra 1:
12.0 Students simplify fractions with polynomials in the numerator and denominator by factoring both and reducing them to the lowest terms.
13.0 Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques.
Total Pages
N/A
N/A
Teaching Duration
N/A
Report this Resource
\$4.00
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.teacherspayteachers.com/Product/Alg-1-Adding-Subtracting-Rational-Expressions-Day-2-419873", "fetch_time": 1542798066000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-47/segments/1542039747665.82/warc/CC-MAIN-20181121092625-20181121114625-00066.warc.gz", "warc_record_offset": 1020055680, "warc_record_length": 20073, "token_count": 619, "char_count": 2716, "score": 3.921875, "int_score": 4, "crawl": "CC-MAIN-2018-47", "snapshot_type": "longest", "language": "en", "language_score": 0.8427140116691589, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"}
|
Question
# An electric dipole, consisting of two opposite charges of magnitude $2\times {{10}^{-6}}~C$ each separated by a distance 3 cm is placed in an electric field of magnitude $2\times {{10}^{5}}~N{{C}^{-1}}$ along the direction making ${{90}^{o}}$ with the field. Torque acting on the dipole is:\begin{align}& \text{A}\text{. }12\times {{10}^{-1}} \\ & \text{B}\text{. }12\times {{10}^{-2}} \\ & \text{C}\text{.}12\times {{10}^{-3}} \\ & \text{D}\text{. }12\times {{10}^{-4}} \\ \end{align}
Hint: Electric dipole has equal charges, but of opposite sign, at both the ends separated by some distance. The force due to the electric field experienced by both the charges will be same in magnitude, but the direction will be opposite. As we know, an equal and opposite couple of forces generate torque and hence, the dipole will experience some torque acting on it while placed in an external electric field. We will calculate the value of this torque by calculating the magnitude of force generated by the electric field and using the relation between the force, torque and the distance of separation.
Formula used:
Force applied by an electric field $\overrightarrow{E}$ on a charge $q$: $\overrightarrow{F}=q\overrightarrow{E}$
Torque on a dipole of charges $q$ and $-q$ separated by a distance $d$ in an electric field $\overrightarrow{E}$: $\tau =qEd$.
An electric dipole is a combination of two equal and opposite charges separated by a finite distance. Direction of dipole is from negative charge to positive charge.
We have an electric dipole, consisting of two opposite charges of magnitude $2\times {{10}^{-6}}~C$ each separated by a distance 3 cm,
Let, $q$ and $-q$ be the charges and $d$ be the distance between them.
So, $q=2\times {{10}^{-6}}~C$
And, $d=3\text{ }cm$
We will convert the distance in SI unit, i.e. meter
$d=3\text{ }cm=3\times {{10}^{-2}}m$
Now, the dipole is placed in an electric field magnitude $2\times {{10}^{5}}~N{{C}^{-1}}$along the direction making ${{90}^{o}}$ with the field.
$E=2\times {{10}^{5}}~N{{C}^{-1}}$
Force applied by an electric field $\overrightarrow{E}$ on a charge $q$ is:
$\overrightarrow{F}=q\overrightarrow{E}$
For a positive charge, direction of force will be the same as for an electric field, but for a negative charge, direction of force will be opposite as for an electric field.
Magnitude of this force will be:
$F=qE$
Where,
$E$ is the magnitude of an electric field.
As we can see the magnitude of both the charges is the same, so magnitude of the force applied on the two charges will be the same, but the direction of forces will be opposite as one charge is positive and other is negative.
So, the two forces on dipole will cancel out each other, but we know that,
Equal and opposite forces separated by a finite distance applied on a body provide torque to the body.
Now, magnitude of torque applied by two equal and opposite forces of magnitude $F$ separated by a distance $d$ is:
$\tau =Fd$
Substituting $F=qE$ ,
We have:
$\tau =qEd$
Now, putting values
$q=2\times {{10}^{-6}}~C$
$d=3\times {{10}^{-2}}m$
$E=2\times {{10}^{5}}~N{{C}^{-1}}$
We get,
\begin{align} & \tau =\left( 2\times {{10}^{-6}}~C \right)\times \left( 2\times {{10}^{5}}~N{{C}^{-1}} \right)\times \left( 3\times {{10}^{-2}}m \right) \\ & \tau =12\times {{10}^{-6+5-2}}Nm \\ & \tau =12\times {{10}^{-3}}Nm \\ \end{align}
The torque applied on the dipole by electric field is $12\times {{10}^{-3}}Nm$
So, the correct answer is “Option C”.
Note:
Students should note that the torque is generated only when the two forces acting on a body are equal in magnitude but opposite in the direction. Also, these forces must be separated by some distance. For a dipole, both the conditions are satisfied as the two charges are equal in magnitude, opposite in sign, and separated by some distance.
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.vedantu.com/question-answer/an-electric-dipole-consisting-of-two-opposite-class-12-physics-cbse-5f624100bc188a78430b9b51", "fetch_time": 1600601755000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-40/segments/1600400197946.27/warc/CC-MAIN-20200920094130-20200920124130-00782.warc.gz", "warc_record_offset": 1110006129, "warc_record_length": 80985, "token_count": 1087, "char_count": 3830, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2020-40", "snapshot_type": "latest", "language": "en", "language_score": 0.8338766694068909, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00062-of-00064.parquet"}
|
# Why does the regression intercept formula include the mean values of both variables?
Hopefully someone good with Quant can help here… For single variable linear regression, the formula for the intercept coefficient is: Intercept = mean Dependent value (y) - (slope*mean independent value(x)) Which is: _ _ Bo = Y - BiX My Question, is why does this formula insist on using the mean values of the dependent and independent variables? If a linear relationship exists with a constant slope, the value of the intercept calculated using the formula above should be the same no matter what point on the line the variable values are taken from. Using the (mean x, mean y) point demonstrates that the regression line passes through this point, but in this case does not seem to serve any other purpose in this formula? Any thoughts? Thanks in advance and keep on plugging!
The formula doesn’t “insist” on using the mean values. In the mathematical derivation of the two coefficients we end up with two equations in two unknowns (intercept and slope). The formula is simply an algebraic result when we solve for the intercept in terms of the slope.
Thanks for replying wyantjs: To clarify, the value of the intercept would come out the same using any point on the regression line correct? I know the formula displays the fact that the mean values are passed through, but I want to be sure going forward that I understand using the mean values is not necessary to arrive at the correct outcome? Thanks for any clarification again~
You are not understanding. Don’t think of it as we are “using” the mean values to derive this coefficient. It just so happens that the formula has the mean values in it as a result of some calculus and algebra. If you insist on thinking of it this way, the answer is that it is necessary to use them arrive at the correct outcome. You cannot just plug any two values for X and Y into this equation and get Bo (assuming you know B1). You will have an error. The point is that you could plug in any two values for X and Y that lie on the line, and not just the mean values, but you have to know the value of the intercept term to know if the two points do in fact lie on the line. Hence, just use the formula as it is.
wyantjs: You are correct, I am still not understanding. To my knowledge, if you have a linear regression line with a known slope, you can pick any point off of this line, plug in the numbers, and arrive at the same intercept value regardless of whether or not the point you chose consisted of the mean variable values or not. I have tested this concept several times and as long as the slope is the same, it works to arrive at the same intercept value. I guess I am still confused as to why they even include the mean values in the formula and why not just say any (X,Y) data point on the regression line? I appreciate the insight and please don’t feel obligated to post again if this seems like an elementry question~ Thanks!!
You cannot know if a point (X,Y) is on the regression line if you do not already know the equation of the line, and therefore already know the value of the intercept. You are using circular logic here. Of course any value on the line is going to give you the same intercept…that’s inbedded in the definition of a line. Try this…take 20 values for X and Y. Do any of them lie on the line? Who the hell knows until you calculate the intercept and slope? You have to find the intercept first. Get it?
wyantjs: I got it now - not sure why that wasn’t sinking in, but it makes sense now! Basically what I need to remember is to use the mean variable values if asked to calc the intercept value on the exam, but I do appreciate the explanations! Have a good week-
Basically, the regression line passes through (X bar, Y bar)
See, the thing is, I’m supposed to be the stats guru here but wyantjs and Hydrogen Rainbow are making me feel marginalized.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.analystforum.com/t/why-does-the-regression-intercept-formula-include-the-mean-values-of-both-variables/29391", "fetch_time": 1716951934000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-22/segments/1715971059167.30/warc/CC-MAIN-20240529010927-20240529040927-00422.warc.gz", "warc_record_offset": 551353007, "warc_record_length": 6038, "token_count": 849, "char_count": 3919, "score": 3.828125, "int_score": 4, "crawl": "CC-MAIN-2024-22", "snapshot_type": "latest", "language": "en", "language_score": 0.9198006391525269, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00032-of-00064.parquet"}
|
# Thread: Combining Algebra Fractions Frustration...
1. ## Combining Algebra Fractions Frustration...
I've been working through an Algebra book trying to re-teach myself...not doing a half-bad job. I've come to a problem I'm not sure I understand. I reach the right answer...but yet it's wrong. I've attached the problem, along with the way I've been working it into a MS Word doc file. Any insight as to where I'm going wrong, ruels I'm forgetting or clarification on how the book comes to the answer it does would be greatly appreciated!
Ev
2. You forgot one rule, not of mathematics but of internet security: Do not download files from an unknown source.
You can use LaTeX to represent equations: http://www.mathhelpforum.com/math-he...-tutorial.html
3. Oooops....
I'll read through how to use the LaTeX tutorial and re-submit.
Sorry and thanks!
Ev
4. My friend, you got the solution correct. The book is simply writing it in another way.
Their answer has the negative sign in front of the entire fraction. To distribute it, you make every TERM in the numerator the opposite of what it is. In this case
-(2x + 11) = -2x - 11
2x + 11 is the same as +2x + 11, right?
The negative of that -2x - 11.
(5 - 3) = 2
So the negative of that should equal negative 2, correct? Let's write:
-(5 - 3) = -5 + 3 = -2
Let me know if you have more questions. Sorry I typed this up very fast.
5. Thank you so much for the reply.
My only question is realy a clarification point...
If I have a -x-23
------
y-z
(just for argument sake)
I can make both terms in the numerator positive by making the entire fraction negative?:
-x-23 = x+23
------- - -------
y - z y - z
If this is correct, I have a new confusion. If I remember right, you can change two signs of the fraction and still keep it equivalent to the original fraction. If I change the two signs in the numerator to positive and then change the sign of the fraction itself...then thats three signs changed, not two. Is it still equivalent?
6. Originally Posted by ejanderson
Thank you so much for the reply.
My only question is realy a clarification point...
If I have a -x-23
------
y-z
(just for argument sake)
I can make both terms in the numerator positive by making the entire fraction negative?:
-x-23 = x+23
------- - -------
y - z y - z
If this is correct, I have a new confusion. If I remember right, you can change two signs of the fraction and still keep it equivalent to the original fraction. If I change the two signs in the numerator to positive and then change the sign of the fraction itself...then thats three signs changed, not two. Is it still equivalent?
That is fine. The original equation is
$\displaystyle \frac{-x-23}{y-z}$
given that any fraction with equal numerator and denominator is 1 then we can multiply by $\displaystyle \frac{-1}{-1}$
$\displaystyle \frac{-x-23}{y-z} \times \frac{-1}{-1} = \frac{-1(-x-23)}{-1(y-z)} = \frac{x+23}{-(y-z)} = -\frac{x+23}{y-z}$
Which is the answer you got.
We can also distribute that minus sign on the denominator instead of leaving it (in the penultimate step):
$\displaystyle \frac{x+23}{z-y}$
You mention changing 3 signs but you have to in order to ensure the equation is still the same. When you're changing the sign you're actually distributing -1 across the term inside the brackets
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/algebra/132472-combining-algebra-fractions-frustration.html", "fetch_time": 1526852861000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-22/segments/1526794863689.50/warc/CC-MAIN-20180520205455-20180520225455-00272.warc.gz", "warc_record_offset": 195798282, "warc_record_length": 11239, "token_count": 847, "char_count": 3331, "score": 4.09375, "int_score": 4, "crawl": "CC-MAIN-2018-22", "snapshot_type": "latest", "language": "en", "language_score": 0.9283151626586914, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00020-of-00064.parquet"}
|
# Negative Correlation Definition
## What Is Negative Correlation?
Negative correlation is a relationship between two variables in which one variable increases as the other decreases, and vice versa.
In statistics, a perfect negative correlation is represented by the value -1.0, while a 0 indicates no correlation, and +1.0 indicates a perfect positive correlation. A perfect negative correlation means the relationship that exists between two variables is exactly opposite all of the time.
### Key Takeaways
• Negative or inverse correlation describes when two variables tend to move in opposite size and direction from one another, such that when one increases the other variable decreases, and vice-versa.
• Negative correlation is put to use when constructing diversified portfolios, so that investors can benefit from price increases in certain assets when others fall.
• Correlation between two variables can vary widely over time. Stocks and bonds generally have a negative correlation, but in the 10 years to 2018, their measured correlation has ranged from -0.8 to +0.2.
## Understanding Negative Correlation
Negative correlation or inverse correlation indicates that two individual variables have a statistical relationship such that their prices generally move in opposite directions from one another. If, for instance, variables X and Y have a negative correlation (or are negatively correlated), as X increases in value, Y will decrease; similarly, if X decreases in value, Y will increase.
The degree to which one variable moves in relation to the other is measured by the correlation coefficient, which quantifies the strength of the correlation between two variables. For example, if variables X and Y have a correlation coefficient of -0.1, they have a weak negative correlation, but if they have a correlation coefficient of -0.9, they would be regarded as having a strong negative correlation.
The higher the negative correlation between two variables, the closer the correlation coefficient will be to the value -1. By the same token, two variables with a perfect positive correlation would have a correlation coefficient of +1, while a correlation coefficient of zero implies that the two variables are uncorrelated and move independently of each other.
The correlation coefficient, usually denoted by “r” or “R”, can be determined by regression analysis. The square of the correlation coefficient (generally denoted by “R2“, or R-squared) represents the degree or extent to which the variance of one variable is related to the variance of the second variable, and is typically expressed in percentage terms.
READ: Inflation & Interest Rates Relationship Explained
For example, if a portfolio and its benchmark have a correlation of 0.9, the R-squared value would be 0.81. The interpretation of this figure is that 81% of the variation in the portfolio (the dependent variable in this case) is related to—or can be explained by—the variation of the benchmark (the independent variable).
## The Importance of Negative Correlation
The concept of negative correlation is a key one in portfolio construction. Negative correlation between sectors or geographies enables the creation of diversified portfolios that can better withstand market volatility and smooth out portfolio returns over the long term.
The building of large and complex portfolios where the correlations are carefully balanced to provide more predictable volatility is generally referred to as the discipline of strategic asset allocation.
Consider the long-term negative correlation between stocks and bonds. Stocks generally outperform bonds during periods of strong economic performance, but as the economy slows down and the central bank reduces interest rates to stimulate the economy, bonds may outperform stocks.
As an example, assume you have a \$100,000 balanced portfolio that is invested 60% in stocks and 40% in bonds. In a year of strong economic performance, the stock component of your portfolio might generate a return of 12%, while the bond component may return -2% because interest rates are on a rising trend. Thus, the overall return on your portfolio would be 6.4% ((12% x 0.6) + (-2% x 0.4).
The following year, as the economy slows markedly and interest rates are lowered, your stock portfolio might generate -5% while your bond portfolio may return 8%, giving you an overall portfolio return of 0.2%.
What if, instead of a balanced portfolio, your portfolio was 100% equities? Using the same return assumptions, your all-equity portfolio would have a return of 12% in the first year and -5% in the second year, which are more volatile than the balanced portfolio’s returns of 6.4% and 0.2%.
## Examples of Negative Correlation
Examples of negative correlation are common in the investment world. A well-known example is the negative correlation between crude oil prices and airline stock prices. Jet fuel, which is derived from crude oil, is a large cost input for airlines and has a significant impact on their profitability and earnings.
If the price of crude oil spikes up, it could have a negative impact on airlines’ earnings and hence on the price of their stocks. But if the price of crude oil trends lower, this should boost airline profits and therefore their stock prices.
Here’s how the existence of this phenomenon can help in the construction of a diversified portfolio. As the energy sector has a substantial weight in most equity indices, many investors have significant exposure to crude oil prices, which are typically quite volatile. As the energy sector, for obvious reasons, has a positive correlation with crude oil prices, investing part of one’s portfolio in airline stocks would provide a hedge against a decline in oil prices.
## Special Considerations
It should be noted that this investment thesis may not work all of the time, as the typical negative correlation between oil prices and airline stocks might occasionally turn positive. For example, during an economic boom, oil prices and airline stocks may both rise; conversely, during a recession, oil prices and airline stocks could slide in tandem.
When negative correlation between two variables breaks down, it can play havoc with investment portfolios. For example, US equity markets experienced their worst performance in a decade in the fourth quarter of 2018, partly fueled by concerns that the Federal Reserve (Fed) would continue to raise interest rates.
Fears of rising rates also took their toll on bonds, leading their normally negative correlation with stocks to fall to the weakest levels in decades. At such times, investors often discover to their chagrin that there is no place to hide.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://seattlecommunitymedia.org/negative-correlation-definition/", "fetch_time": 1638474274000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00159.warc.gz", "warc_record_offset": 537696891, "warc_record_length": 34100, "token_count": 1303, "char_count": 6723, "score": 3.859375, "int_score": 4, "crawl": "CC-MAIN-2021-49", "snapshot_type": "latest", "language": "en", "language_score": 0.9333395957946777, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00062-of-00064.parquet"}
|
Sunday, February 19, 2012
A Bayesian Approach To Kalman Filtering
INTRODUCTION
Ripped straight from Wikipedia...
The Kalman filter is an algorithm, commonly used since the 1960s for improving vehicle navigation (among other applications, although aerospace is typical), that yields an optimized estimate of the system's state (e.g. position and velocity). The algorithm works recursively in real time on streams of noisy input observation data (typically, sensor measurements) and filters out errors using a least-squares curve-fit optimized with a mathematical prediction of the future state generated through a modeling of the system's physical characteristics.
TL;DR
Rockets
There are both linear and non-linear forms of the Kalman filter, with the non-linear forms being the Extended Kalman Filter (EKF), the invariant Extended Kalman Filter, and the Unscented Kalman Filter (UKF). The non-linear versions of this algorithm represent the current "gold standard" in many application areas, including GPS and navigation.
Before jumping in the deep end of the pool, I decided to implement a simple example that shows the ideas and implementation of Kalman filtering, using a recursive Bayesian approach.
TL;DR
Homework
WEIGHTING FUNCTION FOR KALMAN UPDATING
The Kalman filter is often derived from a matrix equation standpoint. The math, at least to me, is long, involved, and fairly nasty to solve without the help of some MATLAB matrix witchery. However, there is also a second, more "gut level" way to approach the Kalman filter - by approaching it as a case of recursive Bayesian filtering.
The derivation for the following equations can be found on pg. 47 of Bayesian Data Analysis, Second Edition, by Gelman, Carlin, Stern, and Rubin.
Basically, it is possible estimate the mean of a normal distribution by following a weighting equation for mean and variance, which can be represented as follows.
$\Large{\mu_1=\frac{\frac{1}{\tau_0^2}\mu_0+\frac{1}{\sigma^2}y}{\frac{1}{\tau_0^2}+\frac{1}{\sigma^2}}}$
$\Large{\frac{1}{\tau_1^2}=\frac{1}{\tau_0^2}+\frac{1}{\sigma^2}}$
Here, $\mu_1$ represents the new estimate for the mean and $\tau_1^2$ is the estimate for the new variance. $\mu_0$ represents the old mean, $\tau_0^2$ represents the old variance, $y$ is the new measured value, and $\sigma^2$ is the measurement variance. I will do an example later on to demonstrate this, but first, let's try to decide what this equation is really saying.
If the old value for variance, $\tau_0^2$, is very large, then $u_0$ is multiplied by a very small number, effectively making the equation...
$\Large{\mu_1 = \frac{\frac{1}{\sigma^2}y}{\frac{1}{\sigma^2}}}$
which can be further reduced to simply...
$\Large{\mu_1 = y}$
What this means, is that if we are not very confident ($\tau_0^2$ is large) of our old mean, $\mu_0$, our best guess for $\mu_1$ is simply the value we measured, $y$. Our new variance also reduces to
$\Large{\frac{1}{\tau_1^2}=\frac{1}{\sigma^2}}$
Conversely, if we are not very confident in the new result $y$, our new mean $\mu_1$ is basically the same as $\mu_0$. The new mean is based largely on the weighting between the new variance and the old - so if we are equally confident in both the previous mean and the new measurement, our best estimate for $\mu_1$ will split the difference.
Using this update function, it is easy to wrap a new estimate back into the prior distribution, which is the core idea behind Bayesian analysis. But how does it come into play with Kalman filtering?
APPLICATION
The Kalman filter can be boiled down to a few basic steps.
1. Predict a future value, based on previous information. (Predict)
2. Compare prediction and measured value, using results as the new previous information. (Update)
3. If you run out of data, or are satisfied with the estimate, exit. Otherwise, GOTO 1.
Pretty simple, right? This method can yield some very interesting results, cutting through measurement error and giving close approximation of the root signal.
From a high level, Bayesian derivation is
$P(X_i|Y_{1:i}) \alpha P(Y_i|X_i,Y_{1:i-1}) P(X_i|Y_{1:i-1})$
Because Bayesian analysis usually wraps it's results back into the prior model,
this reduces to
$P(X_i|Y_{1:i}) \alpha P(Y_i|X_i) P(X_i|Y_{1:i-1})$
What this means is we have a predictive function, $P(Y_i|X_i)$, which will be described a bit later, used in conjunction with a likelihood function, $P(X_i|Y_{1:i-1})$, which was described above. Together these two functions form the core of the Kalman filtering operation: Predict, Measure, Update, Repeat.
SETUP
The key things we need to know are:
The standard deviation of our source. I chose .6
The scale factor of the source. This can be calculated from
$1 = (scale factor)^2+(source std dev)^2$
The standard deviation of the measurements. I chose .5, but this is a parameter you can play around with. The higher the measurement sigma gets, the worse our filter performs, due to signal to noise ratio effects. An intial deviation of .5 is fairly high, so dropping this number to something like .25 will give "prettier" results.
In a real system, you could reasonably expect to have tolerances for both your source equipment and your measurement equipment. Maybe you could make measurements of this, or it could be found in the manufacturer's datasheet.
I will be showing this method through an example, using R. We will create a test case by following a few generic steps:
1. Create a source signal with some noise variance $X$.
2. Create a measurement signal, which will introduce measurement variance $Y$.
3. Kalman filter the measurement signal $k = filt(Y)$.
4. Compare the filtered data $k$ with the original transmission $X$.
So let's begin.
Step 1.
Create a source signal. I chose to generate a random walk.
randomwalk <- function(num, source_scale, source_sigma) {
randy <- c(0,1:num)
for(i in 2:length(randy)) {
randy[i] <- source_scale*randy[i-1]+rnorm(1, sd=source_sigma)
}
return(randy[2:length(randy)])
}
Step 2.
Now, we need a function to create a measurement signal, which will further distort our input.
measured <- sent+rnorm(length(sent), sd=meas_sigma)
return(measured)
}
Step 3.
The signal is pretty badly distorted now... time to write a Kalman filter function.
pred_mean <- function(source_scale, prev_mean) {
return(source_scale*prev_mean)
}
pred_sigma <- function(source_scale, prev_sigma, source_sigma) {
return(sqrt((source_scale**2)*(prev_sigma**2)+source_sigma**2))
}
update_mean <- function(pred_mean, pred_sigma, meas_val, meas_sigma) {
numerator <- (pred_mean/(pred_sigma**2))+(meas_val/(meas_sigma**2))
denominator <- (1/(pred_sigma**2))+(1/(meas_sigma**2))
return(numerator/denominator)
}
update_sigma <- function(pred_sigma, meas_sigma) {
r =(1/(pred_sigma**2))+(1/(meas_sigma**2))
return(1/sqrt(r))
}
filt <- function(y, source_scale, source_sigma, meas_sigma) {
last_mean <- 0
last_sigma <- source_sigma
k <- 1:length(y)
for(i in 1:length(y)) {
est_mean <- pred_mean(source_scale, last_mean)
est_sigma <- pred_sigma(source_scale, last_sigma, source_sigma)
k[i] <- est_mean+rnorm(1, sd=est_sigma)
last_mean <- update_mean(est_mean, est_sigma, y[i], meas_sigma)
last_sigma <- update_sigma(est_sigma, meas_sigma)
}
return(k)
}
A few quick derivations are required for this code to make sense. The update_mean and the update_variance functions were described at the start of the blog, but where on earth did the pred_mean and pred_sigma functions come from?
According to notation of the Kalman filter, we currently have
$X_i = \alpha_s X_{prev}+\omega_s$, where $\alpha_s$ is our source scale, and $\omega_s$ is the source standard deviation. In Kalman filter terms, this equation is the state model, and we got this from the knowledge we have about how the randomwalk values were generated.
The second equation, our space model, is $Y_i = X_i + \omega_m$, where $\omega_m$ is the measurement standard deviation.
In order to get the equations for our pred_mean and pred_sigma we want to find the expected value and variance of the state model, which looks like this.
$E[X_i] = E[\alpha_s X_{prev}+\omega_s]$
using the rules of associativity for expected value...
$E[X_i] = \alpha_s E[X_{prev}]+E[\omega_s]$
The expected value of zero-mean noise is 0, so now we find
$E[X_i] = \alpha_s E[X_{prev}]$
This matches exactly with the code for pred_mean - our new guess is the old mean multiplied by the scale factor.
The pred_sigma function is a little trickier.
$Var(X_i) = Var(\alpha_s X_{prev}+\omega_s)$
Since $Var(Q)$ is the same as $E[(Q-E[Q])^2]$, we now see
$Var(X_i) =$
$E[(\alpha_s X_{prev}+\omega_s)^2]-E[\alpha_s X_{prev}+\omega_s]^2$
Expanding, this now becomes
$Var(X_i) =$
$\alpha_s^2 E[X_{prev}^2]+2\alpha_s E[X_{prev}] E[\omega_s]+$
$E[\omega_s^2] - E[\alpha_s X{prev} + \omega_s]^2$
Continuing, we see that
$- E[\alpha_s X{prev} + \omega_s]^2$ can be expanded to
$- E[\alpha_s X_{prev}+\omega_s]E[\alpha X_{prev}+\omega_s]$
which becomes
$-(E[\alpha_s X_{prev}]+E[\omega_s])(E[\alpha X_{prev}]+E[\omega_s])$
FOIL, and we now see
$-\alpha_s^2E[X_{prev}]^2-2\alpha_s E[\omega_s]E[X_{prev}]-E[\omega_s]^2$
One view of this expression is that $\omega_s$ is a normally distributed random variable with a mean of 0. $E[\omega_s]$ is 0, and squaring that is still 0, so both the $E[\omega_s]$ and $E[\omega_s]^2$ end up going to 0. It is important to note that $E[\omega_s^2]$ will NOT go to zero, as the mean of the squared distribution is no longer centered about 0, and instead becomes $\omega_s^2$.
However, another, more natural view is to note that $E[\omega_s^2]-E[\omega_s]^2$ is identical to $Var(\omega_s)$ which is actually $\omega_s^2$. The $2\alpha_s$ terms cancel.
Either way, the final result is the same.
$Var(X_i) = \alpha_s^2 Var(X_{prev}) + \omega_s^2$
This matches the code for pred_sigma - square root of scale factor squared times the previous sigma + the measurement sigma is our best guess for the new sigma.
See http://mathworld.wolfram.com/Variance.html for an alternate derivation that leads to the same result.
Step 4.
Now we can write a function to tie it all together.
runit <- function() {
source_sigma <- .01
source_scale <- sqrt(1-source_sigma**2)
meas_sigma <- .4
x <- randomwalk(1000, source_scale, source_sigma)
k <- filt(y, source_scale, source_sigma, meas_sigma)
plot(y, type="l", col="red")
lines(k, col="green")
lines(x, col="blue")
}
The blue is the source signal, the red is the signal at measurement, and green is the recovered signal. Even when measurement noise washes out the root signal, we can recover the original fairly well.
Try tuning each of the sigmas, and see how the results change - it's pretty interesting.
There you have it! A simple, logical derivation of the Kalman filter as a recursive Bayesian filter. In the future I plan to write about more complex statistical processing methods as I learn them, such as how to run this simulation with 0 known parameters, or implementation of one of the non-linear Kalman filter algorithms.
As always, critique is both welcome and appreciated.
CODE
To run this code, simply copy and paste into a source file. Then open an R interpreter, do source("path/to/source/file"), then do runit().
randomwalk <- function(num, source_scale, source_sigma) {
randy <- c(0,1:num)
for(i in 2:length(randy)) {
randy[i] <- source_scale*randy[i-1]+rnorm(1, sd=source_sigma)
}
return(randy[2:length(randy)])
}
measured <- sent+rnorm(length(sent), sd=meas_sigma)
return(measured)
}
pred_mean <- function(source_scale, prev_mean) {
return(source_scale*prev_mean)
}
pred_sigma <- function(source_scale, prev_sigma, source_sigma) {
return(sqrt((source_scale**2)*(prev_sigma**2)+source_sigma**2))
}
update_mean <- function(pred_mean, pred_sigma, meas_val, meas_sigma) {
numerator <- (pred_mean/(pred_sigma**2))+(meas_val/(meas_sigma**2))
denominator <- (1/(pred_sigma**2))+(1/(meas_sigma**2))
return(numerator/denominator)
}
update_sigma <- function(pred_sigma, meas_sigma) {
r =(1/(pred_sigma**2))+(1/(meas_sigma**2))
return(1/sqrt(r))
}
filt <- function(y, source_scale, source_sigma, meas_sigma) {
last_mean <- 0
last_sigma <- source_sigma
k <- 1:length(y)
for(i in 1:length(y)) {
est_mean <- pred_mean(source_scale, last_mean)
est_sigma <- pred_sigma(source_scale, last_sigma, source_sigma)
k[i] <- est_mean+rnorm(1, sd=est_sigma)
last_mean <- update_mean(est_mean, est_sigma, y[i], meas_sigma)
last_sigma <- update_sigma(est_sigma, meas_sigma)
}
return(k)
}
runit <- function() {
source_sigma <- .01
source_scale <- sqrt(1-source_sigma**2)
meas_sigma <- .4
x <- randomwalk(1000, source_scale, source_sigma)
k <- filt(y, source_scale, source_sigma, meas_sigma)
plot(y, type="l", col="red")
lines(k, col="green")
lines(x, col="blue")
}
1. Is that pseudocode? If not, what language is it?
1. I guess it is R http://www.r-project.org/
2. It is R - I am trying to cross the gap from MATLAB so I'll probably be doing several more posts with R code.
Though I still L O V E python.
2. Discussion fro reddit which prompted an update in my derivation...
user Kometes_:
I understood everything except the pred_sigma part. To simplify properly shouldn't the last line of the expansion be:
Var(X) = ... E[ω2 ] − α2 E[Xprev]2 - 2αE[Xprev]E[ωs] - E[ω]2
Also this gives the variance of X, so shouldn't pred_sigma be the square root of Var(X)?
pred_sigma <- function...
return sqrt(source_scale^2*prev_sigma^2 + source_sigma^2)
My response:
You are right on the money. I messed up pretty good - time to go correct myself.
Because we are expanding −E[αXprev+ωs]2 , I should have expanded to
−E[Xprev+ωs]E[αXprev+ωs]
which then becomes
−(E[αXprev]+E[ωs])(E[αXprev]+E[ωs])
FOIL, and we see
−α2E[Xprev]2−2αE[ωs]E[Xprev]−E[ωs]2
The reason it works either way is the ωs is a normally distributed random variable with a mean of 0. So E[ωs] is 0, and squaring that is also 0, so both of the bolded terms end up going to 0... I think.
It is important to note that E[ω2s] will NOT go to zero, as the mean of the squared distribution is no longer centered about 0, and instead becomes ω2s.
It was a mistake on my notes that worked out through luck - your expansions are the logical (and correct!) next step. I will correct the blog.
On the pred_sigma part - you are most definitely correct! I have updated my code and things look much better now.
Thanks for helping me get this right.
3. can you explain why in the filt function, k value has a ranif(1) in it? I asked this because i thought noise from the input are from the randomwalk/signal functions.
1. All of my noise functions are a little bit off - I mistranslated form my MATLAB code. I should have been using rnorm, not runif - I will update accordingly. And really, multiplying by the sigma isn't as clean as rnorm(1, sd=sigma) so I will do that as well.
4. For another simple introduction to the Kalman filter, in the setting of learning theory, see the third page of this article:
http://www.indiana.edu/~kruschke/articles/Kruschke2008.pdf
1. Thanks John - Good link! Your book and articles have helped me quite a bit the last few weeks.
5. Thanks a lot for this article. I ported your R code to C# to help me understand how it works:
https://github.com/TaylorClark/KalmanCS
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://kkjkok.blogspot.com/2012/02/bayesian-approach-to-kalman-filtering.html", "fetch_time": 1513246676000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-51/segments/1512948543611.44/warc/CC-MAIN-20171214093947-20171214113947-00481.warc.gz", "warc_record_offset": 143927581, "warc_record_length": 21984, "token_count": 4093, "char_count": 15284, "score": 3.5625, "int_score": 4, "crawl": "CC-MAIN-2017-51", "snapshot_type": "latest", "language": "en", "language_score": 0.9011369943618774, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00013-of-00064.parquet"}
|
# Infinity axiom
The axiom of infinity is an axiom of set theory that postulates the existence of an inductive set . It is called the axiom of infinity, since inductive sets are also infinite sets at the same time . Ernst Zermelo published the first axiom of infinity in 1908 in the Zermelo set theory . It influenced all later set theories, especially the Zermelo-Fraenkel set theory (ZF), the most widespread set theory, which Zermelo's axiom of infinity adopted in a slightly modified form.
## formulation
There is a set that contains the empty set and with each element also the set . ${\ displaystyle A}$${\ displaystyle \ emptyset}$${\ displaystyle x \ in A}$${\ displaystyle x \ cup \ {x \}}$
${\ displaystyle \ exists A \ colon (\ emptyset \ in A \ land \ forall x \ colon (x \ in A \ Rightarrow x \ cup \ {x \} \ in A))}$
The axiom of infinity not only postulates the existence of an infinite set, but also specifies the structure of this infinite set.
## Meaning for math
### Natural numbers
The existence of at least one inductive set , together with the axiom of exclusion, also ensures the existence of natural numbers as a set: ${\ displaystyle I}$
${\ displaystyle \ mathbb {N}: = \ {x \ in I \ mid \ forall z (z \, \, {\ text {inductive}} \ implies x \ in z) \}}$
The natural numbers are defined as the intersection of all inductive sets, as the smallest inductive set.
### Infinite quantities
Without the axiom of infinity, ZF would only ensure that finite sets exist. No statements could be made about the existence of infinite sets. The infinity axiom, together with the power set axiom, ensures that uncountable sets such as B. gives the real numbers.
## Individual evidence
1. Zermelo: Investigations on the fundamentals of set theory , 1907, in: Mathematische Annalen 65 (1908), 261–281; Axiom of the Infinite p. 266f.
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://de.zxc.wiki/wiki/Unendlichkeitsaxiom", "fetch_time": 1696314467000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233511055.59/warc/CC-MAIN-20231003060619-20231003090619-00552.warc.gz", "warc_record_offset": 230108560, "warc_record_length": 7052, "token_count": 455, "char_count": 1854, "score": 3.75, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.843415379524231, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00039-of-00064.parquet"}
|
30 C
Mumbai
Saturday, April 17, 2021
# IPL Team Net Run rate calculation
Let’s check IPL team net run rate calculation formula and method. How they calculate team net run rate? On performance. and this net RR helps teams to stay on top of the point table.
We have a lot of questions about net calculation of cricket matches in because we don’t know the calculation formula of tournament organizers and BCCI. So today I am going to solve all of your questions about net run rate calculation in this article.
## Mathematical formula of net run rate calculation:
Let’s know about each factor in details.
Total score of team: That means total runs made by a one team in a tournament.
Played over: Total over played by team in tournament.
Total score of opposite team: that means total score of opposite teams which is made by an all competitor teams In front of one team while playing against them in tournament.
### For Example:
This is the latest point table of IPL2020 tournament at the date of 28 September 2020. you can see all the team net run rate.
See in above picture where DC has net RR +1.100. lets calculate DC run rate by the above formula.
That means DC played total two match in this two matches DC played total 40 overs and scored 332 runs. That means 332/40 = 8.3
On the other hand, that two team played total 40 overs and scored 288 runs. That means 288/40 = 7.2.
So, the overall net run rate of DC is 8.3 – 7.2 = 1.100. you can see it in above point table.
I hope now you understand how IPL managements calculates the team net run rate. If you got your answer clearly then please share it with others on social media and WhatsApp and comment below.
Follow us for more interesting articles visit our website daily.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "www.ttlyblogs.com", "fetch_time": 1618678697000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618038461619.53/warc/CC-MAIN-20210417162353-20210417192353-00312.warc.gz", "warc_record_offset": 1149433656, "warc_record_length": 78192, "token_count": 407, "char_count": 1752, "score": 3.671875, "int_score": 4, "crawl": "CC-MAIN-2021-17", "snapshot_type": "longest", "language": "en", "language_score": 0.9457815885543823, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00054-of-00064.parquet"}
|
# Thread: Prove the sum of all positive integers k...
1. ## Prove the sum of all positive integers k...
Prove that the sum of all positive integers k where 1 <= k <= n and gcd(k,n) = 1 is (1/2)n*phi(n).
I don't even know where to begin with this. We've been going over primitive roots in class lately so it probably has something to do with them, but I have no idea where they come into the problem.
2. For any n, $\zeta=e^{2\pi i/n}$ is a primitive nth root of unity, and the other primitive roots are exactly the numbers $\zeta^k$, where gcd(k,n)=1. If you picture these on the unit circle and realize that the conjugate of a primitive root is also a primitive root, you can prove it that way.
3. Originally Posted by LoblawsLawBlog
For any n, $\zeta=e^{2\pi i/n}$ is a primitive nth root of unity, and the other primitive roots are exactly the numbers $\zeta^k$, where gcd(k,n)=1. If you picture these on the unit circle and realize that the conjugate of a primitive root is also a primitive root, you can prove it that way.
Uh, we didn't learn anything like that in class.
4. Ok, I tried to explain it using primitive roots because you said you're learning about them. The same proof will carry over to the group $\mathbb{Z}/n\mathbb{Z}$, since this group is isomorphic to the nth roots of unity. But you might not be doing much with groups if this is a number theory course, even though there's a lot of overlap.
I guess the key facts are that $\varphi(n)$ is the number of positive integers less than n that are relatively prime to n and also that gcd(n,k)=gcd(n,n-k). Does that help?
You could probably also prove this using some facts about $\varphi$ listed on its wikipedia page.
edit: Ah, that explains it- there are two uses of the term primitive root. I thought you meant primitive roots of unity and I had never heard of the other meaning. Still, my second paragraph could still be used.
5. Originally Posted by uberbandgeek6
Prove that the sum of all positive integers k where 1 <= k <= n and gcd(k,n) = 1 is (1/2)n*phi(n).
I don't even know where to begin with this. We've been going over primitive roots in class lately so it probably has something to do with them, but I have no idea where they come into the problem.
In any case is $\displaystyle \sum_{k=1}^{n-1} k= \frac{n\ (n-1)}{2}$. If n is prime then $\varphi(n)= n-1$ so that $\displaystyle \sum_{k=1}^{n-1} k= \frac{n}{2}\ \varphi(n)$...
Kind regards
$\chi$ $\sigma$
6. @chisigma: I understand what you mean, but I'm not given that n is prime, so I don't see how that would work.
@Loblawslawblog: I don't understand where you are going with the gcd(n, n-k). I get that it is true, but I don't see how that helps.
7. Originally Posted by uberbandgeek6
@chisigma: I understand what you mean, but I'm not given that n is prime, so I don't see how that would work.
One of Your hypothesis is that $\forall k$ for which $0\le k\le n$ is $\text {gcd} (k,n)=1$ and that means that n is prime...
Kind regards
$\chi$ $\sigma$
8. Originally Posted by chisigma
One of Your hypothesis is that $\forall k$ for which $0\le k\le n$ is $\text {gcd} (k,n)=1$ and that means that n is prime...
Kind regards
$\chi$ $\sigma$
Maybe I'm interpreting it wrong, but I think the problem means that it wants the sum of all k's that are relatively prime with n. So if n was not prime (say n = 10), it would be 1+3+7+9 = 20. Then (1/2)(10)phi(10) = 20 as well.
9. Originally Posted by uberbandgeek6
@Loblawslawblog: I don't understand where you are going with the gcd(n, n-k). I get that it is true, but I don't see how that helps.
Look at your example with n=10. 1 is relatively prime to 10, and so is 10-1=9. Likewise, 3 and 10-3=7 are both relatively prime to 10. Then the sum we're looking for is (1+9)+(3+7)=(4/2)(10).
10. Originally Posted by LoblawsLawBlog
Look at your example with n=10. 1 is relatively prime to 10, and so is 10-1=9. Likewise, 3 and 10-3=7 are both relatively prime to 10. Then the sum we're looking for is (1+9)+(3+7)=(4/2)(10).
The hint, or info, given by LLLB is critical: let
$\Phi(n,k):=\{k\in\mathbb{N}\;;\;1\leq k\leq n\,,\,\,gcd(n,k)=1\}$
1) Prove that $k\in\Phi(n,k)\Longleftrightarrow n-k\in\Phi(n,k)$
2) Thus, we can pair up all the numbers in $\Phi(n,k)$ in pairs $(k,n-k)$ , with
$k\mbox{ and also }n-k\mbox{ in } \Phi(n,k)$
3) Since $|\Phi(n,k)|=\phi(n)$ , there are $\frac{1}{2}\phi(n)$ pairs as above, and
since the sum of each such pair is $n$ we're then done...
Tonio
11. Originally Posted by LoblawsLawBlog
For any n, $\zeta=e^{2\pi i/n}$ is a primitive nth root of unity, and the other primitive roots are exactly the numbers $\zeta^k$, where gcd(k,n)=1. If you picture these on the unit circle and realize that the conjugate of a primitive root is also a primitive root, you can prove it that way.
How can you prove it that way?
12. It's the same proof. A primitive root z^k would correspond to k in Z_n and the pairing (k,n-k) is just the pair z^k and its conjugate. Then the factor of 1/2 comes in because you only need to consider the top half of the unit circle. Add the exponents of the primitive roots and you're done.
So basically there's no reason to go to complex numbers, but I tried to awkwardly jam them in because I thought the OP was learning something similar.
13. That's what I thought. There's no reason to talk about primitive roots here.
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 32, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/number-theory/172931-prove-sum-all-positive-integers-k.html", "fetch_time": 1512993992000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-51/segments/1512948513478.11/warc/CC-MAIN-20171211105804-20171211125804-00344.warc.gz", "warc_record_offset": 177272892, "warc_record_length": 13548, "token_count": 1558, "char_count": 5385, "score": 3.84375, "int_score": 4, "crawl": "CC-MAIN-2017-51", "snapshot_type": "longest", "language": "en", "language_score": 0.9529061317443848, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00007-of-00064.parquet"}
|
## Solve the question: (24/4 ×2/5) -(1/5×-10/3)
Question
Solve the question:
(24/4 ×2/5) -(1/5×-10/3)
in progress 0
11 months 2021-08-11T13:22:03+00:00 1 Answer 0 views 0
46/15
Step-by-step explanation:
(6×2/5)-(1/5×-10/3)
12/5-(-10/15)
12/5+10/15
36+10/15
46/15(ans)
MARK ME.AS BRAINLEIST
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://schoollearningcommons.info/question/solve-the-question-24-4-2-5-1-5-10-3-24984299-15/", "fetch_time": 1656341300000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-27/segments/1656103334753.21/warc/CC-MAIN-20220627134424-20220627164424-00487.warc.gz", "warc_record_offset": 562574751, "warc_record_length": 14155, "token_count": 145, "char_count": 300, "score": 3.5625, "int_score": 4, "crawl": "CC-MAIN-2022-27", "snapshot_type": "latest", "language": "en", "language_score": 0.6308518052101135, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00008-of-00064.parquet"}
|
Search 75,740 tutors
WyzAnt Tutoring Tutor Tutor United States
# Chris S.
## Ivy League Physical Science and Math Tutor
### New York, NY (10026)
Travel radius 3 miles
Hourly fee \$100.00
You don't pay for tutoring unless you find a good fit! You don't pay for tutoringunless you find a good fit!
## Chris's Resources
Ask questions and get free answers from expert tutors
# can you solve this problem?(3/4)(8x+4x)+5y+4=(2/3)(12x+9) (answer)
Starting with the original expression (3/4)(8x+4x)+5y+4=(2/3)(12x+9) combine x's in first part (3/4)(12x)+5y+4=(2/3)(12x+9) multiply factors in first and last terms 9x+5y+4=8x+6 put all y's on left, x's and numbers on right 5y=-1x+2 divide...
# factor the expression (answer)
3d2+23d+14 We can start with one of the factors to have a coefficient of 3 on the variable d and the other coefficient to be one because of the coefficient on d2. (3d+?)(d+?) We know the ?'s have to multiply to 14, so we can try 1 and 14 or 2 and 7. We see that...
# factor the expression (answer)
3d2+23d+14 We can start with one of the factors to have a coefficient of 3 on the variable d and the other coefficient to be one because of the coefficient on d2. (3d+?)(d+?) We know the ?'s have to multiply to 14, so we can try 1 and 14 or 2 and 7. We see that when we...
### Chris's Hourly Rate
Hourly Fee: \$100.00
Save up to 15% with a discount package
### Background Check Status for Chris S.
Chris S. has not yet had a background check. After sending a message to Chris, you will be able to order a background check for \$7.99. The check will be processed by First Advantage. For more information, please review the background check information page.
As part of your tutor selection process, we encourage you to run updated background checks. Please also review the safety tips for hiring tutors.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.wyzant.com/Tutors/NY/New-York/8213998/Resources.aspx", "fetch_time": 1406587115000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2014-23/segments/1406510263423.17/warc/CC-MAIN-20140728011743-00010-ip-10-146-231-18.ec2.internal.warc.gz", "warc_record_offset": 957387375, "warc_record_length": 14680, "token_count": 533, "char_count": 1837, "score": 3.984375, "int_score": 4, "crawl": "CC-MAIN-2014-23", "snapshot_type": "latest", "language": "en", "language_score": 0.8461905121803284, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00005-of-00064.parquet"}
|
Test: Coding - Decoding- 1
# Test: Coding - Decoding- 1
Test Description
## 20 Questions MCQ Test Logical Reasoning for CLAT | Test: Coding - Decoding- 1
Test: Coding - Decoding- 1 for Railways 2023 is part of Logical Reasoning for CLAT preparation. The Test: Coding - Decoding- 1 questions and answers have been prepared according to the Railways exam syllabus.The Test: Coding - Decoding- 1 MCQs are made for Railways 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Coding - Decoding- 1 below.
Solutions of Test: Coding - Decoding- 1 questions in English are available as part of our Logical Reasoning for CLAT for Railways & Test: Coding - Decoding- 1 solutions in Hindi for Logical Reasoning for CLAT course. Download more important topics, notes, lectures and mock test series for Railways Exam by signing up for free. Attempt Test: Coding - Decoding- 1 | 20 questions in 25 minutes | Mock test for Railways preparation | Free important questions MCQ to study Logical Reasoning for CLAT for Railways Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
Test: Coding - Decoding- 1 - Question 1
### If PINK is written as MLKN, how will ROSE be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 1
P - 3 = M
I + 3 = L
N - 3 = K
K + 3 = N
(PINK = MLKN )
Similarly, ROSE = ORPH
Test: Coding - Decoding- 1 - Question 2
### If SELDOM is written as MFTLNC, how will PRINCE be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 2
S E L D O M - M F T L N C
Case 1:
S + 1 = T
E + 1 = F
L + 1 = M
S E L → T F M → M F T
Case 2:
D - 1 = C
O - 1 = N
M - 1 = L
D O M → C N L → L N C
Resulting: SELDOM → MFTLNC
Similarly, PRINCE → JSQDBM
Test: Coding - Decoding- 1 - Question 3
### If GREAT is written as BVBCS, how will UNTIL be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 3
G - 5 = B
R + 4 = V
E - 3 = B
A + 2 = C
T - 1 = S
⇒ GREAT → BVBCS
Similarly,
U - 5 = P
N + 4 = R
T - 3 = Q
I + 2 = K
L - 1 = K
⇒ UNTIL → PRQKK
Test: Coding - Decoding- 1 - Question 4
If OPPOSITE is written as NOQPTJSD, how will STRAIGHT be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 4
O - 1 = N
P - 1 = O
P + 1 = Q
O + 1 = P
S + 1 = T
I + 1 = J
T - 1 = S
E - 1 = D
OPPOSITE → NOQPTJSD
Similarlly,
S - 1 = R
T - 1 = S
R + 1 = S
A + 1 = B
I + 1 = J
G + 1 = H
H - 1 = G
T - 1 = S
Test: Coding - Decoding- 1 - Question 5
If STUDENT is written as RRTBDLS, how will OFFICER be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 5
S - 1 = R
T - 2 = R
U - 1 = T
D - 2 = B
E - 1 = D
N - 2 = L
T - 1 = S
STUDENT → RRTBDLS
Similarly,
O - 1 = N
F - 2 = D
F - 1 = E
I - 1 = G
C - 2 = B
E - 1 = C
R - 2 = Q
OFFICER → NDEGBCQ
Test: Coding - Decoding- 1 - Question 6
In a certain code LAWN is written as JCUP. How will SLIT be coded in that code?
Detailed Solution for Test: Coding - Decoding- 1 - Question 6
Its all in the manner of -2, +2
J = L - 2
C = A + 2
U = W - 2
P = N + 2
and same for SLIT:
Q = S - 2
N = L + 2
G = I - 2
V = T + 2
Test: Coding - Decoding- 1 - Question 7
If TOUR is written in a certain code as 1234, CLEAR as 56784 and SPARE as 90847, what will be the 5th digit for SCULPTURE in the same code?
Detailed Solution for Test: Coding - Decoding- 1 - Question 7
TOUR → 1234
CLEAR → 56784
SPARE → 90847
Common alphabets in the three words are:
(i) R coded with the number 4 in each word.
(ii) E coded with the number 7 in each word.
(iii) A coded with the number 8 in each word.
It is clear from here that each alphabet represents a particular number/code.
In the word, SCULPTURE the 5th digit is P and in the above-given words, P is coded with the number 0.
Test: Coding - Decoding- 1 - Question 8
If BANGALORE is written as CZOFBKPQF, how will CALCUTTA be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 8
B + 1 = C
A - 1 = Z
N + 1 = O
G - 1 = F
A + 1 = B
L - 1 = K
O + 1 = P
R - 1 = Q
E + 1 = F
BANGALORE → CZOFBKPQF
Similarly,
C +1 = D
A - 1 = Z
L + 1 = M
C - 1 = B
U + 1 = V
T - 1 = S
T + 1 = U
A - 1 = Z
CALCUTTA → DZMBVSUZ
Test: Coding - Decoding- 1 - Question 9
If POSTPONE is written as OPTSOPEN, how will RAJASTAN be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 9
Every letter is interchanged by its subsequent second letter.
i.e. POSTPONE → OPTSOPEN
Therefore, RAJASTAN → ARAJTSNA
Test: Coding - Decoding- 1 - Question 10
If EXPLAINING is written as PXEALNIGNI, how will PRODUCTION be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 10
In the code, first three letters are reversed, EXP → PXE
then next two letters, LA → AL
then again next two letters, IN → NI
finally the last three letters are reversed, ING → GNI
Resulting: EXPLAINING → PXEALNIGNI
Similarly,
PRODUCTION → ORPUDTCNOI
Test: Coding - Decoding- 1 - Question 11
If TEACHING is written as CHEATING, how will BANKINGS be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 11
TEACHING:CHEATING ⇔ 12345678:45231678
BANKING:? ⇔ 12345678:45231678
Thus, BANKINGS will become KIANBNGS
Test: Coding - Decoding- 1 - Question 12
In a certain code FIRE is coded as DGPC. What will be the last letter of the coded word for SHOT.
Detailed Solution for Test: Coding - Decoding- 1 - Question 12
F - 2 = D
I - 2 = G
R - 2 = P
E - 2 = C
FIRE →DGPC
Similarly,
S - 2 = Q
H - 2 = F
O - 2 = M
T - 2 = R
QFMR → SHOT.
Hence, the last letter of the coded word is R.
Test: Coding - Decoding- 1 - Question 13
If AWAKE is written as ZVZJD, how will FRIEND be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 13
A - 1 = Z
W - 1 = V
A - 1 = Z
K - 1 = J
E - 1 = D
AWAKE → ZVZJD
Similarly,
F - 1 = E
R - 1 = Q
I - 1 = H
E - 1 = D
N - 1 = M
D - 1 = C
FRIEND → EQHDMC
Test: Coding - Decoding- 1 - Question 14
If in a code language PAINT is written as 74128 and EXCEL as 93596, then how ACCEPT will be written in that language?
Detailed Solution for Test: Coding - Decoding- 1 - Question 14
PAINT - 74128
EXCEL - 93596
Since each letter is assigned a unique code in the given words.
Therefore, the code for ACCEPT is 455978.
Test: Coding - Decoding- 1 - Question 15
If LOGIC is written as BHFNK, how will CLERK be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 15
Write the letters of the given word in reverse order
i.e. LOGIC → CIGOL
Now write the previous letters of the reversed word
i.e. CIGOL → BHFNK
Similarly,
CLERK → KRELC→ JQDKB
Test: Coding - Decoding- 1 - Question 16
If LMNO is written as GHIJ, how will STUV be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 16
L - 5 = G
M - 5 = H
N - 5 = I
O - 5 = J
LMNO → GHIJ
Similarly,
S - 5 = N
T - 5 = O
U - 5 = P
V - 5 = Q
STUV → NOPQ
Test: Coding - Decoding- 1 - Question 17
If TOUR is written as 1234, CLEAR is written as 56784 and SPARE is written as 90847, find the code for CARE
Detailed Solution for Test: Coding - Decoding- 1 - Question 17
TOUR → 1234
CLEAR → 56784
SPARE → 90847
Common alphabets in the three words are:
(i) R coded with the number 4 in each word.
(ii) E coded with the number 7 in each word.
(iii) A coded with the number 8 in each word.
It is clear from here that each alphabet represents a particular number/code.
Using the codes for the letter from given words.
C A R E will become 5 8 4 7
Test: Coding - Decoding- 1 - Question 18
If AVOPDE is written as ATONCE, how will DONE AT GATV AT ZOU DAP be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 18
Here you can see the vowels are not changing, only consonants are changing and they are also changing in the order of -1,-1,-2
Thus,
DONE AT GATV AT ZOU DAP ⇒ COME AS FAST AS YOU CAN
Test: Coding - Decoding- 1 - Question 19
If PREMONITION is written as 68530492904, how will MONITOR be written?
Detailed Solution for Test: Coding - Decoding- 1 - Question 19
Letters of the word PREMONITION are coded as follows:
Letters → Code
P → 6
R → 8
E → 5
M → 3
O → 0
N → 4
I → 9
T → 2
I → 9
O → 0
N → 4
Therefore, code for the word MONITOR is 3049208
Test: Coding - Decoding- 1 - Question 20
How many pairs of letters are there in the word 'SEQUENTIAL' which have as many letters between them as are in the alphabet?
Detailed Solution for Test: Coding - Decoding- 1 - Question 20
Arrange the word in alphabetical order and compare the pair obtained in the word and in the alphabet. The number of letters between should be equal in both. Here, the pairs obtained from the word SEQUENTIAL are SN, SQ, QN SA, and EA. Hence, five is the answer
## Logical Reasoning for CLAT
19 videos|27 docs|67 tests
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code
Information about Test: Coding - Decoding- 1 Page
In this test you can find the Exam questions for Test: Coding - Decoding- 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Coding - Decoding- 1, EduRev gives you an ample number of Online tests for practice
## Logical Reasoning for CLAT
19 videos|27 docs|67 tests
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://edurev.in/course/quiz/attempt/7877_Test-Coding-Decoding-1/cde45763-7ca1-4436-a589-e2843e36fbb8", "fetch_time": 1675521295000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764500140.36/warc/CC-MAIN-20230204142302-20230204172302-00428.warc.gz", "warc_record_offset": 244422842, "warc_record_length": 40428, "token_count": 3162, "char_count": 9167, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.8661766648292542, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00062-of-00064.parquet"}
|
# Weird Trig Graph Question
"The period of function f is 5. If f(1)= 4, f(2)= 5, and f(4)= -2, what is the value of f(7)?"
I'm pretty sure this has something to do with function f(x) as a periodic function when a number p>0 exists, such that for all x in the domain of f, f(x + p) = f(x). I don't understand how this works. Maybe for the first one it sort of makes sense, but the rest of them don't follow the same pattern. I really appreciate any help I could get on this. Thanks.
## Answers and Replies
minger
Science Advisor
From what I can tell, it's a simple problem trying to get you to think about what a period is. Just think, what happens in one period of a repetitive function. Also another hint:
sin 0 = ?
sin 2pi = ?
sin 4pi = ?
FredGarvin
Science Advisor
Along the lines of what Minger said, think about what the definition of "period" is. It is the "time" it takes your function to complete 1 full cycle. So, if the period is 5, that means you can say it has 5 points in it and then it REPEATS. If points 1-5 are the first cycle and then it REPEATS, what can you say about point #7?
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/weird-trig-graph-question.71308/", "fetch_time": 1627589222000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00646.warc.gz", "warc_record_offset": 983594213, "warc_record_length": 14155, "token_count": 306, "char_count": 1101, "score": 3.53125, "int_score": 4, "crawl": "CC-MAIN-2021-31", "snapshot_type": "latest", "language": "en", "language_score": 0.9391707181930542, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00038-of-00064.parquet"}
|
# Quantity Variances Vs. Rate Variances
Save
You need to know if your production and manufacturing facilities are making the most efficient use of their time and materials for the cost expended. The goal is be efficient to maximize profit potential. You can analyze the quantity variance compared to the rate variances for a given period of time. The difference will tell the tale.
## Quantity and Rate Variance Formulas
Begin to compute your company's quantity variance using the following accounting formula: actual quantity of input multiplied by actual price. In other words, the amount of materials purchased at the beginning of the month (or other time period) and the price per unit for the material. An example from the text of Managerial Accounting provides an illustration for a manufacturing facility: 6,500 pounds x \$3.80 per pound. This equals \$24,700.
Compute your company's rate variance using this formula: actual quantity of input multiplied by standard price. Using the above example, the 6,500 pounds of material purchased would typically cost your company \$4 per pound instead of \$3.80. Reflect that in your rate variance and you will arrive at \$26,000.
## Computing Price Variance
After using the quantity variance formula and the rate variance formula you can now compute the price variance this way: \$26,000 minus \$24,700 equals a price variance of \$1,300. In other words, you have a favorable rate variance because your cost was less than it normally would have been for the same amount of material purchased. Next, you need to analyze how your price variance stacks up against your quantity variance to see if your company has been working efficiently or not.
## Computing Quantity Variance
Computing your quantity variance involves one more formula: standard quantity output instead of actual output -- multiplied at the standard price. While our imaginary scenario reflects 6,500 pounds of material, your company might normally produce the same quantity with only 6,000 pounds of material. This would reflect a waste of 500 pounds, unnecessarily, in manufacturing. The quantity variance addresses that discrepancy. The 6,000 pounds x \$4 per pound equals \$24,000. Comparing the actual quantity cost at the standard price you normally pay (\$26,000) to the cost that should have been incurred for the correct amount of material needed for the job (\$24,000) reflects a \$2,000 negative difference.
## Significance
Compare the price variance total of \$1,300 to the quantity variance of \$2,000 and you arrive at a total variance of \$700. It is important to note that the \$1,300 price variance represented a positive number, as this was a profit from paying less for materials. The \$2,000 number, however, was a negative number, as it represented wasted material costs of that amount. Therefore, after deducting your profit of \$1,300 from your excess expense of \$2,000, you are still left with a negative balance of \$700 for the project analyzed.
## References
Promoted By Zergnet
## Related Searches
Check It Out
### Are You Really Getting A Deal From Discount Stores?
M
Is DIY in your DNA? Become part of our maker community.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.ehow.com/info_8695303_quantity-variances-vs-rate-variances.html", "fetch_time": 1505873530000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-39/segments/1505818686117.24/warc/CC-MAIN-20170920014637-20170920034637-00358.warc.gz", "warc_record_offset": 456984570, "warc_record_length": 19187, "token_count": 663, "char_count": 3184, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2017-39", "snapshot_type": "latest", "language": "en", "language_score": 0.8708791732788086, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00020-of-00064.parquet"}
|
# What is the name for this type of constraint?
I have what would be a straightforward mixed-integer linear programming problem, except for the fact that some of the constraints are of the form $$f(x_1,x_2,x_3,\ldots,x_n) < c$$, where $$f$$ is 'take the maximum of the largest coordinate and the sum of all the smaller ones'
In lisp:
(defn f [& l]
(let [sl (reverse (sort l))]
(max (first sl) (reduce + (rest sl)))))
(f 1 2 3 4 5) -> 10
In three dimensions e.g. I think I can rephrase this for e.g. $$f<10$$ as
(defn constraint [x,y,z]
(or
(and (<= x 10)
(<= (+ y z) 10))
(and (<= y 10)
(<= (+ x z) 10))
(and (<= z 10)
(<= (+ x y) 10))))
Which is clearly the union of $$3$$ (or $$n$$) convex objects (prisms).
Is there a name for this type of constraint? Are there techniques and packages for solving these kinds of problems?
• Please use mathematical notation to explain the constraint that you're trying to achieve. Commented Mar 6, 2019 at 0:58
• I don't know how! Even to write the first version. The second one would be easy in TeX, can I just type that in here? Commented Mar 6, 2019 at 15:28
• Stackexchange uses MathJax to implement LaTeX equations. Commented Mar 6, 2019 at 17:04
I would call the constraint "upper- and lower-bounds on the maximum element." Note that you are actually dealing with two separate constraints. Define the max element function as follows $$\max:\mathbb{R}^{n}\to\mathbb{R}\qquad\max(x)\equiv\max_{i\in\{1,\ldots,n\}}x_{n}.$$ Your first constraint is "take the max element and ensure that it is less than $$c$$": $$\max(x) while the second is "take the sum, subtract the max element, and ensure that the result is less than $$c$$": $$\mathbf{1}^{T}x-\max(x) where $$\mathbf{1}=[1,\ldots,1]^{T}$$ is the usual column vector of ones. Observe that (1) is a convex upper-bound on $$\max(x)$$ while (2) is a nonconvex lower-bound.
Such problems with a fixed upper-bound have a very elegant solution via the Big-M method. Let $$M$$ be an arbitrary large number that satisfies $$\max(x)\le M$$ for all possible choices of $$x$$. Then it is easy to verify that $$\alpha=\max(x)\le M\quad\iff\quad\alpha\ge x\ge\alpha-(1-z)M,\quad\mathbf{1}^{T}z=1,\quad z\in\{0,1\}^{n}.$$ Using the above identity, we can implement (1) and (2) exactly as the following mixed integer constraints $$f(x) where we have conveniently used $$c$$ as the Big-M parameter.
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://scicomp.stackexchange.com/questions/31178/what-is-the-name-for-this-type-of-constraint", "fetch_time": 1718484895000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00083.warc.gz", "warc_record_offset": 462015976, "warc_record_length": 40812, "token_count": 718, "char_count": 2387, "score": 3.59375, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.8854254484176636, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00014-of-00064.parquet"}
|
# How do you simplify (2x^2+6x+4)/(4x^-12x-16)?
Jul 5, 2015
I will assume the expression in the question should be
$\frac{2 {x}^{2} + 6 x + 4}{4 {x}^{2} - 12 x - 16} = \frac{x + 2}{2 \left(x - 4\right)}$
with exclusion $x \ne - 1$
#### Explanation:
$\frac{2 {x}^{2} + 6 x + 4}{4 {x}^{2} - 12 x - 16}$
$= \frac{2 \left({x}^{2} + 3 x + 2\right)}{4 \left({x}^{2} - 3 x - 4\right)}$
$= \frac{2 \left(x + 1\right) \left(x + 2\right)}{4 \left(x + 1\right) \left(x - 4\right)}$
$= \frac{x + 2}{2 \left(x - 4\right)}$
with exclusion $x \ne - 1$
If $x = - 1$ the original expression becomes $\frac{0}{0}$ which is not defined.
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://socratic.org/questions/how-do-you-simplify-2x-2-6x-4-4x-12x-16", "fetch_time": 1675545059000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00750.warc.gz", "warc_record_offset": 536432983, "warc_record_length": 5835, "token_count": 287, "char_count": 628, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.4874299168586731, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00059-of-00064.parquet"}
|
# Trigonometry Examples
Convert to Interval Notation x-8/x<2
Find the LCD of the terms in the equation.
Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.
Since contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part then find LCM for the variable part .
The LCM is the smallest number that all of the numbers divide into evenly.
1. List the prime factors of each number.
2. Multiply each factor the greatest number of times it occurs in either number.
The number is not a prime number because it only has one positive factor, which is itself.
Not prime
The LCM of is the result of multiplying all prime factors the greatest number of times they occur in either number.
The factor for is itself.
occurs time.
The LCM of is the result of multiplying all prime factors the greatest number of times they occur in either term.
Multiply each term by and simplify.
Multiply each term in by in order to remove all the denominators from the equation.
Simplify each term.
Multiply by .
Cancel the common factor of .
Move the leading negative in into the numerator.
Cancel the common factor.
Rewrite the expression.
Subtract from both sides of the inequality.
Convert the inequality to an equation.
Factor using the AC method.
Consider the form . Find a pair of integers whose product is and whose sum is . In this case, whose product is and whose sum is .
Write the factored form using these integers.
Set equal to and solve for .
Set the factor equal to .
Add to both sides of the equation.
Set equal to and solve for .
Set the factor equal to .
Subtract from both sides of the equation.
Consolidate the solutions.
Find the domain of .
Set the denominator in equal to to find where the expression is undefined.
The domain is all values of that make the expression defined.
Interval Notation:
Interval Notation:
Use each root to create test intervals.
Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.
Test a value on the interval to see if it makes the inequality true.
Choose a value on the interval and see if this value makes the original inequality true.
Replace with in the original inequality.
The left side is less than the right side , which means that the given statement is always true.
True
True
Test a value on the interval to see if it makes the inequality true.
Choose a value on the interval and see if this value makes the original inequality true.
Replace with in the original inequality.
The left side is not less than the right side , which means that the given statement is false.
False
False
Test a value on the interval to see if it makes the inequality true.
Choose a value on the interval and see if this value makes the original inequality true.
Replace with in the original inequality.
The left side is less than the right side , which means that the given statement is always true.
True
True
Test a value on the interval to see if it makes the inequality true.
Choose a value on the interval and see if this value makes the original inequality true.
Replace with in the original inequality.
The left side is not less than the right side , which means that the given statement is false.
False
False
Compare the intervals to determine which ones satisfy the original inequality.
True
False
True
False
True
False
True
False
The solution consists of all of the true intervals.
or
Convert the inequality to interval notation.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.mathway.com/popular-problems/Trigonometry/991148", "fetch_time": 1582558943000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-10/segments/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00249.warc.gz", "warc_record_offset": 794218684, "warc_record_length": 12001, "token_count": 741, "char_count": 3531, "score": 4.53125, "int_score": 5, "crawl": "CC-MAIN-2020-10", "snapshot_type": "latest", "language": "en", "language_score": 0.9249166250228882, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00014-of-00064.parquet"}
|
## Sunday, June 17, 2018
### Swap Two Integers without using a Temporary Variable
This is an older exercise, but it still comes up.
This falls under things you wouldn't bother worrying about in real life.
## Swap Two Integers without using a Temporary Variable
The idea is to just add the two integer variables and then subtract each piece out in turn.
Say you have x=5 and y=9.
If you are using Python you can just stop right here as "tuple unpacking" will so all the work for you.
x, y = (y, x)
and you are done. x is 9 and y is 5.
Let's assume you want to do it the harder way...
Replace x with the sum of x and y.
x = x + y
x = 5 + 9 = 14
Now you have "lost" x, but you still have y and you can deduce what your old x is.
For the new y take your total and subtract out y, which will leave you with the "old" x in y's place.
y = x - y
y = 14 - 9 = 5
Now to get the new x, take the total and subtract out the new y. which will give you the old y which goes in x's place. It's more confusing to write it out in English that to just see the math.
x = x - y
x = 14 - 5 = 9
To Summarize
x = x + y [x changes]
y = x - y [y changes: Total minus original y]
x = x - y [x changes again]
#### 2 comments:
Eva E. Murphy said...
I used to believe that swapping without using a temporary variable was impossible, but your posts have convinced me it is possible. seriously, your blogs are too good, they always provide me something unique. I wish I could visit more of your blogs, but I am not able to do that right now because I have to find a platform from which I can Buy Compare And Contrast Essay online at an affordable price.
Dissertation Writing Services London said...
Goodness Algebra! That was a tough subject. Back in my school days, I had a friend that helped me with my algebra class. She works in one of the popular Dissertation Writing Services London now and we do not talk a lot anymore. In school, we used to talk and play often but I lost touch with her ever since we entered universities. Maybe I should contact her again. Anyways thank you for bringing back so many memories.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://nondogblog.frap.org/2018/06/swap-two-integers-without-using.html", "fetch_time": 1653631006000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-21/segments/1652662636717.74/warc/CC-MAIN-20220527050925-20220527080925-00423.warc.gz", "warc_record_offset": 504147281, "warc_record_length": 12597, "token_count": 543, "char_count": 2121, "score": 3.828125, "int_score": 4, "crawl": "CC-MAIN-2022-21", "snapshot_type": "longest", "language": "en", "language_score": 0.9546477794647217, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00006-of-00064.parquet"}
|
Terry O.
# How many seconds will it take for the ball to reach a height of 80 feet?
If a ball is thrown upward from the ground with an initial speed of 96 feet per second, its height h, in feet, after t seconds is
h = −16t2 + 96t.
How many seconds will it take for the ball to reach a height of 80 feet? There are two answers. (Enter your answers as a comma-separated list.)
t =
sec
Explain why there are two answers.
By:
Tutor
5.0 (472)
Affordable, Experienced, and Patient Geometry Tutor
Terry O.
I do not understand the answer.
How many seconds will it take for the ball to reach a height of 80 feet?
Report
02/22/18
Terry O.
I do not understand how many seconds or how you are getting to an answer.
Terry
Report
02/22/18
Philip P.
Solve the quadratic equation t2 - 6t + 5 = 0 to get the answers. You solve a quadratic by factoring or by using the quadratic formula (hint: this one factors). Quadratic equations usually have two answers. Can you solve the quadratic?
Report
02/22/18
## Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
#### OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.wyzant.com/resources/answers/454793/how_many_seconds_will_it_take_for_the_ball_to_reach_a_height_of_80_feet", "fetch_time": 1628197475000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-31/segments/1627046157039.99/warc/CC-MAIN-20210805193327-20210805223327-00620.warc.gz", "warc_record_offset": 1118376640, "warc_record_length": 13859, "token_count": 324, "char_count": 1245, "score": 3.6875, "int_score": 4, "crawl": "CC-MAIN-2021-31", "snapshot_type": "latest", "language": "en", "language_score": 0.9202965497970581, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00034-of-00064.parquet"}
|
Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.1 | Set 1
• Last Updated : 02 Jun, 2021
Question 1.
Solution:
We have,
I =
I =
I =
I =
I = 2[√9 – √4 ]
I = 2 (3 − 2)
I = 2 (1)
I = 2
Therefore, the value of is 2.
Question 2.
Solution:
We have,
I =
I =
I = log (3 + 7) − log (−2 + 7)
I = log 10 − log 5
I =
I = log 2
Therefore, the value of is log 2.
Question 3.
Solution:
We have,
I =
Let x = sin t, so we have,
=> dx = cos t dt
Now, the lower limit is,
=> x = 0
=> sin t = 0
=> t = 0
Also, the upper limit is,
=> x = 1/2
=> sin t = 1/2
=> t = π/6
So, the equation becomes,
I =
I =
I =
I =
I =
I = π/6 – 0
I = π/6
Therefore, the value of is π/6.
Question 4.
Solution:
We have,
I =
I =
I =
I =
I = π/4
Therefore, the value of is π/4.
Question 5.
Solution:
We have,
I =
Let x2 + 1 = t, so we have,
=> 2x dx = dt
=> x dx = dt/2
Now, the lower limit is, x = 2
=> t = x2 + 1
=> t = (2)2 + 1
=> t = 4 + 1
=> t = 5
Also, the upper limit is, x = 3
=> t = x2 + 1
=> t = (3)2 + 1
=> t = 9 + 1
=> t = 10
So, the equation becomes,
I =
I =
I =
I = 1/2[log10 – log5]
I = 1/2[log10/5]
I = 1/2[log2]
I = log√2
Therefore, the value of is log√2.
Question 6.
Solution:
We have,
I =
I =
I =
I =
I =
I = 1/ab[tan-1∞ – tan-10]
I = 1/ab[π/2 – 0]
I = 1/ab[π/2]
I = π/2ab
Therefore, the value of is π/2ab.
Question 7.
Solution:
We have,
I =
I =
I = [tan-11 – tan-1(-1)]
I = [π/4 – (-π/4)]
I = [π/4 + π/4]
I = 2π/4
I = π/2
Therefore, the value of is π/2.
Question 8.
Solution:
We have,
I =
I =
I = -e – (-e0)
I = − 0 + 1
I = 1
Therefore, the value of is 1.
Question 9.
Solution:
We have,
I =
I =
I =
I =
I =
I = [1 − 0] − [log(1 + 1) − log(0 + 1)]
I = 1 − [log2 − log1]
I = 1 – log2/1
I = 1 − log 2
I = log e − log 2
I = loge/2
Therefore, the value of is loge/2.
Question 10.
Solution:
We have,
I =
I =
I =
I = [-cosπ/2 + cos0] + [sinπ/2 – sin0]
I = [−0 + 1] + 1
I = 1 + 1
I = 2
Therefore, the value of is 2.
Question 11.
Solution:
We have,
I =
I =
I = log(sinπ/2) – log(sinπ/4)
I = log1 – log1/√2
I =
I = log√2
Therefore, the value of is log√2.
Question 12.
Solution:
We have,
I =
I =
I = log(secπ/4 + tanπ/4 – log(sec0 + tan0)
I = log(√2 + 1) – log(1 + 0)
I =
I = log(√2 + 1)
Therefore, the value of is log(√2 + 1).
Question 13.
Solution:
We have,
I =
I =
I = [log|cosecπ/4 – cotπ/4|] – [log|cosecπ/6 – cotπ/6|]
I = [log|√2 – 1|] – [log|2 – √3|]
I =
Therefore, the value of is .
Question 14.
Solution:
We have,
I =
Let x = cos 2t, so we have,
=> dx = –2 sin 2t dt
Now, the lower limit is,
=> x = 0
=> cos 2t = 0
=> 2t = π/2
=> t = π/4
Also, the upper limit is,
=> x = 1
=> cos 2t = 1
=> 2t = 0
=> t = 0
So, the equation becomes,
I =
I =
I =
I =
I =
Let cos t = z, so we have,
=> – sin t dt = dz
=> sin t dt = – dz
Now, the lower limit is,
=> t = 0
=> z = cos t
=> z = cos 0
=> z = 1
Also, the upper limit is,
=> t = π/4
=> z = cos t
=> z = cos π/4
=> z = 1/√2
So, the equation becomes,
I =
I =
I =
I =
I =
I =
I = -4[(log1/√2 – 1/2(2)) – (log1 – 1/2)]
I = -4[(log1/√2 – 1/4) – (0 – 1/2)]
I = -4[log1/√2 – 1/4 – 0 + 1/2]
I = -4[-log√2 + 1/4]
I = 4log√2 – 1
I = 4 × 1/2log2 – 1
I = 2log2 – 1
Therefore, the value of is 2log2 – 1.
Question 15.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = [tan π – tan0] – [sec π – sec 0]
I = [0 – 0] – [–1 – 1]
I = 0 – (–2)
I = 2
Therefore, the value of is 2.
Question 16.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = [tan π/4 – tan(–π/4)] – [sec π/4 – sec (–π/4)]
I = [1 – (–1)] – [sec π/4 – sec (π/4)]
I = 2 – 0
I = 2
Therefore, the value of is 2.
Question 17.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 1/2[π/2 – 0] + 1/4[sinπ – sin0]
I = 1/2[π/2] + 1/4[0 – 0]
I = π/4
Therefore, the value of is π/4.
Question 18.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 1/12 [-1 – 0] + 3/4[1 – 0]
I = 3/4 – 1/12
I = (9 – 1)/12
I = 8/12
I = 2/3
Therefore, the value of is 2/3.
Question 19.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 1/6[sinπ/2 – sin0] + 1/2[sinπ/6 – sin0]
I = 1/6[1 – 0] + 1/2[1/2 – 0]
I = 1/6 + 1/4
I = (4 + 6)/24
I = 10/24
I = 5/12
Therefore, the value of is 5/12.
Question 20.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 1/2[sinπ/2 – sin0] – 1/6[sin3π/2 – sin0]
I = 1/2[1 – 0] – 1/6[-1 – 0]
I = 1/2 – 1/6(-1)
I = 1/2 + 1/6
I = (6 + 2)/12
I = 8/12
I = 2/3
Therefore, the value of is 2/3.
Question 21.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = 2[-cotπ/2 + cot2π/3]
I = 2[-1/√3 – 0]
I = -2/√3
Therefore, the value of is -2/√3.
Question 22.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I = 1/4[π/2 + π/4 + 0 + 0 – 0 – 0 – 0 – 0]
I = 1/4[3π/4]
I = 3π/16
Therefore, the value of is 3π/16.
My Personal Notes arrow_drop_up
Related Articles
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://origin.geeksforgeeks.org/class-12-rd-sharma-solutions-chapter-20-definite-integrals-exercise-20-1-set-1/?ref=lbp", "fetch_time": 1675134211000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764499842.81/warc/CC-MAIN-20230131023947-20230131053947-00362.warc.gz", "warc_record_offset": 457833330, "warc_record_length": 49493, "token_count": 2328, "char_count": 4996, "score": 4.5625, "int_score": 5, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.8406411409378052, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00043-of-00064.parquet"}
|
# Critical Velocity Calculator, Formula, Critical Velocity Calculation
## Critical Velocity Calculator:
Enter the values of Reynold’s number R , Co-efficient of Viscosity u(m2/s) , Inner Diametre of the pipe D(m) & Density of the liquid p(kg/m3) to determine the value of Critical Velocity Vc(m/s).
Enter Reynold’s number: Enter Co-Efficient of Viscosity: m2/s Enter Inner Diametre of the Pipe: m Enter Density of the Liquid: kg/m3 Result – Critical Velocity: m/s
## Critical Velocity Formula:
The Critical Velocity Vc(m/s) in meter per second is equal to the Reynold’s number R into multiply the Co-efficient of Viscosity u(m2/s) in meter square per second and divided by the Inner Diametre of the pipe D(m) in meter into multiply the Density of the liquid p(kg/m3) in kilogram per meter cube.
The Formula of Critical Velocity can be writtern as,
Vc(m/s) = ( R * u(m2/s) ) / ( D(m) * p(kg/m3) )
Here,
Vc(m/s) = Critical Velocity in meter per second
R = Reynold’s number
u(m2/s) = Co-efficient of Viscosity in meter square per second
Learn More: eV to Velocity Calculator, Formula, eV to Velocity Calculation
D(m) = Inner Diametre of the pipe in meter
p(kg/m3) = Density of the liquid in kilogram per meter cube
### Example Problems for Critical Velocity :
1)Calculate the Critical Velocity and given for Reynold’s number = 3.5 , Co-efficient of Viscosity = 25m3/s , Inner Diametre of the pipe = .30m , Density of the liquid = 1.300kg/m3.
Vc = ( R * u ) / ( D * p )
Vc = ( 3.5 * 25 ) / ( .30 * 1.300 )
Vc = 379.16m/s.
2)Calculate the Density of the liquid and given for Reynold’s number = 3.5 , Co-efficient of Viscosity = 25m3/s , Inner Diametre of the pipe = .30m , Critical Velocity = 379.16m/s.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.electrical4u.net/calculator/critical-velocity-calculator-formula-critical-velocity-calculation/", "fetch_time": 1723218030000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-33/segments/1722640767846.53/warc/CC-MAIN-20240809142005-20240809172005-00399.warc.gz", "warc_record_offset": 571302668, "warc_record_length": 25398, "token_count": 497, "char_count": 1723, "score": 3.828125, "int_score": 4, "crawl": "CC-MAIN-2024-33", "snapshot_type": "latest", "language": "en", "language_score": 0.6459105610847473, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00025-of-00064.parquet"}
|
236 views
Consider the compund propositions given below as:
1. $p \vee \sim (p \wedge q)$
2. $(p \wedge \sim q) \vee \sim (p \wedge q)$
3. $p \wedge (q \vee r)$
Which of the above propositions are tautologies
1. a and c
2. b and c
3. a and b
4. a, b, and c
recategorized | 236 views
$p\vee (p\wedge q)$ = p+p+q = 1+q= 1 so tautology.
C and b cant be tautology so no option is matching.
selected
0
How b. Is tautology
0
only a is tautology but since c nver be tautology so i choose b .
–1 vote
p∨~(p∧q)
p∨~(p∧q) =p ∨ ~p v~q
p ∨ ~p is always true
a is a tautology
(p∧~q)∨~(p∧q)
(p∧~q)∨~(p∧q) = (p∧~q)∨(~pv~q) is also tautology
p∧(q∨r)
p∧(q∨r) = (p^q) v (p^r) is not a tautology
C a and b
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://gateoverflow.in/62228/ugcnet-dec2015-ii-9", "fetch_time": 1521706285000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-13/segments/1521257647782.95/warc/CC-MAIN-20180322073140-20180322093140-00714.warc.gz", "warc_record_offset": 576082249, "warc_record_length": 17614, "token_count": 317, "char_count": 702, "score": 3.515625, "int_score": 4, "crawl": "CC-MAIN-2018-13", "snapshot_type": "latest", "language": "en", "language_score": 0.5964096784591675, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00023-of-00064.parquet"}
|
# Subtracting Mixed Numbers Worksheet Grade 5
A Subtraction Amounts Worksheet can be a useful resource for exercising and testing your child’s subtraction details. Subtraction worksheets should focus on the student’s subtraction information rather than about the connection between addition and subtraction. If you want to use these worksheets with your child, here are some tips you should keep in mind. Utilize a worksheet that can help your kids be aware of the meaning of each and every number along with the technique utilized to conduct subtraction. Subtracting Mixed Numbers Worksheet Grade 5.
## Subtraction
Subtraction amounts worksheets are of help for many class degrees since they aid students process their expertise with the subject. The worksheets have concerns for all digits of your quantity and may aid individuals build a increased knowledge of the notion. If you have questions about the worksheets, feel free to leave a comment or send us an email. We delightful any recommendations for upgrades! The worksheets are meant to provide more practice and support once and for all educating methods.
## Supplement
A math worksheet for introducing solitary-digit figures might help college students find out this crucial ability. Many addition worksheets feature questions written horizontally. Alternatively, from left to right. This permits college students to train emotional remaining-to-proper supplement, which will serve them properly later on. They may also be used as analysis resources. This information will outline for you some great benefits of an supplement amounts worksheet. For a kid to perfect this ability, it is important to exercise with amounts both in recommendations.
## Subtraction of two numbers
A Subtraction of two numbers worksheet is an excellent instrument for teaching individuals relating to this basic math principle. A two-digit amount can have two numbers and another decimal location. The students are taught to approach each place individually, rather than borrowing from the decimal place before. Before they begin the subtraction of two digits, it helps to provide tactile and visual understanding of each place. This type of worksheet may be used by pupils from year one to calendar year three.
## Subtraction of 3 numbers
Third-grade mathematics curriculums often include three-digit subtraction. To teach your individuals the ability, you may make your resources more effective by employing a technique referred to as regrouping. Regrouping involves borrowing a amount through the increased anyone to make the reduce a single smaller sized. These supplies attribute interesting fingers-on actions, in depth training ideas, and computer worksheets. The Educational Library’s solutions make discovering simple and easy , entertaining.
## Subtraction of 4 numbers
To learn how to do the straightforward math concepts dilemma, subtraction of 4 digits, you need to very first know what spot value is. , and tens are common distinct position values.hundreds and Thousands Subtracting coming from a larger sized variety is named minuend, although subtracting coming from a more compact quantity is called subtrahend. The difference is the consequence of the subtraction. Create the situation in posts Minuend – Subtrahend = Difference. You may also use term difficulties to apply 4 digit subtraction.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://numbersworksheet.com/subtracting-mixed-numbers-worksheet-grade-5/", "fetch_time": 1721703654000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00511.warc.gz", "warc_record_offset": 376019912, "warc_record_length": 13868, "token_count": 615, "char_count": 3362, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.9333925843238831, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00060-of-00064.parquet"}
|
1. /
2. CBSE
3. /
4. Class 06
5. /
6. Mathematics
7. /
8. NCERT Solutions for Class...
# NCERT Solutions for Class 6 Maths Exercise 9.4
NCERT Solutions for Class 6 Maths Exercise 9.4 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 6 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free.
NCERT solutions for Maths Data handling
## NCERT Solutions for Class 6 Maths Data handling
###### Question 1. A survey of 120 school students was done to find which activity they prefer to do in their free time:
Preferred activity Number of students PlayingReading story booksWatching TVListening to musicPainting 4530201015
Draw a bar graph to illustrate the above data taking scale of 1 unit length = 5 students
Which activity is preferred by most of the students other than playing?
NCERT Solutions for Class 6 Maths Exercise 9.4
###### Question 2.The number of mathematics books sold by a shopkeeper on six consecutive days is shown below:
Days Sunday Monday Tuesday Wednesday Thursday Friday No. of books sold 65 40 30 50 20 70
Draw a bar graph to represent the above information choosing the scale of your choice.
###### Question 3.Following shows the number of bicycles manufactured in a factory during the year 1998 to 2002. Illustrate this data using a bar graph. Choose a scale your choice.
Years Number of bicycles manufactured 19981999200020012002 20060090011001200
(a)In which year were the maximum number of bicycles manufactures?
(b)In which year were the minimum number of bicycles manufactured?
(a) 2002
(b) 1999
NCERT Solutions for Class 6 Maths Exercise 9.4
###### Question 4.Number of persons in various age groups in a town is given in the following table:
Age Group Number of persons 1 – 1415 – 2930 – 4445 – 5960 – 7475 and above 2 Lakhs1 lakh 60 thousands1 lakh 20 thousands1 lakh 20 thousands80 thousands40 thousands
Draw a bar graph to represent the above information and answer the following questions. (take 1 unit length = 20 thousands)
(a)Which two age groups have same population?
(b)All persons in the age group of 60 and above are called senior citizens. How many senior citizens are there in the town?
(a)Group 30 – 44 and group 45 – 59
80,000 + 40,000 = 1,20,000
## NCERT Solutions for Class 6 Maths Exercise 9.4
NCERT Solutions Class 6 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 6 Maths includes text book solutions from Class 6 Maths Book . NCERT Solutions for CBSE Class 6 Maths have total 14 chapters. 6 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 6 solutions PDF and Maths ncert class 6 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.
## CBSE app for Students
To download NCERT Solutions for Class 6 Maths, Social Science Computer Science, Home Science, Hindi English, Maths Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through the best app for CBSE students and myCBSEguide website.
Test Generator
Create question paper PDF and online tests with your own name & logo in minutes.
myCBSEguide
Question Bank, Mock Tests, Exam Papers, NCERT Solutions, Sample Papers, Notes
### 1 thought on “NCERT Solutions for Class 6 Maths Exercise 9.4”
1. Thank u for helpful solutions
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://mycbseguide.com/blog/ncert-solutions-class-6-maths-exercise-9-4/", "fetch_time": 1695469599000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233506480.7/warc/CC-MAIN-20230923094750-20230923124750-00272.warc.gz", "warc_record_offset": 461002227, "warc_record_length": 21556, "token_count": 918, "char_count": 3833, "score": 3.859375, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.8984665870666504, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00024-of-00064.parquet"}
|
# Fluid statics
Fluid statics is the branch of fluid mechanics which studies fluids at rest.
## Equations
### Pressure variation in a static Fluid
Applying Newton's laws of motions to a fluid at rest, it can be determined that the sum of forces must equal zero throughout the fluid, which means any arbitrary element of the fluid is subjected to forces that sums to zero. Since the fluid is not deforming while it is at rest, the only forces acting on the fluid are those due to gravity and pressure.
For a given fluid element, the pressure is given by the differential equation[1]
where is the density of the fluid, is the gravitational acceleration (vector), and is the pressure at the fluid element.
For small changes in altitude, can be assumed to be constant (A more accurate description for large change in altitude can be found here). Simplifications can also be made for constant density, and reducing the analysis to only the y dimension:
## Examples
For system filled with an isothermal (constant temperature throughout) ideal gas, a relation between pressure altitude can be established using the following method: For ideal gases, the density is given by
Where M is the molar mass, R is the ideal gas constant and T is the temperature (Absolute temperature, in Kelvin or Rankine) of the gas. we can then set up the equation as follows:
Note that we are using the scalar value of gravity (g), so the minus sign is included due to gravity is downwards, in the negative direction of the y-axis. using the method of separation of variables, we can rearrange the equation so
Integrating both sides gives
Where P0 is the reference pressure at point y0 (often taken at the point which P0 is the atmospheric pressure).
Rearranging gives
## References
1. James R. Welty, Charles E. Wicks, Robert E. Wilson, and Gregory L. Rorrer, Fundamentals of Momentum, Heat, and Mass Transfer, 4th Ed. Toronto: John Wiley & Sons Inc, 2001.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.conservapedia.com/index.php?title=Fluid_statics&direction=next&oldid=1243141", "fetch_time": 1571043299000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-43/segments/1570986649841.6/warc/CC-MAIN-20191014074313-20191014101313-00033.warc.gz", "warc_record_offset": 813981556, "warc_record_length": 7680, "token_count": 429, "char_count": 1963, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2019-43", "snapshot_type": "latest", "language": "en", "language_score": 0.912729024887085, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00013-of-00064.parquet"}
|
# Homework Help: The Clam and Its Forces
1. Sep 16, 2008
### Phoenixtears
SOLVED
1. The problem statement, all variables and given/known data
Sea clams move by pushing water in one direction so that the water pushes them in the opposite direction. In doing so a 0.5 kg clam can accelerate 0.6 kg of water from rest to a speed of 1.9 m/s in 0.43 seconds. Calculate the magnitude of the clam's acceleration
2. Relevant equations
F= ma
Vf^2= V0^2 + 2ax
Vf= V0 + at
x= V0*t + .5a(t^2)
3. The attempt at a solution
I'm not sure why my answer isn't checking out. The initial velocity is 0. The final velocity is 1.9. The time is .43. These all fit into the third equation: 1.9= 0 + .43a. a then equals 4.42. This, however, is not working as my answer. What am I missing?
~Phoenix
Last edited: Sep 17, 2008
2. Sep 16, 2008
### ritwik06
Apply conservation of momentum
0.6*1.9=0.5*v
v=2.28
$$a=\frac{\Delta v}{\Delta t}$$
a=2.28/0.43
a=5.3
Does this help u?
3. Sep 17, 2008
### Ygggdrasil
The clam accelerates the water from 0 to 1.9 m/s in 0.43s. You need to calculate the acceleration of the clam, not the water. (hint: remember Newton's third law)
4. Sep 17, 2008
### Phoenixtears
OH! Alrighty, gotcha. That makes sense to me. Thank you so much. I realize now that I was mixing variables that shouldn't be used to solve for this.
Thank you so very much!
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/the-clam-and-its-forces.256946/", "fetch_time": 1540043029000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-43/segments/1539583512750.15/warc/CC-MAIN-20181020121719-20181020143219-00536.warc.gz", "warc_record_offset": 1030359583, "warc_record_length": 12924, "token_count": 452, "char_count": 1365, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2018-43", "snapshot_type": "latest", "language": "en", "language_score": 0.8957920670509338, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00036-of-00064.parquet"}
|
Explore BrainMass
Share
# Trigonometry and Geometry Word Problems
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
3) A building 210 feet tall casts a 90 foot long shadow. If a person stands at the end of the shadow and looks up to the top of the building, what is the angle of the person's eyes to the top of the building (to the nearest hundredth of a degree)? (Assume the person's eyes are 4 feet above ground level.)
A) 66.40° B) 64.09° C) 66.80° D) 25.91°
4) A straight trail with a uniform inclination of 13° leads from a lodge at an elevation of 800 feet to a mountain lake at an elevation of 6000 feet. What is the length of the trail (to the nearest foot)?
A) 23,116 feet B) 26,672 feet C) 6158 feet D) 5337 feet
5) A building 170 feet tall casts a 90 foot long shadow. If a person looks down from the top of the building, what is the measure of the angle between the end of the shadow and the vertical side of the building (to the nearest degree)? (Assume the personts eyes are level with the top of the building.)
A) 28° B) 32° C) 58° D) 62°
Find the missing parts of the triangle.
6) A = 26°30'
B = 25°1O'
a = 11.92
A) C = 129°20', b = 11.36, c = 20.96 B) C = 128°2O b = 11.36, c = 20S6
C) C = 129°20', b = 20.96, c = 11.36 . D) C = 128°20', b = 2O96, c = 11.36
7)B=26.5°
C=114.2°
b=11.34
A)A=37.3°, a=23.18, c= 16.10 B) A=39.3°,a= 16.1O,c=23.18
C)A =39.3°, a = 181O, c=25.18 D)A =37.3°, a =25.18, c= 18.10
Solve the problem.
8) An airplane is sighted at the same time by two ground observers who are 3 miles apart and in front of the airplane. They report the angles of elevation as 13° and 24°. How high is the airplane?
A) 1.44 miles B) 2.6 miles C) 0.67 miles D) 1.22 miles
© BrainMass Inc. brainmass.com October 9, 2019, 6:33 pm ad1c9bdddf
https://brainmass.com/math/trigonometry/91300
#### Solution Summary
It shows how to solve trigonometry and geometry word problems. The response is detailed and received a rating of "5/5" from the student who originally posted the question.
\$2.19
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://brainmass.com/math/trigonometry/91300", "fetch_time": 1582718995000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-10/segments/1581875146342.41/warc/CC-MAIN-20200226115522-20200226145522-00224.warc.gz", "warc_record_offset": 296675607, "warc_record_length": 12610, "token_count": 712, "char_count": 2075, "score": 4.09375, "int_score": 4, "crawl": "CC-MAIN-2020-10", "snapshot_type": "latest", "language": "en", "language_score": 0.8625431060791016, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00043-of-00064.parquet"}
|
On the Ellipsoid and Plane Intersection Equation
Applied Mathematics
Vol. 3 No. 11 (2012) , Article ID: 24506 , 7 pages DOI:10.4236/am.2012.311226
On the Ellipsoid and Plane Intersection Equation
Peter Paul Klein
Computing Center, University of Technology Clausthal, Clausthal-Zellerfeld, Germany
Email: [email protected]
Received August 3, 2012; revised September 10, 2012; accepted September 17, 2012
Keywords: Ellipsoid and Plane Intersection; Identity of Lagrange; Grassmann Expansion Theorem
ABSTRACT
It is well known that the line of intersection of an ellipsoid and a plane is an ellipse. In this note simple formulas for the semi-axes and the center of the ellipse are given, involving only the semi-axes of the ellipsoid, the componentes of the unit normal vector of the plane and the distance of the plane from the center of coordinates. This topic is relatively common to study, but, as indicated in [1], a closed form solution to the general problem is actually very difficult to derive. This is attemped here. As applications problems are treated, which were posed in the internet [1,2], pertaining to satellite orbits in space and to planning radio-therapy treatment of eyes.
1. Introduction
Let an ellipsoid be given with the three positive semiaxes a, b, c
(1)
and a plane with the unit normal vector
which contains an interior point of the ellipsoid. A plane spanned by vectors, and containing the point is described in parametric form by
(2)
Inserting the components of into the equation of the ellipsoid (1) leads to the line of intersection as a quadratic form in the variables and. Let the scalar product in for two vectors and be denoted by
With the diagonal matrix
the line of intersection has the form:
(3)
As is an interior point of the ellipsoid the righthand side of Equation (2) is positive. The matrix in Equation (3) is a Gram matrix. If the vectors and are linearly independent, this is equivalent with the linear independence of the vectors and, the matrix in (3) is positive definite and the line of intersection is an ellipse. In [3] a generalization from three to p-dimensional space is discussed.
Let and be unit vectors orthogonal to the unit normal vector of the plane
(4)
(5)
and orthogonal to eachother
(6)
Furthermore vectors and may be chosen such that
(7)
holds. This will be shown in the next section. Condition (7) ensures that the matrix in (3) has diagonal form. Then the line of intersection reduces to an ellipse in translational form
(8)
with the center
(9)
and the semi-axes
(10)
where
(11)
In order to show that the semi-axes (10) are independent of the choice of this vector may be decomposed orthogonally with respect to:
(12)
where is the distance of plane (2) from the origin. Substituting into (11) one obtains employing (4), (5), (6) and (7)
(13)
The following rules of computation for the cross product in ([4, p.147]) will be applied later on repeatedly. For vectors of the identity of Lagrange holds
(14)
and the Grassmann expansion theorem for the double cross product
(15)
2. Construction of Vectors r and s
Let be a unit vector orthogonal to the unit normal vector of the plane, so that Equations (4) hold. A suitable vector is obtained as a cross product
(16)
Then Equations (5) and (6) are fulfilled: is a unit vector, as can be shown by the identity of Lagrange (14), utilising, and:
Furthermore one obtains according to the rules applying to the spar product:
In case Equation (7) is not fulfilled for the initially chosen vectors r and s, i.e., the following transformation may be performed with
The transformed vectors and satisfy the following conditions:, and, which imply conditions (4)-(6). The expression
becomes zero, when choosing such that
holds. This can be reformulated, in case
to
If
holds, can be chosen, leading to
.
Corollary 1: For the unit vectors and orthogonal to each other and the following statement holds:
(17)
Statement (17) follows by substituting the definition of and utilising, and. For one obtains for instance:
Theorem 1: Let be the unit normal vector of the plane and let vectors and satisfy, , and condition (7). Putting
(18)
and are solutions of the following quadratic equation:
(19)
Proof: Utilising (17) one obtains:
Applying diagonality condition (7) and the identity of Lagrange (14) leads to:
(20)
For the cross product one obtains:
(21)
with the diagonal matrix
(22)
According to Grassmann’s expansion theorem for the double cross product (15)
(23)
follows, since and. Applying (20), (21), (23) one obtains:
(24)
A quadratic equation equivalent to (19) is considered in [5].
Corollary 2: Under the assumptions of Theorem 1 the following three equations are valid:
The first of the three equations was verified in the proof of Theorem 1. The second and the third equation follow analogously.
4. A Formular for d
Theorem 2: Under the assumptions of Theorem 1 the expression for d in (13) is given by:
(25)
where is taken from (12).
Proof: The verification of (25) consists of three steps.
Step 1: Applying the identity of Lagrange (14) the following statements hold:
(26)
With Corollary 2 and the diagonal matrix
(27)
one obtains:
(28)
and it follows by substituting (28) into (26)
(29)
Introducing expressions
(30)
one obtains from (29) using (18) and (30)
(31)
Combining both Equations (31) leads to
(32)
Step 2: Analogously to the verification of (24) the application of the identity of Lagrange (14) yields:
With the diagonal matrix for the cross product holds:
Therefore one obtains
or
(33)
In contrast to the verification of (24), where diagonality condition (7) holds, the analogous expression in (33) need not be zero.
Step 3: Applying the identity of Lagrange (14) again leads to
Substituting the involved cross products according to Corollary 2 and considering diagonality condition (7) one obtains
or
(34)
Squaring both sides of (34) and substituting the expressions from (31) leads to:
Substitution of (33) results in equation
or
(35)
Substitution of (35) in (32) leads to:
(36)
Because of (24)
(37)
holds and with (13) one finally obtains relation (25)
Corollary 3: Under the assumptions of Theorem 1 the area of the ellipse obtained by the intersection of the ellipsoid (1) and a plane with unit normal vector and distance from the origin is given by:
This is proven by the formula for the area of an ellipse:
and by applying (25) and (37). The area of intersection becomes zero in case holds; this corresponds to the limiting case, where the cutting plane becomes a tangent plane. This result has been applied in [6].
5. The Center of the Ellipse
Substituting according to (12) in formulars (9) for the coordinates of the center of the ellipse in the plane spanned by and one obtains:
and
(38)
The center of the ellipse in is given by:
(39)
Theorem 3: Let the assumptions of Theorem 1 be fulfilled. For the center of the ellipse of intersection in holds:
(40)
Proof: With diagonal matrices from (27) and
from (22) utilising and (37) one obtains a representation of equivalent to (40):
(41)
It is sufficient to show that for the difference
holds. Thus the coefficients in the expansion of in with respect to the orthonormal basis are zero, i.e., is the zero vector.
Applying representation (39) one obtains:
The last expression is zero according to (24). Furthermore one obtains:
and by interchanging the roles of and:
Both previous expressions are zero; this follows by applying diagonality condition (7), the identity of Lagrange (14) and Corollary 2:
Interchanging the roles of and leads to:
Corollary 4: The apexes of the ellipse of intersection are given by
where and are denoting the semi-axes according to (10).
Clearly and are points of the plane cutting the ellipsoid. In order to show that they are belonging to the ellipse of intersection, it has to be verified that they are situated on the ellipsoid, i.e. the following equalities hold:
This can be shown using in the form (39) and employing condition (7).
Corollary 5: holds if and only if is an interior point of the ellipsoid (1), because of
In the case of, i.e., for the semi-axes (10) of the ellipse of intersection follows. The center (40) of the ellipse of intersection becomes a tangent contact point
of ellipsoid (1) and a tangent plane with normal vector, since holds.
Corollary 6: Describing the ellipse of intersection (8) in parametric form
with, where and are denoting its semi-axes according to (10), leads to a representation as a curve in three dimensional space as indicated in [7]
This result may be derived substituting the parameters and from the parametric form of the ellipse into Equation (2) of the plane:
or
where is equal to the center of the ellipse as in (39).
6. Applications
As indicated in [2], viewing a section through an ellipsoidal eye from a viewpoint normal to the intersection plane and displaying the intersection on that plane along with a projection of the eye structures and isodose lines, radio-therapy treatment of the eye can be planned. For this purpose the line of intersection of ellipsoid (1) and the plane, having the normal vector and containing the point, situated in the interior of (1), is determined. The plane has the form:
with the unit normal vector:
(42)
The distance of the plane from the origin is given by:
(43)
According to (25) can be written as:
(44)
From (11) it is obvious that holds, as for as an interior point of the ellipsoid is true. Substituting (18) into (10) the semi-axes of the ellipse, the line of intersection of ellipsoid and plane, are given by
(45)
where are solutions of Equation (19):
(46)
With Theorem 3 one obtains by substituting and from (42) and (43) the formular for the center of the ellipse given by:
(47)
Instead of calculating and as solutions of (46) they may be obtained alternatively using the procedure described in 2. Starting with an arbitrary unit vector orthogonal to the unit normal vector given in (42), e.g.
calculating to be orthogonal to both according to and, in case, perform a rotation with angle as described in 2, yielding new vectors and, which are plugged into (18).
A Mathematica program containing both ways of computation of and may be obtained from the author upon request.
In the first special case of a plane containing the origin (see e.g. [1]), i.e. is the zero vector, it follows by (43), (44) and (47) that, and is the zero vector also. Furthermore the semi-axes of the ellipse in (45) reduce to
and from (9) holds. Thus Equation (8) of the line of intersection reduces to
A second special case, where holds, was treated in [2]. Then the above formulas (43), (44) and (47) reduce to:
and
Because of in (12) holds and (38) reduces to
where and are solutions of the quadratic Equation (46) and vectors and have to be determined as described above according to the procedure shown in 2. Thus Equation (8) of the line of intersection turns into:
7. Conclusion
The intention of this paper was, to give an elementary closed form solution to the general problem of the intersection of an ellipsoid and a plane.
REFERENCES
1. The Math Forum, “Intersection of Ellipsoid and Plane,” 2007. http://mathforum.org/library/drmath/view/71275.html
2. The Math Forum, “Ellipsoid and Plane Intersection Equation,” 2000. http://mathforum.org/library/drmath/view/51781.html
3. The Math Forum, “Intersection of Hyperplane and an Ellipsoid,” 2007. http://mathforum.org/library/drmath/view/72315.html
4. A. Korn and M. Korn, “Mathematical Handbook for Scientists and Engineers,” Mc Graw-Hill Book Company, Inc., New York, Toronto, London, 1961.
5. C. C. Ferguson, “Intersections of Ellipsoids and Planes of Arbitrary Orientation and Position,” Mathematical Geology, Vol. 11, No. 3, 1979, pp. 329-336. doi:10.1007/BF01034997
6. M. P. Verma and J. C. Upadhyaya, “On the Electron-Ion Interaction in Hexagonal and Tetragonal Metals,” Journal of Physics F: Metal Physics, Vol. 1, No. 5, 1971, pp. 618- 620. doi:10.1088/0305-4608/1/5/315
7. The Math Forum, “Equation of an Ellipse in 3-Space,” 2003. http://mathforum.org/library/drmath/view/63373.html
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://file.scirp.org/Html/9-7401027_24506.htm", "fetch_time": 1713342036000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00712.warc.gz", "warc_record_offset": 226751311, "warc_record_length": 17537, "token_count": 3002, "char_count": 12214, "score": 3.65625, "int_score": 4, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.9304520487785339, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00043-of-00064.parquet"}
|
# Computing Trendline Values in Excel Charts
SFE , Business Analytics at Sanofi;Merck BioPharma ;Boehringer Ingelheim;Cipla and Emcure
Mar. 21, 2011
1 of 20
### Computing Trendline Values in Excel Charts
• 1. www.LCDing.Com Computing trend line values using excel formulas and functions Trendline values in excel charts Warning: This is for those who wish to explore trend lines in excel. For others, this presentation may be intimidating . Kindly exercise caution!
• 2. Trendlines in charts A trend line is a straight line connecting multiple points on a chart. A trend is a movement in a particular direction The magnitude of the slope of a trend line, or steepness, indicates the strength of the trend. Trend lines can be used to forecast future !
• 3. Six types of trend lines 1. Linear 2. Logarithmic 3. Polynomial 4. Power 5. Exponential 6. Moving Average How to Create a Trendline? – Have a look at my earlier post in LCDing Slide share presentation
• 4. Rule of thumb : Use type of Trend line 1. Linear trend line : use it if data values are increasing or decreasing at a steady rate. 2. Logarithmic trend line : Useful when the rate of change in the data increases or decreases quickly and then levels out. 3. Polynomial trend line : Used when there are data fluctuations like the sales following seasonal trends
• 5. Rule of thumb : use type of Trend line 4. Power trend line : Use with data that has values that increase at specific rate at regular intervals. 5. Exponential trend line : Use when data values increase or decrease rates that are constantly increasing. 6. Moving average trend line : Use it when uneven fluctuations are in data values
• 6. Trend line values using excel formulae You can use the formulas given in the next few slides. Alternatively, you can download and use the excel attached with all formulas. To download excel file, click here
• 7. Linear Trendline equation Equation: y = m * x + b Where, m =SLOPE(y,x) b =INTERCEPT(y,x) RSQ =RSQ(y,x)
• 8. Logarithmic Trendline equation Equation: y = (c * LN(x)) + b Where, c =INDEX(LINEST(y,LN(x)),1) b =INDEX(LINEST(y,LN(x)),1,2) RSQ =RSQ(y,LN(x))
• 9. Power Trendline Equation Equation: y=c*x^b Where, c =EXP(INDEX(LINEST(LN(y),LN(x),,),1,2)) b =INDEX(LINEST(LN(y),LN(x),,),1) RSQ =INDEX(LINEST(LN(y),LN(x),TRUE,TRUE),3,1)
• 10. Exponential Trendline equation Equation: y = c *e ^(b * x) Where, c =EXP(INDEX(LINEST(LN(y),x),1,2)) b =INDEX(LINEST(LN(y),x),1) RSQ =RSQ(LN(y),x)
• 11. Polynomial Trendline(2nd order) Equation Equation: y = (c2 * x ^2) + (c1 * x^1) + b Where, c2 =INDEX(LINEST(y,x^{1,2}),1) c1: =INDEX(LINEST(y,x^{1,2}),1,2) b = INDEX(LINEST(y,x^{1,2}),1,3) RSQ =INDEX(LINEST(y,x^{1,2},TRUE,TRUE),3,1)
• 12. Polynomial Trendline(3rd order) Equation Equation: y=(c3*x^3) + (c2*x^2) + (c1*x^1) +b Where, c3 =INDEX(LINEST(y,x^{1,2,3},1) c2 =INDEX(LINEST(y,x^{1,2}),1) c1 =INDEX(LINEST(y,x^{1,2}),1,2) b = INDEX(LINEST(y,x^{1,2}),1,3) RSQ =INDEX(LINEST(y,x^{1,2,3},TRUE,TRUE),3,1)
• 13. Polynomial Trendline(4th order) Equation Equation: y= (c4*x^4)+(c3*x^3) + (c2*x^2) + (c1*x^1) +b Where, c4 = INDEX(LINEST(y,x^{1,2,3,4},1) c3 =INDEX(LINEST(y,x^{1,2,3},1) c2 =INDEX(LINEST(y,x^{1,2}),1) c1 =INDEX(LINEST(y,x^{1,2}),1,2) b = INDEX(LINEST(y,x^{1,2}),1,3) RSQ =INDEX(LINEST(y,x^{1,2,3,4},TRUE,TRUE),3,1)
• 14. Polynomial Trendline(5th order) Equation Equation: y= (c5*x^5)+(c4*x^4)+(c3*x^3) + (c2*x^2) + (c1*x^1) +b Where, c5 = INDEX(LINEST(y,x^{1,2,3,4,5},1) c4 = INDEX(LINEST(y,x^{1,2,3,4},1) c3 =INDEX(LINEST(y,x^{1,2,3},1) c2 =INDEX(LINEST(y,x^{1,2}),1) c1 =INDEX(LINEST(y,x^{1,2}),1,2) b = INDEX(LINEST(y,x^{1,2}),1,3) RSQ =INDEX(LINEST(y,x^{1,2,3,4,5},TRUE,TRUE),3,1)
• 15. Polynomial Trendline(6th order) Equation Equation: y=(c6*x^6)+(c5*x^5)+(c4*x^4)+(c3*x^3) + (c2*x^2) + (c1*x^1)+b Where, c6 = INDEX(LINEST(y,x^{1,2,3,4,5,6},1) c5 = INDEX(LINEST(y,x^{1,2,3,4,5},1) c4 = INDEX(LINEST(y,x^{1,2,3,4},1) c3 =INDEX(LINEST(y,x^{1,2,3},1) c2 =INDEX(LINEST(y,x^{1,2}),1) c1 =INDEX(LINEST(y,x^{1,2}),1,2) b = INDEX(LINEST(y,x^{1,2}),1,3) RSQ =INDEX(LINEST(y,x^{1,2,3,4,5,6},TRUE,TRUE),3,1)
• 16. www.LCDing.Com Basics of Excel functions used
• 17. Few Basics – Excel functions used - 1 X= Known X's (e.g.: Months on X Axis) Y= Known Y's (e.g.: Sales on Y Axis) Details of Excel functions are taken from Excel Help LN LN(number) Returns the natural logarithm of a number. Natural logarithms are based on the constant e (2.71828182845904). Liniest LINEST(known_y's,known_x's,const,stats) Calculates the statistics for a line by using the "least squares" method to calculate a straight line that best fits your data, and returns an array that describes the line. Because this function returns an array of values, it must be entered as an array formula. Exp EXP(number) Returns e raised to the power of number. The constant e equals 2.71828182845904, the base of the natural logarithm. Index INDEX(array,row_num,column_num) INDEX(reference,row_num,column_num,area_num) Returns a value or the reference to a value from within a table or range. There are two forms of the INDEX() function: array and reference. The array form always returns a value or an array of values; the reference form always returns a reference.
• 18. Few Basics – Excel functions used - 2 RSQ RSQ(known_y's,known_x's) Returns the square of the Pearson product moment correlation coefficient through data points in known_y's and known_x's. For more information, see PEARSON. The r-squared value can be interpreted as the proportion of the variance in y attributable to the variance in x. Slope SLOPE(known_y's,known_x's) Returns the slope of the linear regression line through data points in known_y's and known_x's. The slope is the vertical distance divided by the horizontal distance between any two points on the line, which is the rate of change along the regression line. Intercept INTERCEPT(known_y's,known_x's) Calculates the point at which a line will intersect the y-axis by using existing x-values and y-values. Logest LOGEST(known_y's,known_x's,const,stats) In regression analysis, calculates an exponential curve that fits your data and returns an array of values that describes the curve. Because this function returns an array of values, it must be entered as an array formula.
• 19. Incase you need more details… I have learned these formulas form the following web sites and authors. Click the links below Trendline formulas Bernard Liengme’s website Tushar Mehta’s article Ozgrid Forum Trends-and-forecast-sales-with-charts Choosing-the-best-trendline-for-your-data
• 20. www.LCDing.Com
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.slideshare.net/ksatyamahesh/computing-trendline-values-in-excel", "fetch_time": 1695830304000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00394.warc.gz", "warc_record_offset": 1102168961, "warc_record_length": 42817, "token_count": 2090, "char_count": 6567, "score": 3.828125, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.7202305197715759, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00011-of-00064.parquet"}
|
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A020137 Pseudoprimes to base 8. 8
9, 21, 45, 63, 65, 105, 117, 133, 153, 231, 273, 341, 481, 511, 561, 585, 645, 651, 861, 949, 1001, 1105, 1281, 1365, 1387, 1417, 1541, 1649, 1661, 1729, 1785, 1905, 2047, 2169, 2465, 2501, 2701, 2821, 3145, 3171, 3201, 3277, 3605, 3641, 4005, 4033, 4097 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS This sequence is a subsequence of the sequence A122785. In fact the terms are odd composite terms of A122785. Theorem: If both numbers q & 2q-1 are primes (q is in the sequence A005382) and n=q*(2q-1) then 8^(n-1)==1 (mod n) (n is in the sequence) iff q is of the form 12k+1. 2701,18721,49141,104653,226801,665281,721801,... is the related subsequence. This subsequence is also a subsequence of the sequence A122785. - Farideh Firoozbakht, Sep 15 2006 Composite numbers n such that 8^(n-1) == 1 (mod n). - Michel Lagneau, Feb 18 2012 LINKS Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..613 from R. J. Mathar, terms 614..1000 from T. D. Noe) MATHEMATICA Select[Range[4100], ! PrimeQ[ # ] && PowerMod[8, (# - 1), # ] == 1 &] (* Farideh Firoozbakht, Sep 15 2006 *) CROSSREFS Cf. A001567 (pseudoprimes to base 2), A005382, A122783, A122785. Sequence in context: A110680 A163205 A154862 * A231570 A020190 A225507 Adjacent sequences: A020134 A020135 A020136 * A020138 A020139 A020140 KEYWORD nonn AUTHOR STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified April 12 22:36 EDT 2021. Contains 342933 sequences. (Running on oeis4.)
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://oeis.org/A020137", "fetch_time": 1618287388000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00128.warc.gz", "warc_record_offset": 70982035, "warc_record_length": 3826, "token_count": 701, "char_count": 1913, "score": 3.59375, "int_score": 4, "crawl": "CC-MAIN-2021-17", "snapshot_type": "latest", "language": "en", "language_score": 0.6761623620986938, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00057-of-00064.parquet"}
|
# Binary Shifts
Dive into the fascinating world of Computer Science with an in-depth understanding of Binary Shifts. Explore the significance of this fundamental concept, the different types and techniques, and get to grips with practical examples. This comprehensive guide breaks down the intricate operations of Binary Left and Right Shifts, with an illuminating comparison of Binary Shifters. Moreover, you'll receive insightful tips to master Binary Shift Techniques in the realms of Computer Organisation and Architecture, ensuring you navigate potential pitfalls with ease. Let's unlock the power of Binary Shifts, a must-know for every aspiring computer scientist.
#### Create learning materials about Binary Shifts with our free learning app!
• Flashcards, notes, mock-exams and more
• Everything you need to ace your exams
## Understanding Binary Shifts in Computer Science
In the realm of computer science, binary shifts stand as fundamental operations for data manipulation. They are cornerstones for instruction sets in lower-level coding and data processing situations. Diving deeper into these operations, you'll realize their importance in not just theoretical understandings but also in practical applications.
### Defining Binary Shifts: What are They?
In the heart of binary digits or 'bits', a binary shift moves each digit in a string of binary to the left or the right by a certain position. This operation could imply one of two main types, a left binary shift or a right binary shift.
A left binary shift: This type of shift operation occurs when all bits in a binary number are systematically shifted to the left, resulting in a multiplication by 2.
A right binary shift: This shift is the opposite of the left shift. It involves the displacement of all bits to the right, effectively dividing the number by 2.
For instance, a left shift of a bit string '1011' by one position would result in '0110'.
### Why Binary Shifts are Important in Computer Science
Binary Shifts are paramount in Computer Science because of several reasons:
• They are fundamental for data processing in computers and electronic systems.
• They allow quick multiplication or division of binary numbers by powers of two.
• They help in efficient and fast operations on bits.
• Useful in cryptography and secure data transmission.
#### Binary Shift Technique: How it Works
Binary shift operations function in a simple, efficient, yet effective manner. A left shift by 'n' positions would be analogous to multiplication by $$2^n$$, while a right shift by 'n' would be comparable to division by $$2^n$$. However, you need to note that these simplifications hold true for unsigned integers only.
Code
Binary Shift Left Operation ('<<'): 1011 << 1 => 0110
Binary Shift Right Operation ('>>'):1101 >> 1 => 0110
In most programming languages, such as C++, Java, and Python, shift operators are expressed by '<<' for left shift and '>>' for right shift.
#### Using Binary Shifts in Digital Logic
Binary shifts find indispensable use in digital logic design and electronics, for instance, in shift registers and processors. They serve critical roles for performing arithmetic operations, data serialization, and even manipulation of individual bits in a byte. Consider a scenario where a shift register is storing a binary number '1001'. Applying a shift right operation would output '0100'. Implementing this with a digital logic circuit would involve using flip-flops and logic gates.
Example: Shift Operation -> Shift Right Number Before Shift: 1001 Number After Shift: 0100
## Different Types of Binary Shifts
The world of binary shifts is mainly categorised into two types, namely, Binary Left Shifts and Binary Right shifts. Developing an understanding of these types of shifts is essential to mastering data manipulation in computer science. Both these types carry their unique operations and have significant effects on binary numbers.
### Inside Binary Left Shift Operations
Delving into binary left shift operations, this operation involves shifting all the bits of a number systematically towards the left by a certain number of positions. In more straightforward terms, a binary left shift operation corresponds to multiplying the original number by $$2^n$$, where $$n$$ signifies the number of positions shifted. When it comes to working with binary left shifts, it's vital to understand the displacement patterns of the bits. If you opt for a binary left shift of number 'a' by 'b' bits, the bits of 'a' are moved 'b' places towards the left. Notably, 'b' new bits filled with zeros are added to the right, while the 'b' most significant bits on the left are discarded.
Code
Binary Shift Left Operation ('<<'): a << b => a * $$2^b$$
Example: 0010 << 2 => 1000
Binary Left Shift Operation: Operation where bits in a binary number are shifted to the left by a certain number of positions, effectively multiplying the number by $$2^n$$, where $$n$$ is the number of positions shifted. The leftmost 'n' bits are discarded.
### Working with Binary Right Shifts
Switching gears to binary right shift operations, it's accurate to think of this operation as the opposite of its left shift counterpart. Binary right shift operation involves systematic displacement of the bits towards the right by specific positions. It's like dividing the number by $$2^n$$, where $$n$$ is the number of shift positions. In this case, the 'n' least significant bits are discarded, and 'n' new bits filled with zeroes are inserted at the start of the number. Critical for efficient division and bitwise operations in computer systems, the binary right shift operation proves useful in various programming and networking scenarios.
Code
Binary Shift Right Operation ('>>'): a >> b => a / $$2^b$$
Example: 1100 >> 2 => 0011
Binary Right Shift Operation: Operation which involves shifting all the bits of a number to the right by a certain number of positions, effectively dividing the number by $$2^n$$, where $$n$$ is the number of shift positions. The rightmost 'n' bits are discarded.
#### Understanding Binary Shifter: A Comparative Analysis
Grasping the concept of a binary shifter is a game-changer in getting a firm hold of binary shift operations. A binary shifter essentially executes the binary shifts, be it left or right. It is hardware in a digital system and an integral part of ALUs (Arithmetic Logic Units). Now, putting the spotlight on a comparative analysis between binary left shifts and right shifts, they might seem like two sides of the same coin. However, the results they produce and the operations they conduct set them apart.
Binary Left Shift Multiply the number by $$2^n$$ Binary Right Shift Divide the number by $$2^n$$
While a left shift multiplies the original number by $$2^n$$, a right shift divides it by $$2^n$$. Hence, left shifts are great for rapid multiplication, while right shifts are ideal for swift division of binary numbers. Understanding the difference and using them correctly in your code can significantly boost efficiency and performance. In summary, binary shifts – left or right – are fundamental to efficient data processing and manipulation in computer systems, and furthering your knowledge about these operations is a wise step towards mastering computer science.
## Practical Examples of Binary Shifts
In Computer Science, it's one thing to understand the theory behind binary shifts, but putting such knowledge to practice is crucial for grasping the full implications of these operations. Binary shifts are typically applied in digital logic, binary arithmetic, or even in writing efficient, short programs for mathematical calculations. This section provides detailed practical examples, which aim to solidify your understanding of how binary shifts work in actual application.
### A Closer Look at Binary Shift Operations in Practice
In everyday computing, binary shifts are incredibly important for enhancing performance and optimising code in aspects like reducing time complexity or memory usage. Binary Left Shift in Practice: Practically, you can use binary left shift operations whenever you need to double a number or multiply it by a power of two quickly. For instance, if you want to multiply a number, say 13, by 4, you can achieve this without the need for long multiplication. Instead, you can express 13 in binary as $$1101_{2}$$ and shift it to the left by 2 (since $$2^{2}=4$$). This process gives $$110100_{2}$$, which is the binary equivalent of 52 - the result of 13 multiply by 4. This is a clear demonstration of binary shift usage in unsigned arithmetic.
Code
Binary Shift Left Operation: 13 << 2 => 52
Example: $$1101_{2}$$ << 2 => $$110100_{2}$$ (52 in decimal notation)
Binary Right Shift in Practice: Considering binary right shift operations, they are extremely beneficial when you aim to halve a number or divide it by a power of two swiftly. For example, to halve the number 20, we express it in binary as $$10100_{2}$$, and apply a binary right shift of 1, deriving $$1010_{2}$$ - the binary form of 10, the result of 20 divided by 2.
Code
Binary Shift Right Operation: 20 >> 1 => 10
Example: $$10100_{2}$$ >> 1 => $$1010_{2}$$ (10 in decimal notation)
Pay attention to the way the binary left shift and right shift operations help in swift, efficient calculations. It's clear from these practical applications that binary shift remains pivotal for operations on binary numbers.
#### Binary Shift Example: A Hands-on Approach
Let's dive into an actual problem-solving situation involving binary shifts. Consider that you are building an effective solution for an algorithmic problem in a programming language. The problem is high on arithmetic operations involving multiplication and division by powers of two. Binary shifts present an efficient way to handle these operations. As demonstrated previously, instead of doing multiplication or division in a conventional way, using binary shifts can optimise your solution. Let's go over an illustrative example: Suppose you have an 8-bit binary number '10010011'. Let's perform a left shift and right shift operation on this binary number: - Left Shift, say by 2 bits, would result in: '01001100' - Right Shift, say by 3 bits, would result in: '00010010' Notice how the bits are moved towards the left or right. Remember, in the case of left shifts, the rightmost bits are filled with zeros, while in right shifts, the leftmost bits are replaced with zeros.
Code
8-bit Binary Number: 10010011
Left Shift by 2: 10010011 << 2 => 01001100
Right Shift by 3: 10010011 >> 3 => 00010010
These examples elucidate how binary shifts can affect a binary number, and how they can be used efficiently in problem-solving and code optimisation.
### Binary Shift in Digital Logic: An Illustrative Explanation
Binary shifts prove critical in digital logic and hardware design; for instance, in electronic circuits like shift registers and processors. Think about a shift register, a sequential digital circuit that primarily can shift the binary information present within it towards the left or the right. Say you have a 4-bit shift register, holding a binary number: '1011'. Now, if a shift right operation is to be implemented, it leads to: '0101'. Through these operations, shift registers render beneficial services like memory storage, data transfer, and signal delay.
Example: Shift Register Operation Initial Data: 1011 Shift Right Operation => New Data: 0101
In summary, understanding binary shifts is not just pivotal to computer science education but also proves instrumental in practical applications, offering necessary tools for data manipulation, problem-solving, code optimisation, and digital logic design.
## How to Master Binary Shift Techniques in Computer Organisation and Architecture
Making sense of Binary Shift Techniques in the realm of computer organisation and architecture does look challenging, but with a well-defined learning pathway and plenty of practice, you can master them.
### Essential Learning Steps for the Binary Shift Technique
Learning binary shifts necessitates a systematic approach. Here’s how you can achieve competence in this technique: Understanding Binary Numeral System: Binary number system forms the very foundation of binary shift operations. It involves numbers made up of 0's and 1's. Therefore, the first step in mastering binary shifts is having a solid understanding of the binary number system.
Code
Example: A binary number: 1011
Grasping Bit Manipulation: As binary shifts involve the manipulation of bits, learning the concept of bit manipulation is crucial. It's the process of modifying binary data at the level of individual bits. Learning Binary Left Shift Operation: The next step is to comprehend the binary left shift operation. It involves shifting the bits of binary numbers towards the left, which is equivalent to multiplying the number by $$2^n$$, where $$n$$ is the number of shift positions. Understanding Binary Right Shift Operation: Similar to left shifts, the right shift operation displaces bits towards the right, essentially dividing the number by $$2^n$$.
Code
Left Shift Operation: 2 (10 in binary) << 2 => 8 (1000 in binary)
Right Shift Operation: 8 (1000 in binary) >> 1 => 4 (100 in binary)
Practice: Finally, apply your knowledge and understanding in solving problems. Practice applications of binary shifts in digital logic and binary arithmetic.
### Common Pitfalls and How to Avoid Them When Using Binary Shift Techniques
Using Binary Shift Techniques often invites a few common mistakes. Here's how you can avoid them: Incorrect Number of Shifts: Often, beginners make the error of incorrect number of shifts - either shifting less or more than required. Remember, the number of shifts equates to the power of two with which you're multiplying or dividing.Binary operations are a great way to perform these operations on large numbers swiftly.
Example: Incorrect Implementation: 10 (1010 in binary) << 3 => 80 (1010000 in binary) Correct Implementation: 10 (1010 in binary) << 1 => 20 (10100 in binary)
Wrong Direction of Shift: The direction of shift operation plays a critical role in deriving the result. Often errors occur by applying a right shift instead of left or vice versa. Always remember the basic rule: left shifts multiply whereas right shifts divide. Forgetting Bit Disposal Rule: Remember, in left shifts, the left-most 'n' bits are discarded, whereas in right shifts, it's the right-most 'n' bits. Forgetting this rule can lead to incorrect results. Double-checking your solution can help eliminate such slip-ups.
#### Tuning Your Binary Shift Skills: Pro Tips and Tricks
To enhance understanding and master the binary shift operations, here are some pro tips and tricks: Practise Logic Building: Improving your logical thinking can significantly enhance your capabilities to solve problems involving binary shifts. Regularly practice problems requiring binary operations. Use of Online Tools: Utilise online binary calculators and converters to cross-check your calculations. They can serve as an effective means to understand how these operations work. Code Optimisation: Try writing efficient and short programs for mathematical calculations using binary shifts. This not only saves computational time but also helps you learn and implement binary shifts effectively. Deep Dive into Hardware: Understand how shift registers and processors use these operations. This can provide real-life context and enhance your understanding of the practical implications of binary shifts. Remain committed to practicing and applying the binary shift techniques in solving various problems. This will help you deepen your understanding, enabling you to overcome common pitfalls and ultimately master the binary shift techniques in computer organisation and architecture.
## Binary Shifts - Key takeaways
• Binary shifts are fundamental for data processing in computers and electronic systems, allowing for quick multiplication or division of binary numbers by powers of two.
• A left binary shift results in a multiplication by 2 while a right binary shift effectively divides the number by 2.
• In computer programming languages, such as C++, Java, and Python, shift operators are expressed by '<<' for left shift and '>>' for right shift.
• Binary shifts are integral in digital logic design, aiding in efficient operations on bits, data serialization, data transmission and manipulation of individual bits in a byte.
• The two main types of binary shifts are Binary Left Shifts and Binary Right shifts, each playing a unique role in data manipulation.
#### Flashcards in Binary Shifts 27
###### Learn with 27 Binary Shifts flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
What are the different types of binary shifts in computer science?
There are three types of binary shifts in computer science: the logical right shift, the arithmetic right shift, and the left shift.
How can binary shifts be utilised in computer programming?
Binary shifts can be utilised in computer programming for efficient arithmetic calculations like multiplication or division by powers of 2. They are also used for binary data manipulation, implementing graphics in games, or encryption algorithms.
What is the significance of left and right binary shifts in data processing?
Left and right binary shifts serve as quick methods for multiplying or dividing by two (or powers of two). These shifts change the position of bits in binary data, affecting its numerical value, and are used in various data processing tasks such as encryption and compression.
What are the possible effects of performing binary shifts on a given data set?
Performing binary shifts on a data set can either multiply or divide the value by a power of two. Left shifts effectively multiply, and right shifts divide. However, it can also lead to data loss if the shift moves data off the end of the binary string.
Can binary shifts affect the overall performance of a computer programme?
Yes, binary shifts can affect the performance of a computer programme. They are often used for operations like multiplication and division, because shifting bits is faster than standard arithmetic operations, thus enhancing a programme's performance.
## Test your knowledge with multiple choice flashcards
What is the difference between Logical Shift and Arithmetic Shift in Binary Shift operations?
How does a Circular Shift operation work in Binary Shift operations?
What are the applications of binary left shift in computer architecture?
StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.
##### StudySmarter Editorial Team
Team Computer Science Teachers
• Checked by StudySmarter Editorial Team
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.studysmarter.co.uk/explanations/computer-science/computer-organisation-and-architecture/binary-shifts/", "fetch_time": 1721400767000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00246.warc.gz", "warc_record_offset": 853750691, "warc_record_length": 73988, "token_count": 3921, "char_count": 19578, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.868225634098053, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00001-of-00064.parquet"}
|
Curve Sketching
# Curve Sketching - 18.01 Calculus Jason Starr Fall 2005...
This preview shows pages 1–3. Sign up to view the full content.
18.01 Calculus Jason Starr Fall 2005 Lecture 9. September 29, 2005 Homework. Problem Set 2 all of Part I and Part II. Practice Problems. Course Reader: 2B-1, 2B-2, 2B-4, 2B-5. 1. Application of the Mean Value Theorem. A real-world application of the Mean Value Theorem is error analysis . A device accepts an input signal x and returns an output signal y . If the input signal is always in the range 1 / 2 x 1 / 2 and if the output signal is, 1 y = f ( x ) = 1 + x + x 2 + x 3 , what precision of the input signal x is required to get a precision of ± 10 3 for the output signal? If the ideal input signal is x = a , and if the precision is ± h , then the actual input signal is in the range a h x a + h . The precision of the output signal is f ( x ) f ( a ) . By the Mean Value | | Theorem, f ( x ) f ( a ) = f ( c ) , x a for some c between a and x . The derivative f ( x ) is, f ( x ) = (3 x 2 + 2 x + 1) . (1 + x + x 2 + x 3 ) 2 For 1 / 2 x 1 / 2, this is bounded by, 3(1 / 2) 2 + 2(1 / 2) + 1 | f ( x ) = 7 . 04 . | ≤ [1 + ( 1 / 2) + ( 1 / 2) 2 + ( 1 / 2) 3 ] 2 Thus the Mean Value Theorem gives, f ( x ) f ( a ) = f ( c 7 . 04 x a 7 . 04 h. | | | ) || x a | | | Therefore a precision for the input signal of, h = 10 3 / 7 . 04 10 4 guarantees a precision of 10 3 for the output signal. 2. First derivative test. A function f ( x ) is increasing , respectively decreasing , if f ( a ) is less than f ( b ), resp. greater than f ( b ), whenever a is less than b . In symbols, f is increasing, respectively decreasing, if f ( a ) < f ( b ) whenever a < b, resp. f ( a ) > f ( b ) whenever a < b.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
18.01 Calculus Jason Starr Fall 2005 If f ( a ) is less than or equal to f ( b ), resp. greater than or equal to f ( b ), whenever a is less than b , then f ( x ) is non-decreasing , resp. non-increasing . If f ( x ) is increasing, the
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 3
Curve Sketching - 18.01 Calculus Jason Starr Fall 2005...
This preview shows document pages 1 - 3. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.coursehero.com/file/196138/Curve-Sketching/", "fetch_time": 1513335830000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00555.warc.gz", "warc_record_offset": 717460287, "warc_record_length": 100779, "token_count": 768, "char_count": 2396, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2017-51", "snapshot_type": "latest", "language": "en", "language_score": 0.7394161224365234, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00051-of-00064.parquet"}
|
# Quantitative Aptitude [Advanced Level] For SSC CGL : 21st December
Q1.If p/x+q/y=m and q/x+p/y=n then what is x/y is equal to?
(a)(np+mq)/(mp+nq)
(b)(np+mq)/(mp-nq)
(c)(np – mq)/(mp – nq)
(d)(np–mq)/(mp–nq)
Q2.If the equation x² – px + q = 0 and x² + qx – p = 0 have a common root, then which one of the following is correct?
(a) p – q = 0
(b) p + q – 2 = 0
(c) p + q – 1 = 0
(d) p – q – 1 = 0
Q3.If √(3x^2–7x–30)–√(2x^2–7x–5)=x –5 has α & β as its roots, then the value of αβ is
(a)–15
(b) –5
(c) 0
(d) 5
Q4. If x=2^(1/3)+2^(–1/3), then the value of 2x^3–6x–5 is equal to
(a) 0
(b) 1
(c) 2
(d) 3
Q5. Three equal circles each of diameter d are drawn on a plane in such a way that each circle touches the other two circles. A big circle is drawn in such a manner that it touches each of the small circles internally. The area of the big circle is
(a) πd^2
(b) πd^2 (2 –√3)^2
(c) (πd^2 (√3+1)^2)/2
(d) (πd^2 (√3+2)^2)/12
Q6. In the figure given above, O is the centre of the circle. The line UTV is a tangent to the circle at T, ∠VTR = 52° and ∆PTR is an isosceles triangle such that TP = TR. What is ∠x + ∠y+∠z equal to ?
(a) 175°
(b) 208°
(c) 218°
(d) 250°
Q7.A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man’s eye when at a distance of 60 m from the bottom of tower. After 5 s, the angle of depression becomes 30°. What is the approximate speed of the boat assuming that it is running in still water?
(a) 31.5 km/hr
(b) 36.5 km/hr
(c) 38.5 km/hr
(d) 40.5 km/hr
Q8. In a building there are 24 cylindrical pillars. The radius of each pillar is 28 cm and height is 4 m. Find the total cost of painting the curved surface area of all pillars at the rate of Rs. 8 per square metre.(in Rs)
(a) 1100
(b) 1183
(c) 1232
(d) 1352
Q9. Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m3, find the number of hours it will take to fill the reservoir?
(a) 20
(b) 30
(c) 40
(d) 50
Q10. A round balloon of unit radius subtends an angle of 90° at the eye of an observer. What is the distance of the centre of the balloon from the eye of the observer?
(a) 1/√2
(b) √2
(c) 2
(d) 1/2
Click here for best SSC CGL mock tests, video course, live batches, books or eBooks
×
Thank You, Your details have been submitted we will get back to you.
Join India largest learning distination
What You Will get ?
• Daily Quizes
• Subject-Wise Quizes
• Current Affairs
• previous year question papers
• Doubt Solving session
OR
Join India largest learning distination
What You Will get ?
• Daily Quizes
• Subject-Wise Quizes
• Current Affairs
• previous year question papers
• Doubt Solving session
OR
Join India largest learning distination
What You Will get ?
• Daily Quizes
• Subject-Wise Quizes
• Current Affairs
• previous year question papers
• Doubt Solving session
Enter the email address associated with your account, and we'll email you an OTP to verify it's you.
Join India largest learning distination
What You Will get ?
• Daily Quizes
• Subject-Wise Quizes
• Current Affairs
• previous year question papers
• Doubt Solving session
## Enter OTP
Please enter the OTP sent to
/6
Did not recive OTP?
Resend in 60s
Join India largest learning distination
What You Will get ?
• Daily Quizes
• Subject-Wise Quizes
• Current Affairs
• previous year question papers
• Doubt Solving session
Join India largest learning distination
What You Will get ?
• Daily Quizes
• Subject-Wise Quizes
• Current Affairs
• previous year question papers
• Doubt Solving session
## Almost there
+91
Join India largest learning distination
What You Will get ?
• Daily Quizes
• Subject-Wise Quizes
• Current Affairs
• previous year question papers
• Doubt Solving session
## Enter OTP
Please enter the OTP sent to Edit Number
Did not recive OTP?
Resend 60
## By skipping this step you will not recieve any free content avalaible on adda247, also you will miss onto notification and job alerts
Are you sure you want to skip this step?
## By skipping this step you will not recieve any free content avalaible on adda247, also you will miss onto notification and job alerts
Are you sure you want to skip this step?
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.sscadda.com/quantitative-aptitude-advanced-level-for-ssc-cgl-21st-december", "fetch_time": 1621033747000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-21/segments/1620243991829.45/warc/CC-MAIN-20210514214157-20210515004157-00636.warc.gz", "warc_record_offset": 1060124070, "warc_record_length": 26433, "token_count": 1316, "char_count": 4267, "score": 3.953125, "int_score": 4, "crawl": "CC-MAIN-2021-21", "snapshot_type": "latest", "language": "en", "language_score": 0.8451693058013916, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"}
|
Homogeneous Polynomial
Get Homogeneous Polynomial essential facts below. View Videos or join the Homogeneous Polynomial discussion. Add Homogeneous Polynomial to your PopFlock.com topic list for future reference or share this resource on social media.
Homogeneous Polynomial
In mathematics, a homogeneous polynomial, sometimes called quantic in older texts, is a polynomial whose nonzero terms all have the same degree.[1] For example, ${\displaystyle x^{5}+2x^{3}y^{2}+9xy^{4}}$ is a homogeneous polynomial of degree 5, in two variables; the sum of the exponents in each term is always 5. The polynomial ${\displaystyle x^{3}+3x^{2}y+z^{7}}$ is not homogeneous, because the sum of exponents does not match from term to term. A polynomial is homogeneous if and only if it defines a homogeneous function. An algebraic form, or simply form, is a function defined by a homogeneous polynomial.[2] A binary form is a form in two variables. A form is also a function defined on a vector space, which may be expressed as a homogeneous function of the coordinates over any basis.
A polynomial of degree 0 is always homogeneous; it is simply an element of the field or ring of the coefficients, usually called a constant or a scalar. A form of degree 1 is a linear form.[3] A form of degree 2 is a quadratic form. In geometry, the Euclidean distance is the square root of a quadratic form.
Homogeneous polynomials are ubiquitous in mathematics and physics.[4] They play a fundamental role in algebraic geometry, as a projective algebraic variety is defined as the set of the common zeros of a set of homogeneous polynomials.
## Properties
A homogeneous polynomial defines a homogeneous function. This means that, if a multivariate polynomial P is homogeneous of degree d, then
${\displaystyle P(\lambda x_{1},\ldots ,\lambda x_{n})=\lambda ^{d}\,P(x_{1},\ldots ,x_{n})\,,}$
for every ${\displaystyle \lambda }$ in any field containing the coefficients of P. Conversely, if the above relation is true for infinitely many ${\displaystyle \lambda }$ then the polynomial is homogeneous of degree d.
In particular, if P is homogeneous then
${\displaystyle P(x_{1},\ldots ,x_{n})=0\quad \Rightarrow \quad P(\lambda x_{1},\ldots ,\lambda x_{n})=0,}$
for every ${\displaystyle \lambda .}$ This property is fundamental in the definition of a projective variety.
Any nonzero polynomial may be decomposed, in a unique way, as a sum of homogeneous polynomials of different degrees, which are called the homogeneous components of the polynomial.
Given a polynomial ring ${\displaystyle R=K[x_{1},\ldots ,x_{n}]}$ over a field (or, more generally, a ring) K, the homogeneous polynomials of degree d form a vector space (or a module), commonly denoted ${\displaystyle R_{d}.}$ The above unique decomposition means that ${\displaystyle R}$ is the direct sum of the ${\displaystyle R_{d}}$ (sum over all nonnegative integers).
The dimension of the vector space (or free module) ${\displaystyle R_{d}}$ is the number of different monomials of degree d in n variables (that is the maximal number of nonzero terms in a homogeneous polynomial of degree d in n variables). It is equal to the binomial coefficient
${\displaystyle {\binom {d+n-1}{n-1}}={\binom {d+n-1}{d}}={\frac {(d+n-1)!}{d!(n-1)!}}.}$
Homogeneous polynomial satisfy Euler's identity for homogeneous functions. That is, if P is a homogeneous polynomial of degree d in the indeterminates ${\displaystyle x_{1},\ldots ,x_{n},}$ one has, whichever is the commutative ring of the coefficients,
${\displaystyle dP=\sum _{i=1}^{n}x_{i}{\frac {\partial P}{\partial x_{i}}},}$
where ${\displaystyle \textstyle {\frac {\partial P}{\partial x_{i}}}}$ denotes the formal partial derivative of P with respect to ${\displaystyle x_{i}.}$
## Homogenization
A non-homogeneous polynomial P(x1,...,xn) can be homogenized by introducing an additional variable x0 and defining the homogeneous polynomial sometimes denoted hP:[5]
${\displaystyle {^{h}\!P}(x_{0},x_{1},\dots ,x_{n})=x_{0}^{d}P\left({\frac {x_{1}}{x_{0}}},\dots ,{\frac {x_{n}}{x_{0}}}\right),}$
where d is the degree of P. For example, if
${\displaystyle P=x_{3}^{3}+x_{1}x_{2}+7,}$
then
${\displaystyle ^{h}\!P=x_{3}^{3}+x_{0}x_{1}x_{2}+7x_{0}^{3}.}$
A homogenized polynomial can be dehomogenized by setting the additional variable x0 = 1. That is
${\displaystyle P(x_{1},\dots ,x_{n})={^{h}\!P}(1,x_{1},\dots ,x_{n}).}$
## References
1. ^ D. Cox, J. Little, D. O'Shea: Using Algebraic Geometry, 2nd ed., page 2. Springer-Verlag, 2005.
2. ^ However, as some authors do not make a clear distinction between a polynomial and its associated function, the terms homogeneous polynomial and form are sometimes considered as synonymous.
3. ^ Linear forms are defined only for finite-dimensional vector space, and have thus to be distinguished from linear functionals, which are defined for every vector space. "Linear functional" is rarely used for finite-dimensional vector spaces.
4. ^ Homogeneous polynomials in physics often appear as a consequence of dimensional analysis, where measured quantities must match in real-world problems.
5. ^ D. Cox, J. Little, D. O'Shea: Using Algebraic Geometry, 2nd ed., page 35. Springer-Verlag, 2005.
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 21, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.popflock.com/learn?s=Homogeneous_polynomial", "fetch_time": 1606866975000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-50/segments/1606141685797.79/warc/CC-MAIN-20201201231155-20201202021155-00493.warc.gz", "warc_record_offset": 141671562, "warc_record_length": 16730, "token_count": 1395, "char_count": 5244, "score": 4.1875, "int_score": 4, "crawl": "CC-MAIN-2020-50", "snapshot_type": "latest", "language": "en", "language_score": 0.8867213129997253, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00063-of-00064.parquet"}
|
## Search This Blog
### Professor Riddle
Professor Riddle - 4 july
A professor thinks of two consecutive numbers between 1 and 10.
'A' knows the 1st number and 'B' knows the second number
A: I do not know your number.
B: Neither do I know your number.
A: Now I know.
There are four solution for this.What are they ??
1. 2,3 and 8,9 are possible ansswers.
1. 2+5+7=141711
4+8+1=133344
6+2+4=121614
3+5+4=121917
2+6+8=??????
2. Wats al this shit about?????? Its either...
1,2 or 9,10 easy
2. 2,3
3,4
9,8
8,7
1. It cannot be (2,3) because if person A has number 2, person B cannot have number 1, so person A knows that B would have 3. If person A has number 3, and person B has number 3, person B would know.
3. 7,8
8,7
3,4
4,3
ist sol= 2-3
2nd Sol= 4-5
3rd Sol= 6-7
4th Sol= 8-9
1. 2+5+7=141711
4+8+1=133344
6+2+4=121614
3+5+4=121917
2+6+8=??????
5. plz exlain me the solution of this
6. If A or B had 1 or 10, then they would have solved it straight away. But neither did.
But when B discovered that A didn't know, he went from not knowing to knowing. So B must have had a number where A's answer was crucial.
If B had 2, then A could have 1 or 3 - and the crucial answer "I don't know your number, either" would have ruled out 1, leaving 3 as the other number (Solution: 2 and 3)
Now, if B had 3, then he would expect A to have either 2 or 4. But if A had 2, A could have guessed the answer already (because B had already said he didn't know, so could not have had 1). When B discovers that A also doesn't know, he can rule out 2 as the answer. (Solution: 3 and 4).
Exactly the same arguments work at the other end of the range, providing the other two solutions: (Solution: 9 and 8) and (Solution: 8 and 7)
1. 2+5+7=141711
4+8+1=133344
6+2+4=121614
3+5+4=121917
2+6+8=??????
2. Only two possible solutions: [3,4] and [7,8].
A does not know, thus he has a number [2,9] (at this stage).
B does not know either. This means he has a number from [3,8] and by extension A has a number [4,7]. If A had 2, and B had 1, B would have known A's number, but since they both don't know, B must either have the number 3 or 7 to open the possibility of A having 4 or 8.
Since A now knows, it excludes the possibility of [4,5] and [6,7], leaving [3,4] and [7,8].
3. The Answer is only A-2,B-3 and A-9,B-8. I could not understand your explanation.
a=2 b=3
a's answer cannot be 1 or 10 because otherwise he would know that b was 2. the fact that b doesn't know either means b isn't 1 or 10 either. As a is 2, b's only possible solutions are 1 and 3, and as we know b can't be 1, a would now know that b is 3.
a=9 b=8
Same as before, a is 9 and doesn't know whether b is 10 or 8. But as b doesn't know either, it can't be 10, therefore it is 8
a=8 b=7
Now this and the next one are a lot more difficult to solve; you need to see the crucial thing between a and b; a originally tells b that it doesn't know b's number, and with this knowledge b still is unable to work out a's number. However, a is able to work out b's number with the knowledge that b doesn't know its number. When a is 4, 5, 6, or 7 there is no possible way for a to know b's number without more information. However, with 8 we can use the solution we had for the first two answers to prove that b cannot be 9; if it were it would be able to work out what a was when a originally told b that it didn't know its number. So the fact that telling b that a didn't know b's number didn't make b know a's number meant that b must 7.
a=3 b=4
Same solution as the previous one, but using the fact that b would work out that a was 3 if b was 2, meaning it has to be 4.
5. But it said "between 1 and 10"...surely that does not include 1 and 10 as possibilities?
6. Gaping flaw in this logic: "Now, if B had 3, then he would expect A to have either 2 or 4. But if A had 2, A could have guessed the answer already (because B had already said he didn't know, so could not have had 1). When B discovers that A also doesn't know, he can rule out 2 as the answer. (Solution: 3 and 4)."
Description in the riddle only states that A knows the answer, not B. This whole (4,3) and (8,7) answer relies on the fact that B's number MUST be sequentially follow A's number which the logic to the valid answers (2,3) and (8,9) proves is not the case. Additionally the riddle never states this. The only four valid answers are (A,B) = (2,3),(3,2),(8,9),(9,8).
7. The logic is solid, Admin just swapped "A" and "B" in the explanation. This is a good one! Hard to wrap the head around
8. I understand 2 and 3, 3 and 4, 7 and 8, 8 and 9, but why can't we use (4 and 5), (5 and 6), (6 and 7)?
9. Solutions: A8, B7 : A9, B8 : A2, B3 : A3, B4
A9 B8 and A2, B3 DO work because…
If A 9 B8- A says I don’t know, B would know if he had 10, meaning he can only have 8, so A says he knows. The same applies for A2, B3
A8 B7 and A3, B4 DO work because…
If A 8, B7 - A says “I don’t know”. At this point, if B had 9 he would know what A is, but since B does not know after hearing A’s first statement, then A can deduce that B does not have 9 and therefore must have 7. The same applies for A 3, B 4
A8 B9 and A3, B2 DON’T work because…
If A 8 B9- when A says “I don’t know”, then B would know A had 8. The same applies for A3, B2.
Too many variables exist for any other answer.
10. "...numbers between 1 and 10." are: 2,3,4,5,6,7,8,9 . One and Ten should not be included. That's what my math teacher told me long time ago.
A: I do not know your number.
(Saying this A is neither 2 nor 9. If A is 2, the consecutive number is only 3 . If A is 9, the consecutive number is only 8.)
B: Neither do I know your number.
(Saying this, B is neither 2 nor 9. B is neither 3 nor 8. If B is 3, he will know A, since the consecutive number is only 4 (since A is not 2). If B is 8, the consecutive number is only 7 (since A is not 9).)
A: Now I know.
(From B statement, A realized that possibility of B are: 4,5,6,7 only.)
So if
A: 3, B: 4 , then A may say "Now I know"
A: 4, B: 5 , then A may say "Now I know"
A: 5, B: 4 or 6. A may not yet say: "Now I know"
A: 6, B: 5 or 7. A may not yet say: "Now I know"
A: 7, B: 6 then A may say "Now I know"
A: 8, B: 7 then A may say "Now I know"
Solutions:
A: 3 and B: 4
A: 4 and B: 5
A: 7 and B: 6
A: 8 and B: 7
1. very gud interpretation
11. The only possible answer is 2,3 coz ur informed that from the beginning A knew the first number and B knew the second number...since B knows the second number all she has to do is subtract 1from her number to get A's number...but since she admits she doesnt know then it means that her number is 2 because A cannot have 1 as his number coz the question says between 1 and 10...so since they are consecutive numbers...A's number must be 3..but if A's number is 3 then chronologically..his number is the second number..so unless the professor lied to them which is highly improbable...so the only possible solution is 3,4...but if this is the solution with respect to the alphabet..then how comes B didnt know A's number..since they are consecutive she could have simply subtracted 1 to know A's number since her number was allegedly the second ....therefore...any two digit combo between 1 and 10 could be the answer...all we know is that either the professor or B is a liar...the professor could have lied about who's number came first chronologically or B could have lied about not knowing A's number...either way one of them foiled the experiment for us.
12. A: I do not know your number.
B: Neither do I know your number.
A: Now I know.
Ans : The soln is b/w 1 & 10. So 1 & 10 is eliminated from our calculations. then If 'A' is Having 3 means. A dont know the value of B. Now B is saying 'Neither do I know ur number', means it realized the soln is only b/w 1 to 10. so the next value of 1 is 2 only. Now 'A' realized that B might have 2. So A=3, B=2. Vice versa for 8&9. A=8,B=9.
Possible Ans is : 2,3 & 8,9
13. between 1 to 10 means ,
Number 1 and 10 not included
so ur ans is incorrect.
14. between 1 to 10 means ,
Number 1 and 10 not included
so ur ans is incorrect.
15. odd number and even number
16. Isn't it 5 solutions?
3-4
4-5
5-6
6-7
7-8
17. Either the riddle is badly constructed and not explained clearly or its very BS riddle.
if all 3 of them know that professor thought consecutive numbers and A knows his number so where is the scope of a riddle in that? B has A+1.
18. From first statement of A
....DAT means A Choose numbers from 3 to 8.
From second statement by B....B numbers are 4567.
So if A is 3 den B is 4.
And if A is 4 den B is 5.
IF A is 5 we can't say which number is B.( but A says I know now)
Same with 6.
If A is 7 den B is 6.
If A is 8 den B is 7.
Ans:
34
45
76
87
19. From first statement of A
....DAT means A Choose numbers from 3 to 8.
From second statement by B....B numbers are 4567.
So if A is 3 den B is 4.
And if A is 4 den B is 5.
IF A is 5 we can't say which number is B.( but A says I know now)
Same with 6.
If A is 7 den B is 6.
If A is 8 den B is 7.
Ans:
34
45
76
87
20. if A is 2 then B is 3.
if A is 9 B then is 8.
Ans 10 is absolutely wrong .Had one number be 10 ,he should have known the other consecutive number as 9 which was not so.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.briddles.com/2011/07/professor-riddle.html", "fetch_time": 1503174812000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-34/segments/1502886105922.73/warc/CC-MAIN-20170819201404-20170819221404-00257.warc.gz", "warc_record_offset": 503254394, "warc_record_length": 31306, "token_count": 2982, "char_count": 9233, "score": 4.125, "int_score": 4, "crawl": "CC-MAIN-2017-34", "snapshot_type": "longest", "language": "en", "language_score": 0.9617621898651123, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00037-of-00064.parquet"}
|
# Why is the period of a pendulum on the Moon $\sqrt{6}$ times its period on the Earth? [closed]
I came across this equation: $$T_m= \sqrt{6}T_e$$.
Can anyone tell me how this equation is derived? This is how I tried to, but I got stuck after some time.
So the time period of a simple pendulum on the earth= $$2\pi\sqrt{l/g}$$
The time period of a pendulum on the moon is $$2\pi\sqrt{l/(g/6)}$$
Now how do I create an equation which shows the time period of a pendulum on the moon with respect to the time period of a pendulum on the earth.
And please be as detailed as possible!
• The first time period is $T_e$ and the second is $T_m$, so what problem are you having? – G. Smith Jul 10 at 20:20
• Just divide one expression by the other, keeping in mind that $\sqrt {ab} =\sqrt{a}\sqrt{b}$ – Agnius Vasiliauskas Jul 10 at 20:24
• The 6 is approximate, not exact. Writing $\sqrt 6$ wrongly suggests that it is exact. – G. Smith Jul 10 at 20:33
$$T_m = 2 \pi \sqrt{\frac{l}{\frac{g}{6}}} = 2 \pi \sqrt{\frac{6l}{g}} = 2 \pi \sqrt{6\frac{l}{g}} = \sqrt{6} \left(2 \pi \sqrt{\frac{l}{g}}\right) = \sqrt{6} T_e$$
\begin{align*} T_e&=2\pi\sqrt{\frac lg}\\ T_m&=2\pi\sqrt{\frac l{\frac g6}}=\sqrt6\cdot2\pi\sqrt{\frac lg}=\sqrt6\cdot T_e\\ \end{align*}
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://physics.stackexchange.com/questions/564943/why-is-the-period-of-a-pendulum-on-the-moon-sqrt6-times-its-period-on-the-e/564946", "fetch_time": 1604023444000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-45/segments/1603107906872.85/warc/CC-MAIN-20201030003928-20201030033928-00472.warc.gz", "warc_record_offset": 470217843, "warc_record_length": 28720, "token_count": 429, "char_count": 1255, "score": 4.46875, "int_score": 4, "crawl": "CC-MAIN-2020-45", "snapshot_type": "latest", "language": "en", "language_score": 0.8129599094390869, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00017-of-00064.parquet"}
|
Solution For Maximum Depth Of Binary Tree
Problem Statement:
Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Solution:
This problem can be solved easily using a recursive approach. We can use Depth First Search (DFS) to calculate the maximum depth of the binary tree. We will calculate the depth of the left subtree and the depth of the right subtree recursively. Then we will take the maximum of these two values and add one to it, as the root node is at depth 1. The resulting value will be the maximum depth of the binary tree.
Here is the algorithm for calculating the maximum depth of a binary tree using DFS:
1. If the given binary tree is empty, return 0.
2. Calculate the depth of the left subtree recursively by calling the maximumDepth function on the left child of the root node.
3. Calculate the depth of the right subtree recursively by calling the maximumDepth function on the right child of the root node.
4. Take the maximum of the two depths and add one to it to get the maximum depth of the binary tree.
5. Return the maximum depth.
Here is the Python code for the above algorithm:
“`
class TreeNode:
def init(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def maximumDepth(root: TreeNode) -> int:
if root is None:
return 0
left_depth = maximumDepth(root.left)
right_depth = maximumDepth(root.right)
return max(left_depth, right_depth) + 1
“`
We have defined the TreeNode class to represent each node of the binary tree. The `maximumDepth` function takes the root node of the binary tree as input and returns the maximum depth of the binary tree.
We start by checking if the root node is empty. If it is, we return 0 as the depth.
Next, we calculate the depth of the left subtree recursively by calling the `maximumDepth` function on the left child of the root node. Similarly, we calculate the depth of the right subtree recursively by calling the `maximumDepth` function on the right child of the root node.
Once we have calculated the depths of the left and right subtrees, we take the maximum of the two values and add one to it to get the maximum depth of the binary tree.
Finally, we return the maximum depth.
This solution has a time complexity of O(n), where n is the number of nodes in the binary tree, as we visit each node once. The space complexity is O(h), where h is the height of the binary tree, as the maximum depth of the recursive call stack is equal to the height of the tree.
Step by Step Implementation For Maximum Depth Of Binary Tree
```public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}
int left = maxDepth(root.left);
int right = maxDepth(root.right);
return Math.max(left, right) + 1;
}```
```def maxDepth(root):
if root is None:
return 0
else:
# Compute the depth of each subtree
l_depth = maxDepth(root.left)
r_depth = maxDepth(root.right)
# Use the larger one
if (l_depth > r_depth):
return l_depth+1
else:
return r_depth+1```
```/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
// recursive solution
var maxDepth = function(root) {
if (root === null) return 0;
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
};```
```/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int maxDepth(TreeNode* root) {
if(root==NULL)
return 0;
int lh=maxDepth(root->left);
int rh=maxDepth(root->right);
return max(lh,rh)+1;
}
};```
```/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int MaxDepth(TreeNode root) {
//if the tree is empty, return a depth of 0
if(root == null){
return 0;
}
//if the tree is not empty, recursively call MaxDepth on the left and right subtrees
//and return the maximum depth of the two
else{
return Math.Max(MaxDepth(root.left), MaxDepth(root.right)) + 1;
}
}
}```
Scroll to Top
Top 100 Leetcode Practice Problems In Java
Get 30% Off Instantly!
[gravityforms id="5" description="false" titla="false" ajax="true"]
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://prepfortech.in/leetcode-solutions/maximum-depth-of-binary-tree/", "fetch_time": 1701191604000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00297.warc.gz", "warc_record_offset": 557218392, "warc_record_length": 36690, "token_count": 1116, "char_count": 4616, "score": 4.375, "int_score": 4, "crawl": "CC-MAIN-2023-50", "snapshot_type": "latest", "language": "en", "language_score": 0.7485269904136658, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00028-of-00064.parquet"}
|
# DC Circuits Homework
1. Sep 27, 2009
### mustang1988
1. Eight lights are connected in parallel to a 119V source by two leads of total resistance 2.5 ohms. If 240mA flows through each bulb what is the resistance of each? What fraction of the total power is wasted in the leads?
2. V=IR, P=I^2(R)
3. V=.24(119)=495.83/8=62ohms
P=(.24)^2(62)=3.57W
These are the answers i got but according to my book they are wrong. Can anyone tell me what im doing wrong? thanks
2. Sep 27, 2009
### 206PiruBlood
Well as for your first question you seem to have V=iV.
Now you know the resistors are connected in parallel.
To find the total resistance of all components, add the reciprocals of the resistances R of each component and take the reciprocal of the sum. :D.
Last edited: Sep 27, 2009
3. Sep 27, 2009
### mustang1988
2.5ohms is the total resistence? I thought that was just the resitence in the wire going from the battery and the wire going to the battery
Last edited: Sep 27, 2009
4. Sep 27, 2009
### gnurf
1) An equivalent circuit where you include a 2.5 ohms series resistor to represent the loss in the wires.
2) Kirchoff's Current Law
3) Kirchoff's Voltage Law
4) Ohm's Law
5) The expression for power.
Use them in that order, and see how that goes.
5. Sep 27, 2009
### mustang1988
what is Kirchoff's current law and voltage law? they're not in my book
6. Sep 28, 2009
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/dc-circuits-homework.340858/", "fetch_time": 1508276937000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220905-00830.warc.gz", "warc_record_offset": 977558035, "warc_record_length": 15309, "token_count": 424, "char_count": 1388, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2017-43", "snapshot_type": "longest", "language": "en", "language_score": 0.9387100338935852, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00045-of-00064.parquet"}
|
# Building Numbers
Knowing how to build numbers is an important lesson for kindergarteners just getting their feet wet with counting and numbers. These guided exercises can give kids a deeper understanding of number sense, numerical order, and values. Kindergarteners will begin to associate values with numbers and comprehend the concepts of greater than and less than, building upon skills they will need in the later grades.
This lesson includes 5 printable learning activities.
## Building Numbers Song
There's never just one way to make a number—you can build each out in so many ways! Learn all about how to making numbers with the buttons on Grandma's sweater in this funky composing and decomposing numbers song. Get down with Grandma as your child learns all about building numbers in different ways.
Roly loves hats, and they look great on him too! Have your little mathematician help this oh-so stylish cat count the hats in his collection by using his addition skills to make sure the right amount of hats are on the table. This game is a good way to have your kid practise representing addition facts with objects and it'll test her understanding of cumulative addition as she builds upon previous sums to satisfy Roly's need to keep tabs on all his hats.
## Building Numbers Pizza Party
Help Muggo, Cuz-Cuz, and Penelope make their favorite kinds of pizza! The maths game Building Numbers Pizza Party lets kids use their number sense and counting skills to help their hungry friends make pizza with exactly the right toppings. Designed with kindergarteners in mind, this educational game asks young mathematicians to add the correct number and type of toppings to each pizza. The activity’s narrator also reads out the addition happening in each round for added reinforcement of maths concepts.
## Number Pair Staircase 2
This awesome game adds a little creativity into learning early maths skills. Your students will play build a strong foundation in addition by helping Roly the cat re-construct this staircase using their understanding of addition, and the numbers 1-10. After successfully grouping two sets of blocks together, students get an automatic review of the mental addition they just performed.
## The Chocolate Shop
Two of our Brainzy characters team up to pack chocolate boxes for customers at a chocolate shop.
## Ten Frame Signs
Get practise counting numbers 11–20 with the help of your favorite Brainzy characters in this maths game for early learners. Perfect for kindergarteners and first graders, children will build their understanding of place value and other number sense skills as they practise composing and decomposing numbers 11–20 using ten frame signs. This place value practise will help them understand how to quickly identify a "ten" and master more complex maths concepts.
## Patterns Puzzle
Jigsaw puzzles are a great way to practise beginning maths skills -- plus,they're just plain fun! Take a crack at this online jigsaw puzzle with four different skill levels for learners of all ages.
Create new collection
0
### New Collection>
0Items
What could we do to improve Education.com?
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://nz.education.com/guided-lesson/composing-decomposing/", "fetch_time": 1571376465000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-43/segments/1570986677884.28/warc/CC-MAIN-20191018032611-20191018060111-00119.warc.gz", "warc_record_offset": 642369981, "warc_record_length": 45357, "token_count": 616, "char_count": 3151, "score": 4.375, "int_score": 4, "crawl": "CC-MAIN-2019-43", "snapshot_type": "longest", "language": "en", "language_score": 0.9477338790893555, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00055-of-00064.parquet"}
|
# AP Statistics : How to interpret dotplots
## Example Questions
### Example Question #1 : Graphing Data
A basketball coach wants to determine if a player's height can be used to predict the number of points that player scores in a season. Before using a statistical test to determine the precise relationship of the variables, the coach wants a visual of the data to see if there is likely to be a relationship. Which of the following should the coach create?
Histogram
Bar chart
Bell curve
Z-score
Scatterplot
Scatterplot
Explanation:
A scatterplot is a diagram that shows the values of two variables and provides a general illustration of the relationship between them.
### Example Question #1 : Using Scatter Plots
Based on the scatter plot below, is there a correlation between the and variables? If so, describe the correlation.
Yes; positive linear relationship
Yes; negative exponential relationship
No; there is no correlation
Yes; negative linear relationship
Yes; negative linear relationship
Explanation:
The data points follow an overall linear trend, as opposed to being randomly distributed. Though there are a few outliers, there is a general relationship between the two variables.
A line could accurately predict the trend of the data points, suggesting there is a linear correlation. Since the y-values decrease as the x-values increase, the correlation must be negative. We can see that a line connecting the upper-most and lower-most points would have a negative slope.
An exponential relationship would be curved, rather than straight.
### Example Question #2 : Graphing Data
Order the correlation coefficients to fit the order of the following graphs (two coefficients will not be used)
, , , ,
, ,
, ,
, ,
, ,
, ,
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.varsitytutors.com/ap_statistics-help/how-to-interpret-dotplots", "fetch_time": 1485154875000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-04/segments/1484560282140.72/warc/CC-MAIN-20170116095122-00093-ip-10-171-10-70.ec2.internal.warc.gz", "warc_record_offset": 756747807, "warc_record_length": 33442, "token_count": 362, "char_count": 1775, "score": 3.84375, "int_score": 4, "crawl": "CC-MAIN-2017-04", "snapshot_type": "latest", "language": "en", "language_score": 0.9054749608039856, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00014-of-00064.parquet"}
|
# What surface is represented by the equation z=3 ?
justaguide | Certified Educator
The surface represented by z = 3 is a plane consisting of all point that have the z co-ordinate as 3.
The x and y co-ordinates can take on any value.
The only constraint we have here is that all points that form the plane have the z-coordinate as 3.
The surface represented by z = 3 consists of all points that have the value of the z-coordinate as 3.
giorgiana1976 | Student
In general, an equation x = k, represents a plane, parallel to y-z plane.
The equation y = k, represents a plane, parallel to x-z plane.
The equation z = k, represents a plane, parallel to x-y plane.
The equation z = 3 represents the set of all points whose z coordinate is 3.
Therefore, the equation z = 3 is the horizontal plane, that is parallel to x-y plane and it is located 3 units above x-y plane.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.enotes.com/homework-help/what-surface-represented-by-equation-z-3-250564", "fetch_time": 1516140805000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-05/segments/1516084886739.5/warc/CC-MAIN-20180116204303-20180116224303-00739.warc.gz", "warc_record_offset": 921890372, "warc_record_length": 9708, "token_count": 218, "char_count": 877, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2018-05", "snapshot_type": "latest", "language": "en", "language_score": 0.9657808542251587, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00037-of-00064.parquet"}
|
### meooow's blog
By meooow, 6 years ago,
### The problem
Let us consider the problem where we need to quickly calculate the following over some set S of j for some value x.
Additionally, insertion of new j into S must also be efficient.
This will most likely be encountered with DP problems. For example, the recent problem 1083E - The Fair Nut and Rectangles from Round #526 has the following DP formulation after sorting the rectangles by x.
f(i) = xi·yi - ai + max1 ≤ j < i{ - xj·yi + f(j)}
The problem requires quick calculation of the above define maximum for each index i. How can this be done?
### The idea
Notice the special form of mj·x + cj. This is identical to the equation of a straight line with slope mj and Y-intercept cj. So the problem is equivalent to being given a set of lines and asked for the maximum y value any of those lines can give at a particular x. If you draw a bunch of straight lines on a plane, you'll notice that the maximum values are along what appears to be a convex hull.
Some observations:
• Every line on the hull provides the maximum value on some contiguous range of x values. Conversely, every line not on the hull is useless and never provides the maximum.
• All the lines on the hull have different slopes. The order of slopes also determines their position on the hull. A line with lower slope appears on the hull to the left of one with a higher slope.
So, a possible strategy can be to only maintain the convex hull and not keep the useless lines .
### A specific problem
Let us further consider the rectangle problem mentioned above.
For clarity, let's substitute x and y of the problem statement with p and q, and allow x and y to only refer to coordinates of the 2D plane where we consider the lines.
f(i) = pi·qi - ai + max1 ≤ j < i{ - pj·qi + f(j)}
In this problem the slope of the lines mj is given by - pj. Due to the nature of the constraints (no rectangles are nested), after sorting rectangles by increasing p we will find they are also sorted by decreasing q.
The idea:
1. We'll keep the lines of the hull, in sorted order of slope.
2. We iterate over the rectangles from i = 1 to n, and pi < pj, qi > qj for i < j.
3. When we have to get the maximum at some x = qi, all lines that provide the maximum at positions > qi are now useless, since further maximum queries are guaranteed to occur at < qi.
4. When a new line is inserted, the slope of this line - pi is guaranteed to be lesser than all lines in the hull. Therefore, this line is also guaranteed to provide the maximum in some range ( - ∞, x]. However, adding this line may make it so that some lines previously on the hull are no longer on it. These lines need to be removed to maintain the hull.
For this particular problem:
• The lines are inserted in sorted order of slope
• The query positions are also in sorted order
Implementing query and insert:
Query
When querying at x = qi, just compare the value at x of the rightmost line with that of the line next to it. If it is lower, remove it and repeat. When done, get the value at x of the rightmost line as the answer to the query.
Insert
When inserting a line, if the intersection point of this line and the leftmost line lies to the right of that of the leftmost line and the line to the right of it, the leftmost line is no longer on the hull. Remove it, and repeat. Parallel lines pose an exception to this since they will never intersect, and must be handled separately if such a situation is possible in the problem.
And that's it... since we add lines at one end and remove at both ends, the data structure for the job is a deque.
Solution for the rectangle problem:
Complexity is if N lines are inserted and Q queries are made.
##### A few what ifs
What if slopes are sorted in increasing order instead?
You can modify the logic accordingly.... or you can observe that negating the slope has the effect of mirroring lines about the Y-axis, so you can use one implementation for both.
What if slopes are sorted but in reverse order of the query positions?
Both adding and removing will be done at one end, so a stack is required. Of course a deque can also do the job of a stack.
What if minimum is required instead of maximum?
Again, you can modify the logic... or you can observe that negating both slope and Y-intersect has the effect of mirroring about the X-axis. You can use the same implementation.
### A more general problem
Let us consider a problem where
• The lines are inserted in sorted order of slope
• The query positions are in arbitrary order
To tackle this problem nothing needs to be changed for insertion. However we can no longer remove lines when answering queries. In order to answer queries, notice that each line provides the maximum in some range which is defined by its intersection point with the previous and next line. Given a particular x we can use binary search to find the line where the value will be maximum.
Solution for the rectangle problem using this method:
Complexity is .
### The original problem
Coming back to the general version,
• The lines are inserted in arbitrary order of slope
• The query positions are in arbitrary order
This is referred to as the "fully dynamic" version of CHT. A convenient way to implement this is using a sorted set, such as std::set in C++ or TreeSet in Java. The idea is to maintain the set sorted by slope. To query, binary search is used as before. To insert, the position at which the line should be inserted is located. If this line does not appear on the hull, it is not inserted. If it does, useless lines are removed from both the left and right of the inserted line. The complexity using this method is .
You can find a neat implementation here (thanks to Chilli for the link). A couple more can be found here and here. I do not want to go into further details about this method, because I personally find using Li Chao tree much simpler if the fully dynamic version is required. Li Chao tree is a specialized segment tree that also deals with the convex hull trick, and there exists a nice tutorial for it on cp-algorithms. This page also contains an alternate interpretation of CHT.
### Conclusion
Although this tutorial focuses on the technique of CHT, it is worth mentioning that in contests CHT will almost always be intended as a way to optimize DP. You can refer to link titled "Dynamic Programming Optimizations" below to check out the forms of DP recurrences that can be optimized this way.
Note about precision: You may have noticed that the function intersectX in the code uses long double to find the coordinate. Is this good enough? Since queries are (usually) at integer x, the lines which provide the maximum in a range completely contained in interval between two consecutive integers are useless since they never provide a maximum at any integer coordinate. So we actually do not even need long double, floor/ceil division will do just fine. Thanks to tmwilliamlin168 for pointing this out to me. Note that integer division is not the same as floor division in C++ for negative numbers.
##### Problems
Ordered approximately by difficulty:
That concludes my first tutorial on Codeforces. Thanks for reading and I hope it was useful. Any suggestions or improvements are welcome.
The nice images above were made with Desmos.
If you want other links/problems on CHT to be added, comment below and I will add them.
• +274
| Write comment?
» 6 years ago, # | +5 Great tutorial! Another good resource for those who prefer to learn from videos is Algorithms Live — Convex Hull Optimization.
• » » 6 years ago, # ^ | ← Rev. 2 → +8 I've added the link. One thing that irked me, in the first part the author says that (x - y)2 + prevCost is not really CHT because the functions are parabolic and not straight lines, but the expression can be expanded to y2 - 2xy + x2 + prevCost which needs to be minimized for fixed y over some x, so it actually can be solved in the normal way with a convex hull of lines.
» 6 years ago, # | +11 I think PDELIV deserves a mention in the problem list. It requires you to use it in a way I personally hadn't considered before.
• » » 6 years ago, # ^ | 0 Nice problem, added to the list!
» 6 years ago, # | ← Rev. 2 → +23 I like the implementation created by simonlindholm, found in the KTH notebook. I originally saw ksun48 use it here: https://mirror.codeforces.com/contest/1083/submission/46863810.I think it's a lot less magic than the other 2 implementations linked (no mutable member functions/closures), and I believe it's also substantially faster.
• » » 6 years ago, # ^ | +5 This implementation appears short and neat. Added to the blog. Have you also compared the performance?
• » » » 6 years ago, # ^ | ← Rev. 2 → +5 The primary thing that differentiates this implementation is that it stores the intersection point during insertion. This makes the implementation a lot shorter as well as the queries somewhat faster. Overall, compared to the other 2 implementations linked (called HullDynamic and chtDynamic respectively), it's somewhat slower at insertion than the other two, significantly faster at querying than HullDynamic, and slightly faster at querying than chtDynamic. Overall, it's very competitive in performance.Insertions (1e7 insertions) LineContainer: 0.588702 HullDynamic: 0.480461 chtDynamic: 0.513528 Queries (1e5 insertions, 1e7 queries) LineContainer: 0.109056 HullDynamic: 0.385707 chtDynamic: 0.146591 Overall (1e7 insertions, 1e7 queries) LineContainer: 0.784574 HullDynamic: 0.995786 chtDynamic: 0.768847 And finally, line count :^) LineContainer: 36 HullDynamic: 44 chtDynamic: 75 I think the KTH implementation is clearly the winner. Benchmarks can be found here: https://ideone.com/caYDdF
• » » 6 years ago, # ^ | +20 Starting with C++14, std::less is transparent, so you don't even need the hack with the global bool Q. Instead, you can use different operator< for lines and query points. submission
• » » » 6 years ago, # ^ | 0 Oh, that's nice. Is there any reason you made p mutable?
• » » » » 6 years ago, # ^ | +5 p is the x-coordinate of the intersection with the next line and you need to update that when inserting new lines. A line inside the set is const, so you need mutable to make p modifiable. (k and m don't need to be changed, so they're not mutable.)
• » » » 6 years ago, # ^ | +10 Oh, neat! I've made that change to KACTL: https://github.com/kth-competitive-programming/kactl/commit/165807e28402c9be906f6e6a09452431787bb70d
• » » » » 5 years ago, # ^ | 0 Is it possible to remove lines from the struct?
• » » » » » 5 years ago, # ^ | +4 You can technically remove lines from the structure, but you cannot bring back the lines you previously discarded for the purpose on maintaining only the hull instead of all lines. One or more of those discarded lines may have the second largest value at some $x$ where the removed line had the max value, which you cannot recover. So you will be having an incomplete hull.
• » » » » » » 5 years ago, # ^ | 0 Yeah, that makes sense. So is there any other way which allows remove or update queries on the line parameters while maintaining the complete hull?
• » » » » » » » 5 years ago, # ^ | +3 Not that I know of, assuming you want to keep the same or close enough complexity.
• » » » » » » » 5 years ago, # ^ | ← Rev. 2 → 0 If queries is offline I think Divide & Conquer O(n * log^2) helps like in Dynamic Connectivity (easy google). Online harder, idk maybe some kind of SQRT decomposition on queries. Smth like keep last B queries and proceed in stupid way, for other queries there is built CHT. So number of stupid asks will be B * q, number of CHT rebuilds will be q / B. To avoid sorting we can merge, so if B = sqrt(n), and for simplicity q = n. Complexity is O(n * sqrt(n) + q * log(n))
• » » » » » 5 years ago, # ^ | 0 You can do it using treap
• » » » » 4 years ago, # ^ | 0 Why do you need this 'while' in add function? while ((y = x) != begin() && (--x)->p >= y->p) { intersect(x, erase(y)); } I deleted it and got AC. Maybe it's useful for different problems?
• » » » » » 4 years ago, # ^ | 0 I guess it's perhaps unnecessary when the lines you're adding are increasing in some manner? KACTL's stress tests fail without those two lines, though, so in general they are necessary.
• » » 5 years ago, # ^ | ← Rev. 2 → 0 How do I make it query the minimum value instead of the maximum? I'm just starting to learn this, so sorry for the dumb question.Edit: I figured it out, you're supposed to insert the negatives.
• » » 17 months ago, # ^ | 0 Can you tell me what changes must be made to work for the min convex hull?
• » » » 17 months ago, # ^ | -8 You can do this - Code//declaring the max convex hull LineContainer cht; //adding a line with slope m and constant c cht.add(-m, -c); //querying min value at x int ans = -cht.query(x);
» 6 years ago, # | +5 Here's another related problem: YATP.
• » » 6 years ago, # ^ | +5 Nice problem. If anyone needs hints... SpoilerCentroid decomposition.Further explanation in this video: Algorithms Live — YATP w/ Lewin Gan
» 6 years ago, # | +5 Great Tutorial!!!
» 6 years ago, # | ← Rev. 2 → +5 This problem POLY can also be added here.
• » » 6 years ago, # ^ | 0 Thanks, updated.
» 5 years ago, # | 0 Also, Lena and Queries
» 5 years ago, # | 0 Great Tutorial! I tried solving the problem 1083E - The Fair Nut and Rectangles but for some reason my code is giving WA on test 5. Can someone please help me. 57194241
• » » 5 years ago, # ^ | +3 You're forcibly including the first rectangle always. The optimal solution might leave it out.Fix is that when in ll m = get_max(lines, v[i].q); you find m < 0 you should not add it to dp[i].
» 5 years ago, # | ← Rev. 2 → 0 How do I modify the data structure so it gets the minimum at a point instead of the maximum?Edit: I figured it out, you should insert the negatives of the slopes and constants.
» 5 years ago, # | 0 which order of the slopes or queries are relevant? decreasing or increasing. Or both? I'll be appreciated if you answer this comment :3
» 5 years ago, # | 0 The only difference between my AC code 69191641 and my WA on test 6 code for problem E — The Fair Nut and Rectangles was the "long double" used for comparing in fuction check(), which i put there because I saw that in many other's code. However, I didn't used any division, and the problem statement clearly said that xi, yi, ai are all int, so I'm very confused. Can someone please explain ?
• » » 4 years ago, # ^ | 0 $b$ can be up to $10^{18}$ and $m$ can be up to $10^6$, so this multiplication overflows 64bit integers.
» 4 years ago, # | 0 can some please tell me how to solve this problem https://mirror.codeforces.com/contest/319/problem/C using li chao tree....
» 4 years ago, # | 0 In the implementation of "A More General Problem", how are you using lower bound for deque. You are doing lower bound for vector but in comparator using deque. I am not getting it. Can you explain it or share some links from where I can read about it? meooow. Thank You
• » » 4 years ago, # ^ | +8 The vector has integers $0, 1, 2, 3, 4, ...$ so this is just a clever way to not code his own binary search to find the index of the optimum line for a particular $x$.It would be a bit tricky to use lower_bound over the deque because we have to find the intersection with the next line. To do it you can keep the intersection with the next line in the struct and update it on insert.
• » » » 4 years ago, # ^ | 0 Thank you for answering it.I got it for deque.How will we write lower bound for a set (In Full Dynamic Version) for query part? Nson
• » » » » 4 years ago, # ^ | +8 Here is the kactl versionThe $p$ in the line struct represents the $x$ coordinate of the intersection with the next line. You can see it is modified upon insertion. This way you can do the same lower_bound without knowing the next line.
• » » » » » 4 years ago, # ^ | 0 Got it. Thank you.
• » » 4 years ago, # ^ | +8 Nson is correct, it is just to avoid writing binary search code.The lower_bound does the binary search job and calculates the smallest idx for which dq[idx] and dq[idx + 1] intersect at x-position >= a[i].q.
• » » » 4 years ago, # ^ | 0 Got it.Thank you.
» 4 years ago, # | 0 Hi, was looking at the Li Chao tree method and it seems a lot simpler. Would just like to know, does Li Chao tree have any limitations? Is it possible to use it even in a non-dynamic version (lines are sorted by slope, query not arbitrary)?
• » » 4 years ago, # ^ | +5 Yes, if it works as fully dynamic, that means you can insert and query in any order. That includes sorted order.Limitations of Li Chao tree that I can think of are (1) it only supports integer queries, and(2) operations take logarithmic time with respect to the query domain size rather than the current size of the hull.
» 4 years ago, # | 0 Hi. Can anyone help my with problem Avoiding airports?My main idea: For every outgoing edge u->v (s,e) find the min cost that we can reach v at time e:From all incoming edges into u, earlier than s, find the min:F(v,e) = min{e_i < s}(F(u, e_i) + (s — e_i) ^ 2) = s^2 + min(F(u,e_i) + e_i ^ 2 — 2 * e_i * s)This is standard CHT. For every vertex I keep CHT and process time points:1) if at time T I have outgoing edge from u to v (T, T2) — I say add to CHT[v] at time T2 line (-2*T2, Query(CHT[u], T) + T2^2)2) if at time T I have incoming edge (e.g. the promise from step 1): I just added that line to CHT of a given vertex.See my code: https://pastebin.com/RPnrHGJ0It fails on test 5, where 50 < n < 100, m ~200000, and there is at least one pair (u,v) with the same departure and landing time (s = e)Any ideas?
• » » 4 years ago, # ^ | 0 I found the solution with LiChao tree that is very similar to mine. Can anyone explain why LiChao is needed?
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://mirror.codeforces.com/blog/entry/63823", "fetch_time": 1722762549000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-33/segments/1722640393185.10/warc/CC-MAIN-20240804071743-20240804101743-00362.warc.gz", "warc_record_offset": 308954266, "warc_record_length": 33851, "token_count": 4658, "char_count": 17985, "score": 4.375, "int_score": 4, "crawl": "CC-MAIN-2024-33", "snapshot_type": "latest", "language": "en", "language_score": 0.9111645817756653, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00039-of-00064.parquet"}
|
59150
59,150 (fifty-nine thousand one hundred fifty) is an even five-digits composite number following 59149 and preceding 59151. In scientific notation, it is written as 5.915 × 104. The sum of its digits is 20. It has a total of 6 prime factors and 36 positive divisors. There are 18,720 positive integers (up to 59150) that are relatively prime to 59150.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 20
• Digital Root 2
Name
Short name 59 thousand 150 fifty-nine thousand one hundred fifty
Notation
Scientific notation 5.915 × 104 59.15 × 103
Prime Factorization of 59150
Prime Factorization 2 × 52 × 7 × 132
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 910 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 59,150 is 2 × 52 × 7 × 132. Since it has a total of 6 prime factors, 59,150 is a composite number.
Divisors of 59150
1, 2, 5, 7, 10, 13, 14, 25, 26, 35, 50, 65, 70, 91, 130, 169, 175, 182, 325, 338, 350, 455, 650, 845, 910, 1183, 1690, 2275, 2366, 4225, 4550, 5915, 8450, 11830, 29575, 59150
36 divisors
Even divisors 18 18 9 9
Total Divisors Sum of Divisors Aliquot Sum τ(n) 36 Total number of the positive divisors of n σ(n) 136152 Sum of all the positive divisors of n s(n) 77002 Sum of the proper positive divisors of n A(n) 3782 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 243.208 Returns the nth root of the product of n divisors H(n) 15.6399 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 59,150 can be divided by 36 positive divisors (out of which 18 are even, and 18 are odd). The sum of these divisors (counting 59,150) is 136,152, the average is 3,782.
Other Arithmetic Functions (n = 59150)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 18720 Total number of positive integers not greater than n that are coprime to n λ(n) 780 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5966 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 18,720 positive integers (less than 59,150) that are coprime with 59,150. And there are approximately 5,966 prime numbers less than or equal to 59,150.
Divisibility of 59150
m n mod m 2 3 4 5 6 7 8 9 0 2 2 0 2 0 6 2
The number 59,150 is divisible by 2, 5 and 7.
• Arithmetic
• Abundant
• Polite
Base conversion (59150)
Base System Value
2 Binary 1110011100001110
3 Ternary 10000010202
4 Quaternary 32130032
5 Quinary 3343100
6 Senary 1133502
8 Octal 163416
10 Decimal 59150
12 Duodecimal 2a292
20 Vigesimal 77ha
36 Base36 19n2
Basic calculations (n = 59150)
Multiplication
n×i
n×2 118300 177450 236600 295750
Division
ni
n⁄2 29575 19716.7 14787.5 11830
Exponentiation
ni
n2 3498722500 206949435875000 12241059132006250000 724058647658169687500000
Nth Root
i√n
2√n 243.208 38.9629 15.5951 9.00308
59150 as geometric shapes
Circle
Diameter 118300 371650 1.09916e+10
Sphere
Volume 8.66868e+14 4.39662e+10 371650
Square
Length = n
Perimeter 236600 3.49872e+09 83650.7
Cube
Length = n
Surface area 2.09923e+10 2.06949e+14 102451
Equilateral Triangle
Length = n
Perimeter 177450 1.51499e+09 51225.4
Triangular Pyramid
Length = n
Surface area 6.05997e+09 2.43892e+13 48295.8
Cryptographic Hash Functions
md5 86db75f9eb404bc1bf9c5665983b4036 058ebde254f3eb85ab4864606d77509c6c861295 b9ef1b0e8f9ebdeabc82fab1152dd9bf6b3e7e97bc8d4592eea68fcbf55a57c3 4c820bf60f963bd605763d17c0a04b3e16e71d47521423312fdd0056b6fb0c77b73ca5e9eed86631b7e1901b300d0f3431df13ab6e52b1dbf1eda88787a191c6 901f0bfe209c82b469c5ee8899039b1fcde1ddfe
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://metanumbers.com/59150", "fetch_time": 1597511972000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-34/segments/1596439740929.65/warc/CC-MAIN-20200815154632-20200815184632-00377.warc.gz", "warc_record_offset": 401318189, "warc_record_length": 7568, "token_count": 1520, "char_count": 4141, "score": 3.765625, "int_score": 4, "crawl": "CC-MAIN-2020-34", "snapshot_type": "latest", "language": "en", "language_score": 0.7786079049110413, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00005-of-00064.parquet"}
|
# A hot air balloon is rising upward with a constant speed of 2.50m/s. When the balloon is 3.0 m above the ground, the balloonist drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?PS. Do you mind explaining why you chose a certain formula or why you plugged in certain values?
2
by cbooey
2014-09-23T19:59:28-04:00
Let's assume that the balloonist dropped the compass over the the side as the balloon still rising upward at 2.5m/s
So you have:
Vi = -2.5 m/s (because the compass is still going with the balloon as it starts to fall down)
d = 3 m
a = 9.81 m/s^2
At first you use this formula to find the Vf (finally velocity):
Vf^2 = 6.25 + 2(9.81)(3)
Vf^2 = 65.11
Vf = 8.07 m/s
Finally you use this formula to find the time:
Vf = Vi +at
8.07 = -2.5 + (9.81)t
10.57 = (9.81)t
t = 1.08 s
2014-09-23T20:38:47-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
In the first answer submitted, Samuel got a good answer. But for the life of me,
I'm having trouble following his solution. I'm not that great at math and physics,
so here's how I did it :
I always call 'up' positive and 'down' negative, and I use one single
formula for the height of anything that's tossed or dropped:
H = Height at any time
H₀ = height when it's released = 3m
V₀ = vertical speed when it's released = 2.5 m/s
T = time after it's released
G = acceleration of gravity = -9.8 m/s²
H = H₀ + V₀T - 1/2 G T²
H = 3 + 2.5T - 4.9T²
That's the height of the compass at any time after it's dropped, and
we simply want to know the time ' T ' when H = 0 (it hits the ground).
- 4.9T² + 2.5 T + 3 = 0
That's a perfectly good quadratic equation, which you can solve for ' T ' .
The solutions are T = -0.567sec and T = 1.078sec.
The one with physical significance in the real situation is T = 1.078sec.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://brainly.com/question/127586", "fetch_time": 1484906157000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-04/segments/1484560280825.87/warc/CC-MAIN-20170116095120-00381-ip-10-171-10-70.ec2.internal.warc.gz", "warc_record_offset": 798113156, "warc_record_length": 9662, "token_count": 654, "char_count": 2109, "score": 4.0625, "int_score": 4, "crawl": "CC-MAIN-2017-04", "snapshot_type": "latest", "language": "en", "language_score": 0.9143683910369873, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00054-of-00064.parquet"}
|
# Question 1 (Hessenberg Form, Eigenvalues, QR Algorithm) Let A be t...
## Transcribed Text
Question 1 (Hessenberg Form, Eigenvalues, QR Algorithm) Let A be the inverse Hilbert matrix 16 -120 240 -140 -120 1200 -2700 1680 A = 240 -2700 6480 -4200 -140 1680 -4200 2800 Using SciPy 0.10.0 and later, you should be able to create A by using the command A = scipy linalg. invhilbert (4) Otherwise, you will need to create A by specifying its components. (a) Find the two vectors V1, V2 and their corresponding Householder transformations Q1.Q2 which will reduce A to upper Hessenberg form H: A = Q2Q1HQ;Q2 Show each of the two steps of the reduction. You can perform these steps by hand or by using python. (b) Use the function scipy linalg. eig in PYTHON to obtain the spectra of each of A and H. You must produce two sorted vectors of eigenvalues (i.e. one vector for A and one for H) and display them in long format. Use numpy . sort as appropriate. Find the relative error between these two vectors of eigenvalues. (c) Use PYTHON (or any other programming language) to write a function which uses the unshifted QR algorithm to produce the Schur decomposition of a tridiagonal matrix. You must submit your code in electronic form SO that it can be run, if necessary. Use your function to obtain the spectrum of H, sort the resulting vector, and compare your answer to that of b) above. You can use the command numpy set_printoptions (precision=16) to set the default output precision to 16 significant figures (this is somewhat like the format long in matlab). (d) Explain, in general terms, any differences between the eigenvalues produced by the eig function in b) above, and the eigenvalues produced by your Schur decomposition in c) above. Use your observations of the code and the outputs, as well as the theory covered in the Notes. Remember that both vectors of eigenvalues are approximations. Recall the significance of size of the matrix A in relation to methods for obtaining the roots of polynomials.
## Solution Preview
These solutions may offer step-by-step problem-solving explanations or good writing examples that include modern styles of formatting and construction of bibliographies out of text citations and references. Students may use these solutions for personal skill-building and practice. Unethical use is strictly forbidden.
import numpy as np
import scipy.linalg as lg
from scipy.linalg import *
#Question a)
A=invhilbert(4)
def computeHessenberg(A):
n=A.shape[1] ; H=A.copy() ; Qs=[];Vs=[];
for k in range(0,n-2):
x=H[0:n,k:(k+1)].copy()
y=H[0:n,k:(k+1)].copy()
y[k+1]=lg.norm(x[k+1:n])
y[k+2:n]=0
v=(x-y)/lg.norm(x-y)
Q=np.eye(n)-2*v.dot(v.T)
Vs.append(v);Qs.append(Q)
H=Q.dot(H).dot(Q)
return (H,Qs,Vs)
H,Qs,Vs = computeHessenberg(A)
#Question b)
np.set_printoptions(precision=16)
eig_A=np.sort(eig(A)[0]).real
eig_H=np.sort(eig(H)[0]).real...
By purchasing this solution you'll be able to access the following files:
Solution.zip.
\$45.00
for this solution
PayPal, G Pay, ApplePay, Amazon Pay, and all major credit cards accepted.
### Find A Tutor
View available Numerical Analysis Tutors
Get College Homework Help.
Are you sure you don't want to upload any files?
Fast tutor response requires as much info as possible.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.24houranswers.com/college-homework-library/Mathematics/Numerical-Analysis/44108", "fetch_time": 1670506590000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00875.warc.gz", "warc_record_offset": 629145163, "warc_record_length": 36989, "token_count": 838, "char_count": 3267, "score": 3.59375, "int_score": 4, "crawl": "CC-MAIN-2022-49", "snapshot_type": "latest", "language": "en", "language_score": 0.8200528621673584, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00033-of-00064.parquet"}
|
5 February, 12:22
# A compound is 24.7% calcium, 1.2% hydrogen, 14.8% carbon, and 59.3% oxygen. write the empirical formula and name the compound.
+2
1. 5 February, 12:39
0
For simplicity, assume that the total mass of the compound is 100 g.
Therefore, by mass
Ca = 24.7 g, H = 1.2 g, C = 14.8 g, and O = 59.3 g
Converting these to moles
Mol = mass*1/atomic mass
Ca = 24.7*1/40.078 = 0.6163 mol Ca
H = 1.2*1/1.01 = 1.1881 mol H
C = 14.8*1/12.01 = 1.2323 mol C
O = 59.3/16 = 3.7062 mol O
Next, divide all the mols by the smallest value obtained.
Ca: 0.6163/0.6163 = 1 mol Ca
H: 1.1881/0.6163 = 2 mol H
C: 1.2323/0.6163 = 2mol C
O: 3.7062/0.6163 = 6 mol O
Therefore, empirical formula of the compound is
CaH2C2O6
This compound is referred to as Calcium Bicarbonate
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://cpep.org/physics/1863244-a-compound-is-247-calcium-12-hydrogen-148-carbon-and-593-oxygen-write-.html", "fetch_time": 1656496707000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00411.warc.gz", "warc_record_offset": 220551165, "warc_record_length": 7933, "token_count": 330, "char_count": 780, "score": 3.6875, "int_score": 4, "crawl": "CC-MAIN-2022-27", "snapshot_type": "latest", "language": "en", "language_score": 0.8326594829559326, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"}
|
Finding value of $\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n^3}}{\ln(n)}$
Find the value of $$\lim_{n\rightarrow \infty}\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n^3}}{\ln(n)}$$
My Try: Using Stolz-Cesaro,
Let $$\displaystyle a_{n} = 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{n}$$ and $$b_{n} = \ln(n)$$
So $$\displaystyle \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = \lim_{n\rightarrow \infty}\frac{1}{{(n+1)^3}}\cdot \frac{1}{\ln\bigg(1+\frac{1}{n}\bigg)} = 0$$
Please explain if what I have done above is right.
• I haven't heard of Cizero stolza but I would bound the numerator by $\int_1^{n} \frac 1 {x^{3}} \, dx =\frac 1 {2n^{2}}$ to show that the limit is $0$. – Kavi Rama Murthy Sep 10 '18 at 8:51
• I wonder why so many downvotes. There was a simple mistake in asker's application of Cesaro-Stolz, but he did show effort. +1 to compensate. – Paramanand Singh Sep 10 '18 at 13:26
Let $H_n=\sum_{k=1}^n \frac 1k$. Since $H_n=\log n +O(1)$, $$H_{n^3} = 3\log n +O(1)$$ thus$$\frac{H_{n^3}}{\log n} = 3+O\left(\frac{1}{\log n} \right)=3+o(1)$$
The limit is $3$.
Use of Cesaro-Stolz is tricky here. Let's first evaluate the limit of $a_n$ given by $$a_n=\frac{1}{\log n} \sum_{k=1}^{n}\frac{1}{k}$$ Using Cesaro-Stolz we have $$\lim_{n\to \infty} a_n=\lim_{n\to\infty} \frac{1/n}{\log n-\log(n-1)}=-\lim_{n\to \infty} \frac{1}{n\log(1-(1/n))}=1$$ and then the sequence in question, say $b_n$, is given by $b_n=3a_{n^3}$ so $b_n\to 3$ as $n\to\infty$.
• I like this creativity! (+1) – Shashi Sep 10 '18 at 11:08
• Great simplification! – gimusi Sep 10 '18 at 12:15
Your $a_n$ should go up to $\frac1{n^3}$. Then $a_{n+1}-a_n=\frac{1}{(n+1)^3}+\frac{1}{(n+1)^3-1}\dots+\frac1{n^3+2}+\frac{1}{n^3+1}$.
I don't know whether you can arrive there by your method, but you can find the limit by using the asymptotics of harmonic numbers / definition of the euler $\gamma$ constant. Then the limit should be $3$.
Note that using Stolz-Cesaro we obtain
$$\frac{\sum_{k=1}^{(n+1)^3}\frac1k-\sum_{k=1}^{n^3}\frac1k}{\ln(n+1)-\ln n}\sim\frac{\ln(n+1)^3-\log n^3}{\ln(1+\frac1n)}=3\frac{\ln(1+\frac1n)}{\ln(1+\frac1n)}=3$$
but, as already noted by Gabriel Romon, this is not necessary and it is sufficient use that by Harmonic series
$$\frac{\sum_{k=1}^{n^3}\frac1k}{\ln n}\sim\frac{\ln n^3}{\ln n}=3$$
• A direct use of Cesaro-Stolz leads to complications. I have tried a slight variation using subsequences. – Paramanand Singh Sep 10 '18 at 11:05
To answer your question directly. The mistake is made in $a_{n+1}-a_n$. For example, if $n = 10$, then this is given by $$a_{11}-a_{10} = 1/1331 + 1/1330 + \cdots + 1/1001$$ since $11^3 = 1331$ and $10^3=1000$.
If you insert this, you still are going to need to do something similar as Gabriel suggested by viewing the summation as upper and lower Riemann sums.
Edit2: if you now use Stolz Cesaro, which is possible, it works out as Gimusi did.
• Why cannot we apply CS? – gimusi Sep 10 '18 at 10:04
• I am sorry you're right. Made a mistake here. – Stan Tendijck Sep 10 '18 at 10:14
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/2911688/finding-value-of-lim-n-rightarrow-infty-frac1-frac12-frac13-cd", "fetch_time": 1561542968000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-26/segments/1560628000266.39/warc/CC-MAIN-20190626094111-20190626120111-00056.warc.gz", "warc_record_offset": 486510355, "warc_record_length": 37670, "token_count": 1238, "char_count": 3087, "score": 4.0625, "int_score": 4, "crawl": "CC-MAIN-2019-26", "snapshot_type": "latest", "language": "en", "language_score": 0.6292011141777039, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00055-of-00064.parquet"}
|
# How to show that $A_1 = A_2$?
supposition: $A_1=\{k \in \mathbb{Z}\ \colon\ k | (bc +a) \text{ and }k |b\}$, $A_2=\{k \in \mathbb{Z}\ \colon\ k|a \text{ and }k|b\}$
claim: $A_1=A_2$
(my) proof: Let's show that $A_1 \implies A_2$. Let $k \in A_1$, so $k \mid (bc +a)$ and $k \mid b$. Now if $k \mid (bc+a)$, then $k \mid 1 \cdot (bc+a) -bc$ $\implies$ $k \mid a$. So $A_1 \implies A_2$.
I'm not sure how to show $k \mid 1 \cdot (bc+a) -bc$?
-
One does not usually write that a set implies another set. Rather, one write "$A_1\subseteq A_2$", and "$A_2\subseteq A_1$". – Arturo Magidin Oct 14 '11 at 16:26
Since $k|b$, then $k|bc$. Since $k|(bc+a)$ and $k|bc$, then $k|(bc+a)-bc$. Will you be accepting any more of the 45 questions you've asked so far, or just the 7 you accepted so far (including today's acceptance of a question that was last modified on November 2, 2010)? – Arturo Magidin Oct 14 '11 at 16:29
If $k \mid x$ and $k\mid y$ then $k\mid (x-y)$. Let $x=bc+a$, $y=bc$. We are told that $k \mid x$. We are also told that $k\mid b$, which implies $k\mid bc$, that is, $k \mid y$. – André Nicolas Oct 14 '11 at 16:31
Explicitly, what is needed is the following: If $k|u$ and $k|v$, then $k |(au + bv)$, for all integers $a,b$. – JavaMan Oct 14 '11 at 18:01
@alvoutila: Are you really having a hard time figuring out how $k|b$ implies $k|bc$? $k|b$, and $b|bc$, so by transitivity $k|bc$. Yes, you could go by the definition, but you are just repeating the proof that "divides" is transitive. – Arturo Magidin Oct 15 '11 at 19:23
I think you just needed a reminder of the definition of $|$, but here are the ideas in a more spelled-out form than the comments had:
Showing $A_1\subseteq A_2$
For this part, we just have to show $\left(k∣bc+a \text{ and } k∣b\right)\Rightarrow k|a$, since the other condition for $A_2$, $k|b$, is included in the definition of $A_1$.
If $k∣bc+a$ and $k∣b$, then $(bc+a)/k$ and $b/k$ are integers, by the definition of "$|$". Multiplying the second of those by the integer $c$ tells us that $cb/k=bc/k$ is an integer, too.
But then $$\frac{bc+a}k-\frac{bc}k=\frac{a}k$$ is an integer because it's just a difference of integers.
Finally, $a/k$ being an integer means $k|a$, by definition.
Showing $A_2\subseteq A_1$
For this part, we just have to show $k|b\text{ and }k|a\Rightarrow k∣bc+a$, because $k|b$ is included in the defintion of $A_2$.
If $k∣b$ and $k∣a$, then $b/k$ and $a/k$ are integers, so "$c$ times the first plus the second" is an integer, too. But that number is $(bc+a)/k$, so $k|(bc+a)$.
Conclusion
Since $A_1\subseteq A_2$ and $A_2\subseteq A_1$, $A_1=A_2$. (If you had an element in one set but not the other to contradict the equality, it would contradict one of the two subset statements, too.)
-
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://math.stackexchange.com/questions/72633/how-to-show-that-a-1-a-2", "fetch_time": 1469606162000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-30/segments/1469257826736.89/warc/CC-MAIN-20160723071026-00183-ip-10-185-27-174.ec2.internal.warc.gz", "warc_record_offset": 162715640, "warc_record_length": 18407, "token_count": 1010, "char_count": 2775, "score": 4.25, "int_score": 4, "crawl": "CC-MAIN-2016-30", "snapshot_type": "latest", "language": "en", "language_score": 0.7939064502716064, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"}
|
# Find Max and Min Value of Numpy Array with its index ( 1D & 2D)
Find Max and Min Value of Numpy Array
Beginners always face difficulty in finding the max and min Value of Numpy. In fact, the max and min are very useful in finding statistics. Therefore in this entire tutorial, you will know how to find the max and min value of Numpy and its index for both the one-dimensional and multi-dimensional array.
## 1. Max and Min Value for the One Dimensional (1-D) array
Let’s create a 1-D NumPy array using the numpy.array() method. The array() method accepts the list of all values you want to create NumPy array as an argument.
``````#1-D array
import numpy as np
array = np.array([19,5,10,1,9,17,50,19,25,50])
print(array)``````
### Finding the Maximum Value
To find the max value you have to use the max() method. Just pass the input array as an argument inside the max() method.
``````max = np.max(array)
print("The maximum value in the array is :",max)``````
### Index for the Maximum Value
To find the index for the maximum value you have to pass a specific condition as the argument inside the numpy.where() method.
``````#Index of the Maximum element
conditon = (array == max)
result = np.where(conditon)
print("Arrays for the max element:",result)
print("List for the maximum value indexes:",result[0])``````
Please note that. If there are duplicate values in the array then the output will be a list of the same maximum values.
### Finding the Minimum Value
In the same way, You can find the minimum value using the numpy.min() method.
``````# Minimum Value
min = np.min(array)
print("The minimum value in the array is :",min)``````
### Index of the Minimum Value
The concept to find the index of the min value is the same as finding the maximum values. You have to just change the condition inside the numpy.where() method.
``````# Index of the Minimum element
conditon = (array == min)
result = np.where(conditon)
print("Arrays for the min element:",result)
print("List for the min value indexes:",result[0])``````
## 2. Max and Min Value for the Two Dimensional (2-D) array
There are three ways you can find the max and min value of the NumPy array.
1. Maximum and Minumum in the entire array
2. Max and Min values in each column
3. Maximum and Minimum values in each row.
Let’s create a two-dimensional before finding max and min values. I am randomly generating a 2-D array of size 4×3.
``````#2-D Array
array_2d = np.arange(12).reshape(4,3)
print(array_2d)``````
### Finding the Max Value in the entire array
You can find the maximum value in the entire array using the same numpy.max() method just like you have used in finding the max in 1D. It will find the lowest element and gives the output.
``````max_2d = np.max(array_2d)
print("The maximum value for the 2D-array:",max_2d)``````
### Maximum Value in Each Column and Row
Here you have to use an extra argument and it is axis = 0 or axis =1.
#### Max Value in Column
``````# maximum value in each column
max_in_column = np.max(array_2d,axis=0)
print(max_in_column)``````
#### Max Value in Row
``````# maximum value in each row
max_in_row = np.max(array_2d,axis=1)
print(max_in_row)``````
Here I am using the same method max() but now I am passing axis =0 to tell the interpreter to traverse along with the columns and axis =1 to traverse along the columns.
### Finding the Index for the Max Value in 2D
You can easily find the index of the max value in a 1-dimensional NumPy array. But for the 2D array, you have to use Numpy module unravel_index. It will easily find the Index of the Max and Min value.
``````from numpy import unravel_index
result = unravel_index(np.max(array_2d),array_2d.shape)
print("Index for the Maximum Value in the 2D Array is:",result)``````
Here I am passing the two arguments inside the unravel_index() method one is the maximum value of the array and the shape of the array. Here In our case, the shape of the array is 4 rows and 3 columns.
### Finding the Min Value in the entire array
To find the minimum value inside the array you have to use the numpy.min() method and pass the array.
``````#Minimum Element in the 2D- Array
min_2d = np.min(array_2d)
print("The minimum value for the 2D-array:",min_2d)``````
### Minimum Value in Each Column and Row
#### Min Value in Column
``````# minimum value in each column
min_in_column = np.min(array_2d,axis=0)
print(min_in_column)``````
#### Min Value in Row
``````# minimum value in each row
min_in_row = np.min(array_2d,axis=1)
print(min_in_row)``````
To find the min value in each column and row you have to just change the value of the axis, axis = 0 for the column, and axis =1 for the row in the min() method.
### Index for the Min Value in 2D
Just like you have learned how to find the index of max value in 2D using the unravel_index() method. Here you will also use that. You have to pass the minimum value and the shape of the array as the arguments.
```index_of_min = unravel_index(np.min(array_2d),array_2d.shape)
print("Index for the Minimum Value in the 2D Array is:",index_of_min)```
## Other Question
Question: Difference between amax() and max()
Many data science learner readers have asked us what is the difference between the two methods amax() and max(). The answer is that np.max() is an alias of the method amax(). They both are able to compute the maximum elements of any input array. You can also find the maximum value along the axis.
If you run the method then you can see both are calling the same function.
``````import numpy as np
print(np.max)``````
Output
`<function numpy.core.fromnumeric.amax>`
``````import numpy as np
print(np.amax)``````
Output
`<function numpy.core.fromnumeric.amax>`
Source:
numpy.ndarray.min
numpy.ndarray.max
Subscribe to our mailing list and get interesting stuff and updates to your email inbox.
Something went wrong.
Meet Sukesh ( Chief Editor ), a passionate and skilled Python programmer with a deep fascination for data science, NumPy, and Pandas. His journey in the world of coding began as a curious explorer and has evolved into a seasoned data enthusiast.
Thank you For sharing.We appreciate your support. Don't Forget to LIKE and FOLLOW our SITE to keep UPDATED with Data Science Learner
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.datasciencelearner.com/find-max-and-min-value-of-numpy-array/", "fetch_time": 1701474140000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00666.warc.gz", "warc_record_offset": 817376473, "warc_record_length": 61028, "token_count": 1608, "char_count": 6268, "score": 4.0625, "int_score": 4, "crawl": "CC-MAIN-2023-50", "snapshot_type": "latest", "language": "en", "language_score": 0.6985865831375122, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00012-of-00064.parquet"}
|
Address Mobile, AL 36601 (251) 721-3605
# margin of error equation statistics Bucks, Alabama
That is, the critical value would still have been 1.96. A 95% level of confidence has α = 0.05 and critical value of zα/2 = 1.96.A 99% level of confidence has α = 0.01 and critical value of zα/2 = 2.58.A HP39GS Graphing CalculatorList Price: $79.99Buy Used:$24.28Buy New: \$34.45Approved for AP Statistics and Calculus5 Steps to a 5 AP Statistics, 2014-2015 Edition (5 Steps to a 5 on the Advanced Placement Popular Articles 1.
The condition you need to meet in order to use a z*-value in the margin of error formula for a sample mean is either: 1) The original population has a normal Otherwise, use the second equation. Calculating a Confidence Interval for a Mean When we Know the Standard Deviation Examples of Confidence Intervals for Means Calculating a Confidence Interval for a Mean What Is a Confidence Interval? The symbol E denotes the margin of error of the unknown population mean.
Rumsey When a research question asks you to find a statistical sample mean (or average), you need to report a margin of error, or MOE, for the sample mean. It is used to denote the level of confidence that we are working with. Post a comment and I'll do my best to help! These two may not be directly related, although in general, for large distributions that look like normal curves, there is a direct relationship.
Definition The margin of error for a particular statistic of interest is usually defined as the radius (or half the width) of the confidence interval for that statistic.[6][7] The term can Thank you,,for signing up! As a rough guide, many statisticians say that a sample size of 30 is large enough when the population distribution is bell-shaped. Linearization and resampling are widely used techniques for data from complex sample designs.
For example, if your CV is 1.95 and your SE is 0.019, then: 1.95 * 0.019 = 0.03705 Sample question: 900 students were surveyed and had an average GPA of 2.7 By using this site, you agree to the Terms of Use and Privacy Policy. Also from About.com: Verywell & The Balance This site uses cookies. What Sample Size Do You Need for a Certain Margin of Error?
Note that there is not necessarily a strict connection between the true confidence interval, and the true standard error. Thanks, You're in! For example, a 95% confidence interval with a 4 percent margin of error means that your statistic will be within 4 percentage points of the real population value 95% of the The larger the margin of error, the less confidence one should have that the poll's reported results are close to the true figures; that is, the figures for the whole population.
A school accountability case study: California API awards and the Orange County Register margin of error folly. In other words, the range of likely values for the average weight of all large cones made for the day is estimated (with 95% confidence) to be between 10.30 - 0.17 Multiply by the appropriate z*-value (refer to the above table). Otherwise, use a z-score.
Retrieved on 2 February 2007. ^ Rogosa, D.R. (2005). What is a Survey?. gives you the standard error. Along with the confidence level, the sample design for a survey, and in particular its sample size, determines the magnitude of the margin of error.
The chart shows only the confidence percentages most commonly used. Refer to the above table for the appropriate z*-value. Solution The correct answer is (B). The true standard error of the statistic is the square root of the true sampling variance of the statistic.
Check out the grade-increasing book that's recommended reading at Oxford University! Check out our Statistics Scholarship Page to apply! Sampling: Design and Analysis. The margin of error of an estimate is the half-width of the confidence interval ... ^ Stokes, Lynne; Tom Belin (2004). "What is a Margin of Error?" (PDF).
The critical value for a 90% level of confidence, with corresponding α value of 0.10, is 1.64. The terms statistical tie and statistical dead heat are sometimes used to describe reported percentages that differ by less than a margin of error, but these terms can be misleading.[10][11] For That means if the poll is repeated using the same techniques, 98% of the time the true population parameter (parameter vs. For example, the area between z*=1.28 and z=-1.28 is approximately 0.80.
FPC can be calculated using the formula:[8] FPC = N − n N − 1 . {\displaystyle \operatorname {FPC} ={\sqrt {\frac {N-n}{N-1}}}.} To adjust for a large sampling fraction, the fpc In other words, the maximum margin of error is the radius of a 95% confidence interval for a reported percentage of 50%. AP Statistics Tutorial Exploring Data ▸ The basics ▾ Variables ▾ Population vs sample ▾ Central tendency ▾ Variability ▾ Position ▸ Charts and graphs ▾ Patterns in data ▾ Dotplots This level is the percentage of polls, if repeated with the same design and procedure, whose margin of error around the reported percentage would include the "true" percentage.
The critical t statistic (t*) is the t statistic having degrees of freedom equal to DF and a cumulative probability equal to the critical probability (p*). In addition, for cases where you don't know the population standard deviation, you can substitute it with s, the sample standard deviation; from there you use a t*-value instead of a When working with and reporting results about data, always remember what the units are. Skip to main contentSubjectsMath by subjectEarly mathArithmeticAlgebraGeometryTrigonometryStatistics & probabilityCalculusDifferential equationsLinear algebraMath for fun and gloryMath by gradeK–2nd3rd4th5th6th7th8thHigh schoolScience & engineeringPhysicsChemistryOrganic chemistryBiologyHealth & medicineElectrical engineeringCosmology & astronomyComputingComputer programmingComputer scienceHour of CodeComputer animationArts
In practice, researchers employ a mix of the above guidelines. The presence of the square root in the formula means that quadrupling the sample size will only half the margin of error.A Few ExamplesTo make sense of the formula, let’s look For example, suppose we wanted to know the percentage of adults that exercise daily. The margin of error for the difference between two percentages is larger than the margins of error for each of these percentages, and may even be larger than the maximum margin
Test Your Understanding Problem 1 Nine hundred (900) high school freshmen were randomly selected for a national survey.
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.sceneadvisor.com/Alabama/margin-of-error-equation-statistics.html", "fetch_time": 1566598282000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-35/segments/1566027319082.81/warc/CC-MAIN-20190823214536-20190824000536-00039.warc.gz", "warc_record_offset": 314372541, "warc_record_length": 8416, "token_count": 1469, "char_count": 6535, "score": 3.6875, "int_score": 4, "crawl": "CC-MAIN-2019-35", "snapshot_type": "latest", "language": "en", "language_score": 0.9057639837265015, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00006-of-00064.parquet"}
|
### Understanding the Fibonacci Sequence and Golden Ratio
The Fibonacci Sequence
The Fibonacci sequence is possibly the most simple recurrence relation occurring in nature. It is 0,1,1,2,3,5,8,13,21,34,55,89, 144… each number equals the sum of the two numbers before it, and the difference of the two numbers succeeding it. It is an infinite sequence which goes on forever as it develops.
The Golden Ratio/Divine Ratio or Golden Mean -
The quotient of any Fibonacci number and it’s predecessor approaches Phi, represented as ϕ (1.618), the Golden ratio. The Golden Ratio is best understood geometrically by the golden rectangle. A rectangle unevenly divided resulting into one square and one rectangle, the square’s sides would have the ratio of 1:1, and the new rectangle would be exactly proportionate to the original rectangle – 1:1.618.
This iteration can continue both ways, infinitely. If you plot a quarter circle inside each of the squares as they reiterate, the golden spiral is formed. The golden spiral is possibly the most simple mathematic pattern that occurs in nature like shells of snails, sea shells, horns, flowers, plants. Numbers are only what we use to organize quantitative information.
The Golden Ratio can be applied to any number of geometric forms including circles, triangles, pyramids, prisms, and polygons. The golden ratio is formed by thirds within thirds, sixths, the connection between two and three, including every even and odd number itself. The ratio itself represents the transcendence of numbers, understanding our world is not numbers, but what numbers represent. Through the spiral, the ratio illustrates how the numbers, all quantities, are quality. Eventually, all quality can be represented through quantity. Properties qualitative and quantitative are just labels of information, our gathered indisputable fact.
If you graph any number system, eventually patterns appear. In mathematics, numbers and their patterns do not only continue infinitely linear, but in all directions. For example, considering infinite decimal expansion, even the shortest segments have an infinite amount of points.
Our universe and the numbers not only go on infinitely linear, but even it’s short segments have infinite points.
The golden ratio is not the only mathematical pattern that reaches infinity, there are many other patterns as well that reach infinity. Knowing this, ask yourself, how could infinity occur twice? If something were to happen infinitely, how could it happen twice? The answer is simple, infinity represents what is eternal, what is truly whole. For example, if infinity were to be used as a variable in mathematics like all other numbers, it would be denoted as 1∞, 2∞, 3∞, 4∞, etc.
The oneness of everything factual is what you know, what you perceive, what you are aware of, is all the universe looking at itself. This is the universe, even you are the universe, us and everything we know is all the same thing.
Since the numbers are everywhere, everything is a part of a pattern. Reflections of reflections, wheels within wheels. Life itself is a Fractal. (Another video on the Egyptian ‘Temple of Man’ is a symbolic representation of the Fibonacci series of balanced organic expansion and an expression of the underlying principles of life in the universe)
Thanks to Fractal Enlightenment!
http://www.spiritscienceandmetaphysics.com/understanding-the-fibonacci-sequence-and-golden-ratio/
You need to be a member of Ashtar Command - Spiritual Community to add comments!
• So true, Hellen...glad you enjoyed the video...it's all so fascinating.
• I love this subject too, Stick...I find it utterly fascinating. Thank you for mentioning the Drunvalo Melchizedek books...the more that sheds light on our origins the better for all of us. We need to open our eyes and see what's truly right in front of them. ;)
• ~Love this subject, Avatar... thanks for posting. I've studied with Drunvalo Melchizedek & consider his 'Flower of Life' books to be a great entry point for those wishing to 'remember' more about their true origins. Sacred Geometry is a profound gateway to deepening our understanding of this experience we call, Life. Seek & you shall find, Ashtar Massive ;-) ~InLight555
( ( ( It's ALL Happening Right Now!!! ) ) )
• Thank you for sharing this info, Drekx and I wish you would do it more often and even though I'm sure you have in the past we are gaining new members everyday who need to be made aware of this valuable and important information. You have so much to offer that's of value to us.
I am a great respecter of P'taah who is very loving and cares deeply for humans...what a great inspiration he is for me and many others. I once asked him for a sign in the night sky and he provided it which humbled and encouraged me to know he was listening and there for me.
It's sad what 'they've' done to keep these beneficial devices from us but it will not always be this way and one day soon the earth will be changed for the better.
• Thanks Avatar.....It always brings tears of joy to my eyes when I can at the very least, inspire and assist my fellows...You will always find me a great advocate for aetheric energy, as I have been so honoured to be taught in it's application myself....It is a humbling science, which is so amazingly different to what we know as commonplace technology.......It feels right when I am permitted to utilise such devices, in service to the divine plan....In the 1970s, the Plejaren Federation attempted to present aetheric energy devices for public scrutiny, via their genuine contactee, Billy Meier.....Unfortunately, friend Billy and his beloved family, became targets for assassination, by the cabal and their agents.....thus, henceforth, our Plejaren collegues have advised the Sirians (who aquired the responsibility for earth from Ptaah) to only allow the loan of such special devices....As to keep an aetheric energy device at home, or at a business location, can lead to a black ops raid....or execution of Ground Crew/contactees....
I will provide a link, which illuminatingly describes some of our rew work.....But from the perspectives of those who are not fully aware of what our GFL ground crew purpose is....They describe it as political and mysterious....maybe it is, to some degree... ;-) But I'll share it with anyone who opens their minds to aetheric energy tech applications....which is the future science of earth....and the only science.
Kind regards, Drekx
http://www.enterprisemission.com/Norway-Message3.htm
• Unfortunately, Thori many of us have had unsatisfying relationships. I never until just recently thought of my twin flame but now do...I'm so happy for those who find theirs. I agree it's always best to take a step back and re-evaluate our relationships and try to figure out what has gone wrong and what we need to do to make them work. It takes a while for many of us to figure out who or what we really want in this life...it's often better to get to know ourselves first but it's not until we're older that we figure that out.
Nice getting to know you, Thori...<3
• Very interesting, Drekx. I agree that any advancement that would allow us more efficient energy and of course, healing modalities would be confiscated...the dark cannot allow that. However, I have complete faith that one day soon that will all change.
Great comment, Drekx...thanks!
• Sounds very romantic, Feather. ;) My guy was a whiz in math and since I'm not much for math I was very impressed. LOL
• Actually I'm always very happy when you guys bring up the Fibonacci series, the golden mean phi, divine proportion and sacred harmonics.......This is the very portal that connects the dense physical plane with the higher aethers.....It actually enables us to use a higher science, which is in effect based upon the LIFE FORCE itself....Namely, VRIL, ORGONE, PRANA.....I usually describe this simply as aetheric energy.....and it is the polar opposite of the current combustion tech, we see everywhere on surface earth at present...
I would like to state, that aetheric energy devices are currently prohibited by the dark cabal and if any individual were to possess such a device, usually through scientific investigations and re-discoveries made, then they would not be permitted a patent, and their notes would automatically be confiscated under "national security" laws...
However, the GFL do possess aetheric energy devices and do loan them out to GFL Ground Crew for specific purposes only....They are not permitted to be retained by said crew, because it would be a threat to us and also our familes...
YET, we can be taught the principles and operational applications of this tech and it is superb...It is based on the sacred harmonics of nature and links electro-magnetism with GRAVITY.....
As a means for propulsion and numerous other effective applications.....It is SAFE & CLEAN, as it is not entropic....at is based upon life force, so is the polar opposite of entropic decay....
You could call it a polar opposite in scientific principles.....
Radiation and combustion inherent in the science of today, furnish us with such applications as the nuclear reactor, the rocket/missile, the internal combustion engine and many others....
This is the science of entropy...decay, friction, centrigugal, destruction and degradation.....We do see it in the universe, when stars decay.......The energy which sustains stellar, planetary and human life, is actually hidden from the scientific model known......
It is the life principle and based upon the harmonic centripetal spiral...which evades friction and offers the line of least resistance in motion...
I will encourage all to study the works of Viktor Schauberger...an Austrian scientist who re-discovered this principle, among several others...
• Well, Feather, I must admit the fibonacci spiral 'is' quite impressive. LOL
This reply was deleted.
### Serious Copyright infringement from an Author (important)
Hi all!
Today, i have some sad news =(
I received a serious © infringement notice from someone that is involved with a certain channeling website.
This notice was SENT via NING…
Views: 192
Views: 935
## Introduction
Greetings, Dear Family!,…
Views: 3631
### Ashtar – “Who is Ashtar? Mission and Purpose”
Greetings Family,
My…
Views: 1618
Copyright Policy: Always Include 30-50% of the source material and a link to the original article. You may not post, modify, distribute, or reproduce in any way any copyrighted material, trademarks, or other proprietary information belonging to others without obtaining the prior written consent of the owner of such proprietary rights. If you believe that someone's work has been copied and posted on Ashtar Command in a way that constitutes copyright infringement, please Contact Us and include the links to these pages and relevant info.
### Latest Activity
Krishna Kalki replied to Krishna Kalki's discussion Covid-19 Vaccines Are Poison! - Karen Kingston, Top Pharmaceutical Analyst - Must Video
"Dangerous stuff"
3 hours ago
motherbg shared their blog post on Facebook
5 hours ago
James McConnell posted a discussion
ANCIENT AWAKENINGS Sunday Call 9/12/2021 (Archangel Michael, OWS, Shoshanna)James & JoAnna McConnell A GLORIOUS VISTA SHALL OPEN BEFORE YOU Arch Angel Michael and One Who Serves channeled by James McConnellShoshanna - Joanna’s Higher Self These…
6 hours ago
motherbg shared their blog post on Facebook
6 hours ago
Drekx Omega replied to Krishna Kalki's discussion Covid-19 Vaccines Are Poison! - Karen Kingston, Top Pharmaceutical Analyst - Must Video
"No big pharma contract can ever beat the Human immune system.....I never took the shot, never will take the shot, nor any offered booster, or any offered bribe....I remain well...My health is perfect.....It's an attitude of mind, as well as…"
6 hours ago
Drekx Omega left a comment on Comment Wall
"GB News - Mark Dolan slams Macron's outburst against AUKUS alliance.....
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.ashtarcommandcrew.net/profiles/blogs/understanding-the-fibonacci-sequence-and-golden-ratio", "fetch_time": 1632029596000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00484.warc.gz", "warc_record_offset": 690237107, "warc_record_length": 40247, "token_count": 2628, "char_count": 12025, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2021-39", "snapshot_type": "latest", "language": "en", "language_score": 0.9205794334411621, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00040-of-00064.parquet"}
|
101 102 CHAPTER 4 APPLICATIONS OF THE NAVIERSTOKES EQUATIONS 411 Equations of
# 101 102 chapter 4 applications of the navierstokes
• 164
This preview shows page 107 - 110 out of 164 pages.
101
102 CHAPTER 4. APPLICATIONS OF THE NAVIER–STOKES EQUATIONS 4.1.1 Equations of fluid statics The equations of fluid statics are usually derived by a simple force balance under the assumption of mechanical equilibrium (pressure independent of time in a motionless— i.e. , static—fluid) for the fluid system. Here, however, we obtain them as a special case of the N.–S. equations. This is actually much easier; and now that we have these equations available, such an approach, among other things, emphasizes the unity of fluid mechanics in general as embodied in the N.–S. equations. Thus, we begin by writing the 3-D incompressible N.–S. equations in a Cartesian coordinate system with the positive z direction opposite that of gravitational acceleration. This gives u x + v y + w z = 0 , (4.1a) u t + uu x + vu y + wu z = 1 ρ p x + ν u , (4.1b) v t + uv x + vv y + wv z = 1 ρ p y + ν v , (4.1c) w t + uw x + vw y + ww z = 1 ρ p z + ν w g . (4.1d) Now because the fluid is assumed to be motionless, we have u = v = w 0, implying that all derivatives of velocity components also are zero. In particular, there can be no viscous forces in a static fluid. Clearly, the continuity equation is satisfied trivially, and the momentum equations collapse to p x = 0 , (4.2a) p y = 0 , (4.2b) p z = ρg = γ , (4.2c) where γ is the specific weight defined in Chap. 2. The first two of these equations imply that in a static fluid the pressure is constant throughout planes aligned perpendicular to the gravitational acceleration vector; i.e. , p x = 0 implies p is not a function of x , and similarly for y . We now integrate Eq. (4.2c) to formally obtain p ( x, y, z ) = γz + C ( x, y ) , where C ( x, y ) is an integration function . (When integrating a partial differential equation we usually obtain integration functions instead of integration constants.) But we have already noted that Eqs. (4.2a) and (4.2b) imply p cannot be a function of either x or y ; so in this special case C is a constant, and we then have p ( z ) = γz + C . (4.3) We remark here that the first liquid considered in this context was water, and as a result the contribution γz = ρgz to the pressure in a static fluid is often termed the hydrostatic pressure . The value of C must now be determined by assigning a value to p , say p 0 , at some reference height z = h 0 . This leads to p 0 = p ( h 0 ) = γh 0 + C C = p 0 + γh 0 , and from this it follows that p ( z ) = p 0 + γ ( h 0 z ) . (4.4) This actually represents the solution to a simple boundary-value problem consisting of the first- order differential equation (4.2c) and the boundary condition p ( h 0 ) = p 0 , and we will emphasize this viewpoint in the sequel as we introduce several examples to demonstrate use of the equations of fluid statics.
4.1. FLUID STATICS 103 Pascal’s Law One of the first results learned about fluid statics in high school physics classes is Pascal’s law . We will state this, provide physical and mathematical explanations of it, and consider an important application.
#### You've reached the end of your free preview.
Want to read all 164 pages?
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern
Stuck? We have tutors online 24/7 who can help you get unstuck.
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
Ask Expert Tutors
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.coursehero.com/file/pb7cch4/101-102-CHAPTER-4-APPLICATIONS-OF-THE-NAVIERSTOKES-EQUATIONS-411-Equations-of/", "fetch_time": 1582984570000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-10/segments/1581875149238.97/warc/CC-MAIN-20200229114448-20200229144448-00059.warc.gz", "warc_record_offset": 688909842, "warc_record_length": 78603, "token_count": 1137, "char_count": 4427, "score": 3.53125, "int_score": 4, "crawl": "CC-MAIN-2020-10", "snapshot_type": "latest", "language": "en", "language_score": 0.854377031326294, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00046-of-00064.parquet"}
|
0 like 0 dislike
Which expression is equivalent to n2 + 26n + 88 for all values of n?
(n + 8)(n + 11)
(n + 4)(n + 22)
(n + 4)(n + 24)
(n + 8)(n + 18)
0 like 0 dislike
(n + 4)(n+ 22)
Step-by-step explanation:
One way to determine which factored expression is accurate is by asking the question: "Which two numbers multiply to 88, but add to 26?" If you are having difficulty using this method, you can always try the guess-and-check method with a multiple choice question. This way involves multiplying the terms inside one set of parentheses with the terms inside the other set of parentheses.
Step #1: Multiply the first terms
(n + 4)(n + 22) ----> n x n = n²
Step #2: Multiply the inside terms
(n + 4)(n + 22) -----> 4 x n = 4n
Step #3: Multiply the outside terms
(n + 4)(n + 22) -----> n x 22 = 22n
Step #4: Multiply the last terms
(n + 4)(n + 22) -----> 4 x 22 = 88
Step #5: Write the new expanded equation and simplify
(n + 4)(n + 22) -----> n² + 4n + 22n + 88 -----> n² + 26n + 88
by
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://askthetask.com/111500/which-expression-equivalent-26n-for-all-values-of-11-22-24-18", "fetch_time": 1679455364000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-14/segments/1679296943749.68/warc/CC-MAIN-20230322020215-20230322050215-00410.warc.gz", "warc_record_offset": 150717726, "warc_record_length": 7407, "token_count": 332, "char_count": 1003, "score": 4.46875, "int_score": 4, "crawl": "CC-MAIN-2023-14", "snapshot_type": "latest", "language": "en", "language_score": 0.7551628947257996, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00021-of-00064.parquet"}
|
# Describe how the concepts from limits, continuity of functions, an intermediate value of Theorem, and vertical asymptotes can be applied in this scenario.
This assignment is for Calculus I.Please review the below requirements and the attached graph for scenario#.CALCULUS I – ASSIGNMENT REQUIREMENTS:Scenario#1 (2 Pages, 1 APA citation)A curvewith equation y=−(x−1)^2 + 4 andsome rectangles shaded underneath. (Supposewe want to estimate the area under the curve y=−(x−1)^2+ 4 on a certain interval using the rectangles provided.Choose your own intervalbased on scenario #1 above and address the following:MUST RESPOND TO EACH POINT SEPARATELY1. Describehow the concepts from limits, continuity of functions, an intermediate value of Theorem, and vertical asymptotes can be applied in this scenario.2. Estimatethe area under the curve y=−(x−1)^2+4on the intervalfrom x=0 to x=3 using the rectangles provided (see attached graph).3. Provideanother example of a scenario that involves the same concept.
Scenario#2 (2 Pages, 1 APA citation)Avehicle is driving along a road. Its position function is given by a functions(t), where s is measured in feet and t is measured in seconds. Createyour own function based on scenario#2 and address the following:MUST RESPOND TO EACH POINT SEPARATELY1.Draw a graph or figure to represent this situation.2.Describe how the concepts “Derivative of a function at a point using limits, the slope of a tangent line, piecewise function, and higher-order derivative of a function” can be applied in scenario#2.3.Find the instantaneous velocity of the vehicleat seconds.4.Provide another example of a scenario that involves the sameconcept.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://hirequalitywriter.com/describehow-the-concepts-from-limits-continuity-of-functions-an-intermediate-value-of-theorem-and-vertical-asymptotes-can-be-applied-in-this-scenario/", "fetch_time": 1660556113000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-33/segments/1659882572163.61/warc/CC-MAIN-20220815085006-20220815115006-00180.warc.gz", "warc_record_offset": 293675429, "warc_record_length": 9619, "token_count": 387, "char_count": 1671, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2022-33", "snapshot_type": "latest", "language": "en", "language_score": 0.8348302245140076, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00050-of-00064.parquet"}
|
# What is an algorithm?
An algorithm is a sequence of steps to solve a problem. Its purpose is to find the solution to a particular problem. The scope of an algorithm is not limited to mathematical or logical concepts.
Algorithms can be thought of as a procedure to a solve problem. Listed below are the criteria for a procedure to be an algorithm:
1. The procedure must ultimately provide the solution to the problem being addressed. It should be finite.
2. Each step in the sequence must be clear-cut and understandable.
3. The algorithm must be effective.
Let’s learn by example. Here’s a trivial and simple example of an algorithm to make toast:
2. Remove toaster from drawer
4. Turn on and sent temperature / timer
5. When toast pops up add butter
6. Eat and enjoy
When it comes to finding the solution to a problem, there can be multiple approaches. For instance, to find the GCD (Greatest Common Divisor) of two numbers, we can use the Euclidean algorithm, the Lehmer’s GCD algorithm and many other methods. Since a given problem can be solved in multiple ways, people often spend a great deal of time comparing algorithms to determine which one works best in a given situation.
In their book, Algorithms for Dummies, John Mueller and Luca Massaron provide a list of some common use cases of algorithms:
1. Searching
eg: finding a particular web page in the sea of web pages on the internet
2. Sorting
Sorting information or data is arranging in an ordered sequence.
eg: when you arrange items on e-commerce sites in increasing order of price
3. Transforming:
Conversion of one sort of data to another
eg: Shunting Yard Algorithm is used to convert an infix expression to postfix expression
4. Scheduling
According to Wikipedia, scheduling is the method by which work is assigned to resources that complete the work. If it weren’t for scheduling algorithms, then your operating system wouldn’t have been able to efficiently allocate resources for specific tasks at the right times or run multiple tasks at the same time.
eg: in the Shortest Remaining Time algorithm, the process will be allocated to the task which is closest to its completion
5. Graph Analysis
A graph is a non-linear data structure comprised of nodes and edges (which connect a pair of nodes). Graphs have innumerable real-life applications: social networks, recommendation engines, etc.
eg: The breadth-first-search algorithm is used to find the shortest path between two nodes. An application of this could be the shortest route between a source and destination on Google Maps.
6. Cryptography
Encryption is a process of encoding messages to keep them secret, so only “authorized” parties can access the message. This is an important concept for the safe transfer of data between systems and protection against attacks and malicious penetration. Algorithms exist which analyze information and encrypt/decrypt the information.
eg: Message Authentication Codes (MACs) can be used in providing authentication for the origin/source and integrity of messages.
7. Pseudorandom Number Generation
These algorithms use mathematical formulas to produce sequences of random numbers.
eg: electronic games and simulations
This list is a tiny glimpse into the types of algorithms out there. New algorithms are made everyday and existing algorithms are continuously modified. The job of computer programmers is to identify which algorithms are suited for their projects or work.
*************************
Important resource: Here’s a list of algorithms: https://en.wikipedia.org/wiki/List_of_algorithms
Want to join on my learning journey? Feel Free to subscribe!
This site uses Akismet to reduce spam. Learn how your comment data is processed.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://aarushinema.com/2021/01/03/what-is-an-algorithm/?shared=email&msg=fail", "fetch_time": 1618847576000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618038887646.69/warc/CC-MAIN-20210419142428-20210419172428-00611.warc.gz", "warc_record_offset": 192541853, "warc_record_length": 56395, "token_count": 742, "char_count": 3720, "score": 4.3125, "int_score": 4, "crawl": "CC-MAIN-2021-17", "snapshot_type": "latest", "language": "en", "language_score": 0.909996747970581, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00045-of-00064.parquet"}
|
# If f is a continuous real valued function and |f| is constant then show that f is constant.
If $f$ is a continuous real valued function and $|f|$ is constant, then show that $f$ is constant.
I tried using Intermediate Value Theorem for the same, but I'm unable to proceed.
Kindly help.
• Suppose $|f| = c > 0$. Now suppose there were $a, b$ such that $f(a) = c, f(b) = -c$. Can you derive a contradiction? Jan 11, 2015 at 13:12
## 2 Answers
If $|f|=0$ then $f=0$ is constant.
If $|f|=c>0$ then $f$ does not take the value $0$. Since a continuous function taking both positive and negative values also takes the value $0$ by the intermediate value theorem, this means that $f$ either is positive everywhere or negative everywhere. In the first case, we find $f=c$, in the second case we find $f=-c$. In conclusion, $f$ is constant.
$\exists c \in \mathbb{R} \text{ s.t. } \forall x \in \mathbb{R}, \left|f(x)\right|=c$
$$\implies \forall x \in \mathbb{R}, f(x) = c \text{ or } - c$$
If $c=0$ then $f$ constant-valuated. Else suppose that $f$ is not constant. Then :
$$\exists x_0 \in \mathbb{R} \text{ s.t. } \lim\limits_{x_0^-}f \neq \lim\limits_{x_0^+}f$$
Which contradicts continuity.
OR
Using the intermediate value theorem (even just Cauchy theorem), you can say that that if $f$ isn't constant then : $$\exists(x_0,x_1)\in\mathbb{R}^2 \text{ s.t. } f(x_o) = c \text{ and } f(x_1) = -c$$ The intermediate value theorem then allows you to state that : $$\exists{x_c}\in]\min{(x_0,x_1), \max{(x_0,x_1)}[} \text { s.t. } f(x_c)=0$$ And then $\left|f(x_c)\right| = 0 \neq c$ which contradicts the hypothesis.
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/1099974/if-f-is-a-continuous-real-valued-function-and-f-is-constant-then-show-that-f-i", "fetch_time": 1708757370000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00388.warc.gz", "warc_record_offset": 388249283, "warc_record_length": 34776, "token_count": 539, "char_count": 1624, "score": 4.34375, "int_score": 4, "crawl": "CC-MAIN-2024-10", "snapshot_type": "latest", "language": "en", "language_score": 0.7625232934951782, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00027-of-00064.parquet"}
|
# What happens when a ball is thrown vertically in the air?
• stephi
In summary, the conversation discusses the question of which quantities are zero when a ball reaches maximum height after being thrown vertically in the air. The answer is that both speed and velocity are zero at the top, as velocity is the vector quantity and speed is its magnitude. The conversation also mentions the possibility of asking for help with a difficult math problem.
## Homework Statement
A ball is thrown vertically in the air, upward from the surface of the Earth. Which of the following quantities are zero when the ball reaches maximum height?
a)speed b)velocity c)acceleration
## The Attempt at a Solution
i answered just b, but i got it marked wrong so i am unsure. any thoughts?
because when the ball reached maximum height, the ball stops and then changes direction, so then it would have v=0 and the top. but would the speed be 0 as well?
You tell me. What's the relationship between speed and velocity?
velocity has to have direction while speed does not.
stephi said:
velocity has to have direction while speed does not.
Good. Speed is just the magnitude of the velocity. So how would you revise your first answer?
both speed and velocity will be zero?! :)
stephi said:
both speed and velocity will be zero?! :)
Of course.
thank you so much!
is there any chance you could also help me with a math problem that is giving me lots of trouble?
stephi said:
is there any chance you could also help me with a math problem that is giving me lots of trouble?
Did you post it?
yes but no replies yet,.
## 1. What is the initial velocity of the ball when thrown vertically in the air?
The initial velocity of the ball when thrown vertically in the air is the speed at which it is thrown upwards. This velocity is dependent on the force applied by the person throwing the ball and the mass of the ball itself.
## 2. What is the maximum height the ball can reach when thrown vertically in the air?
The maximum height the ball can reach when thrown vertically in the air is determined by the initial velocity and acceleration due to gravity. At the highest point, the velocity of the ball will be zero before it starts to fall back down.
## 3. How long does it take for the ball to reach its maximum height when thrown vertically in the air?
The time it takes for the ball to reach its maximum height when thrown vertically in the air is dependent on the initial velocity and acceleration due to gravity. It can be calculated using the formula t = v/g, where t is time, v is initial velocity, and g is the acceleration due to gravity.
## 4. What is the acceleration of the ball when thrown vertically in the air?
The acceleration of the ball when thrown vertically in the air is due to the force of gravity pulling it downwards. This acceleration is constant and is equal to 9.8 m/s^2 on Earth.
## 5. What is the final velocity of the ball when it hits the ground after being thrown vertically in the air?
The final velocity of the ball when it hits the ground after being thrown vertically in the air is dependent on the initial velocity and the distance it has fallen. It can be calculated using the formula v^2 = u^2 + 2as, where v is final velocity, u is initial velocity, a is acceleration due to gravity, and s is the distance fallen.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/what-happens-when-a-ball-is-thrown-vertically-in-the-air.470424/", "fetch_time": 1708902023000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-10/segments/1707947474643.29/warc/CC-MAIN-20240225203035-20240225233035-00538.warc.gz", "warc_record_offset": 963909631, "warc_record_length": 16663, "token_count": 714, "char_count": 3346, "score": 4.15625, "int_score": 4, "crawl": "CC-MAIN-2024-10", "snapshot_type": "latest", "language": "en", "language_score": 0.9611231088638306, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00053-of-00064.parquet"}
|
Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (anonymous):
Without dividing, do you expect 7/8 divided by 11/16 to be greater than or less than 1? Explain.
OpenStudy (anonymous):
(7/8)/(11/16), using the chain rule = = (7*16)/(8*11) = 112/88 > 1
OpenStudy (hoblos):
7/8 > 11/16 => 7/8 divided by 11/16 must be greater than 1
OpenStudy (anonymous):
Well, it says without actually dividing. The answer should be as follows. 7/8 is almost 1. 11/16 is significantly below 1. Dividing anything close to one by something significantly below 1 is likely to result in a number greater than 1.
OpenStudy (anonymous):
without dividing, doesnt mean you cant multiply. Use the chain rule as i say. It's the best way for you to make sure
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
brianagatica14: Any artist on here?
1 day ago 7 Replies 2 Medals
Uniful3: Art dump..
5 hours ago 18 Replies 1 Medal
mimix: hii i miss everyone
4 hours ago 5 Replies 1 Medal
hellokitty265: Art
3 hours ago 14 Replies 1 Medal
TarTar: Any Songwriters here???
1 day ago 26 Replies 3 Medals
YouMyPlug: Made some art to, took me 2 years to finish, ud83dudc4cud83dude29
4 hours ago 15 Replies 3 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://questioncove.com/updates/4ea8598ee4b0a7d51421fede", "fetch_time": 1721887381000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00687.warc.gz", "warc_record_offset": 408468284, "warc_record_length": 5354, "token_count": 431, "char_count": 1556, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.9192426800727844, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00045-of-00064.parquet"}
|
14209an8.1-3
# 14209an8.1-3 - c Kendra Kilmer Section 8.1 Functions of...
This preview shows pages 1–3. Sign up to view the full content.
c Kendra Kilmer April 23, 2009 Section 8.1 - Functions of Several Variables Definition: An equation of the form z = f ( x , y ) describes a function of two independent variables if for each permissible ordered pair ( x , y ) , there is one and only one value of z determined by f ( x , y ) . Example 1: Given f ( x , y ) = 2 x 2 - 3 xy + y 2 - 4, find the following: a) f ( 3 , 0 ) b) f ( 1 , 2 ) c) f ( 4 , 1 ) Example 2: A company manufactures ten and three speed bicycles. The weekly demand and cost equations are p = 230 - 9 x + y , q = 130 + x - 4 y , C ( x , y ) = 200 + 80 x + 30 y where \$ p is the price of a ten speed bicycle, \$ q is the price of a three speed bicycle, x is the weekly demand for ten speed bicycles, y is the weekly demand for three speed bicycles, and C ( x , y ) is the cost function. a) Find the weekly revenue function and R ( 10 , 15 ) . b) Find the weekly profit function and P ( 10 , 15 ) 1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
c Kendra Kilmer April 23, 2009 Definition: The Cobb-Douglas production function is defined as f ( x , y ) = kx m y n where k , m , and n are positive constants with m + n = 1. Economists use this function to describe the number of units f ( x , y )
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.coursehero.com/file/6861389/14209an81-3/", "fetch_time": 1529449938000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00374.warc.gz", "warc_record_offset": 791921721, "warc_record_length": 51220, "token_count": 649, "char_count": 2456, "score": 3.9375, "int_score": 4, "crawl": "CC-MAIN-2018-26", "snapshot_type": "latest", "language": "en", "language_score": 0.8446406722068787, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00002-of-00064.parquet"}
|
Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
You are not logged in.
## #1 2005-09-14 14:59:37
daisy
Guest
### Help in Math Induction Question
prove that for every non-negative integer n , n^3 mod 6 = n mod 6.
(use mathematical induction)
## #2 2005-09-14 16:16:20
ganesh
Moderator
Registered: 2005-06-28
Posts: 21,812
### Re: Help in Math Induction Question
Put n=7.
7mod6=1
7^3mod6=343mod6=1
Put n=8
8mod6=2
8^3mod6=512mod6=2.
Assume this is true for k.
Therefore, kmod6=k^3mod6.
Try for k+1.
Lets say (k+1)mod6=m
(k+1)^3mod6 = (k^3 + 3k^2+3k+1)mod6.
= (k^3+3k^2+2k+k+1)mod6= (k^3+2k^2+k^2+2k+k+1)mod6
= [k(k^2+2)+k(k+2)+k+1]mod6...
Running out of time...gotta leave....
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
Offline
## #3 2005-09-14 16:43:41
John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588
### Re: Help in Math Induction Question
P(n+1) is (n^3 + 3n^2 + 3n + 1)%6 = (n+1)%6
and assume we start with n^3%6 = n%6, where % means mod (like in C language).
(n^3 + 3n^2 + 2n)%6 + (n+1)%6 = (n+1)%6
(n^3 + 3n^2 + 2n)%6 = 0
Substitute n%6 in place of n^3%6 and get:
(3n(n+1))%6=0
For even numbers, 3n is a multiple of 6, so that works.
For odd numbers, the (n+1) part is even, so that works.
igloo myrtilles fourmis
Offline
## #4 2005-09-15 02:04:44
ganesh
Moderator
Registered: 2005-06-28
Posts: 21,812
### Re: Help in Math Induction Question
ganesh wrote:
Put n=7.
7mod6=1
7^3mod6=343mod6=1
Put n=8
8mod6=2
8^3mod6=512mod6=2.
Assume this is true for k.
Therefore, kmod6=k^3mod6.
Try for k+1.
Lets say (k+1)mod6=m
(k+1)^3mod6 = (k^3 + 3k^2+3k+1)mod6.
= (k^3+3k^2+2k+k+1)mod6= (k^3+2k^2+k^2+2k+k+1)mod6
= {[k²(k+2)+k(k+2)]+k+1}mod6
= {[(k²+k)(k+2)] +k+1}mod6
={[(k(k+1)(k+2) + k+1}mod6
We know that k(k+1)(k+2) is divisble by 6 for any k>1, k∈N,
Hence the above is reduced to
(k+1)mod6
It is seen that it is true for k+1, hence, it is true for any value of k.
q.e.d
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
Offline
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.mathisfunforum.com/viewtopic.php?pid=14333", "fetch_time": 1485072715000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-04/segments/1484560281419.3/warc/CC-MAIN-20170116095121-00351-ip-10-171-10-70.ec2.internal.warc.gz", "warc_record_offset": 574195661, "warc_record_length": 3664, "token_count": 1002, "char_count": 2361, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2017-04", "snapshot_type": "latest", "language": "en", "language_score": 0.7067350745201111, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00044-of-00064.parquet"}
|
# Do non-symmetric “strongly positive definite” matrices have unique positive definite square roots?
It is well known that if $$A$$ is a symmetric positive definite matrix, then it has a unique square root which is positive definite. My question is whether this result extends to a strongly positive definite nonsymmetric matrix.
More precisely, let $$A$$ be a real nonsymmetric $$n\times n$$ matrix, which satisfies the following strong positive definite condition: there exists $$a>0$$ such that for each $$x\in\mathbb R^n$$, the estimate $$\langle Ax, x\rangle\geq a|x|^2$$ holds. Is it true then that there exists a unique (edit: strongly positive definite) matrix $$B$$ such that $$B^2=A$$? I would be very interested in knowing the answer to this result, and reference to a proof. Thanks!
• Did you mean for $B$ to be strongly positive definite as well? It’s not hard to construct a $2\times2$ matrix $A$ that satisfies your conditions but doesn’t have a unique square root. – amd Jul 16 '19 at 0:27
• Sure, thanks for your comment. I want B to be strongly positive definite too. – Lentes Jul 16 '19 at 15:00
Your condition "$$A$$ strongly PD" is equivalent to "$$A+A^T$$ is symmetric $$>0$$". According to,
Largest eigenvalues of matrix and its doubled symmetric part
every $$\lambda\in spectrum(A)$$ satisfies $$Re(\lambda)>0$$.
Thus $$A$$ admits $$\log(A)$$ as its principal logarithm and the principal square root $$A^{1/2}$$ is well defined; cf the first part of my post in
When is square root of transpose and transpose of square root of a matrix are equal?
Moreover, $$A^{1/2}$$ is the unique square root of $$A$$ whose all the eigenvalues have a positive real part. Thus, if $$A$$ admits a strong square root, then it is unique.
EDIT. The difficult part is to see if $$A^{1/2}+{A^{1/2}}^T$$ is $$>0$$.
That is true; cf. Corollary 8 in
https://www.sciencedirect.com/science/article/pii/S0024379500002433
cf. also
Square root of positive definite nonsymmetric matrix
where this question was studied.
• Awesome, thanks a lot for your very complete answer and for linking the paper. – Lentes Jul 16 '19 at 19:41
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/3294294/do-non-symmetric-strongly-positive-definite-matrices-have-unique-positive-defi", "fetch_time": 1611194331000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-04/segments/1610703522150.18/warc/CC-MAIN-20210121004224-20210121034224-00653.warc.gz", "warc_record_offset": 449583122, "warc_record_length": 34221, "token_count": 559, "char_count": 2137, "score": 3.53125, "int_score": 4, "crawl": "CC-MAIN-2021-04", "snapshot_type": "latest", "language": "en", "language_score": 0.888071596622467, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00001-of-00064.parquet"}
|
# Determine the angle between two intersecting lines - Counter-Clockwise direction
Question asked by jdavid05 on Aug 2, 2016
Latest reply on Aug 2, 2016 by jdavid05
Hey everyone:
I am trying to determine the angle between two intersecting lines. However, I need the angle to be from the start point of line1 to the start- or end-point of line2 - whichever is the next point going in a counter-clockwise direction around the circle's perimeter (see diagram at bottom of question).
My steps so far have been:
1) I found the intersection points and saved them to a point file
2) I created a 10cm buffer around the points (0.1 metres)
3) I clipped line1 and line2 to the buffer
4) I calculated the start- and end-points of line1 and line2 as:
with arcpy.da.UpdateCursor("line", ('X0a','Y0a','X1a','Y1a', "SHAPE@")) as cursor:
for row in cursor:
row[0] = row[4].firstPoint.X
row[1] = row[4].firstPoint.Y
row[2] = row[4].lastPoint.X
row[3] = row[4].lastPoint.Y
cursor.updateRow(row)
5) I created separate fields and calculated their values as:
dx = !X1a!-!X0a!
dy = !Y1a!-!Y0a!
dxa= !X1b!-!X0b!
dya= !Y1b!-!Y0b!
r = math.sqrt(math.pow(!dx!, 2) + math.pow(!dy!, 2))
ra = math.sqrt(math.pow(!dxa!, 2) + math.pow(!dya!, 2))
c = math.asin(abs((dxdya - dydxa))/(r*ra)) / math.pi * 180
c is the angle between the points, however, it is not always the angle I need. I need the angle from a point on line1 to the nearest point on line2 going counter-clockwise around the circle's perimeter.
There is probably a really easy solution for this, but, I just haven't been able to figure it out. Any help is greatly appreciated.
I am using ArcGIS 10.2.
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://community.esri.com/message/625409?commentID=625409", "fetch_time": 1575875161000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-51/segments/1575540518337.65/warc/CC-MAIN-20191209065626-20191209093626-00056.warc.gz", "warc_record_offset": 325385802, "warc_record_length": 25254, "token_count": 502, "char_count": 1659, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2019-51", "snapshot_type": "latest", "language": "en", "language_score": 0.8741960525512695, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00045-of-00064.parquet"}
|
^ T
« B
Main » Kinematics & Dynamics » Distance & Displacement » Online Activities I.3.1
### Position and Displacement - Online Activity I.3.1
1. Graph to describe the motion of an object along a co-ordiante line.
1. What is the position of the object at t = 0s ?
3 m 3 m to the right 3 m to the left 0 m
2. How long was the object there (part A)?
0 s 1 s 2 s 3 s
3. At what time does it reach its farthest point away from the object's starting position?
10 s 13 s 15 s 18 s
4. What is the object's displacement between t = 2s and t= 10s ?
3 m 3 m to the right 6 m to the right 10 m to the right
5. Describe the motion between t = 10s to t = 15s.
Object is moving downward. Object is moving backwards. Object is moving to the left. Object is moving to the right.
6. What is the object's displacement between t = 15s to t = 18s?
3 m 3 m to the left 3 m to the right 3m upward
7. At what point(s) does the object stop for 2 seconds?
A B, and G A, and D A, D, and F
8. Which part of the trip is the object's speed the fastest?
B C E F
2. Express the change in the following quantities.
a. T1 = 15°C and T2 = 25°C has a change of ***. - 10°C 10°C
b. W1 = 80 kg and W2 = 70 kg has a change of ***. - 10 kg 10 kg
3. Which displacements are equal?
# d1 (m) d2 (m) 1 5 8 2 7 -2 3 -5 -2 4 15 12 5 0 2 6 -5 -8
1 & 2 2 & 3 3 & 4 4 & 5 1 & 3 3 & 5
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://physics30.edcentre.ca/kindyn/onlinei_3_1.html", "fetch_time": 1553442038000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-13/segments/1552912203462.50/warc/CC-MAIN-20190324145706-20190324171706-00363.warc.gz", "warc_record_offset": 167813408, "warc_record_length": 9497, "token_count": 485, "char_count": 1352, "score": 3.625, "int_score": 4, "crawl": "CC-MAIN-2019-13", "snapshot_type": "latest", "language": "en", "language_score": 0.8757269978523254, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00052-of-00064.parquet"}
|
## Conversion formula
The conversion factor from grams to pounds is 0.0022046226218488, which means that 1 gram is equal to 0.0022046226218488 pounds:
1 g = 0.0022046226218488 lb
To convert 5.3 grams into pounds we have to multiply 5.3 by the conversion factor in order to get the mass amount from grams to pounds. We can also form a simple proportion to calculate the result:
1 g → 0.0022046226218488 lb
5.3 g → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 5.3 g × 0.0022046226218488 lb
M(lb) = 0.011684499895799 lb
The final result is:
5.3 g → 0.011684499895799 lb
We conclude that 5.3 grams is equivalent to 0.011684499895799 pounds:
5.3 grams = 0.011684499895799 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 85.583466037736 × 5.3 grams.
Another way is saying that 5.3 grams is equal to 1 ÷ 85.583466037736 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that five point three grams is approximately zero point zero one two pounds:
5.3 g ≅ 0.012 lb
An alternative is also that one pound is approximately eighty-five point five eight three times five point three grams.
## Conversion table
### grams to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from grams to pounds
grams (g) pounds (lb)
6.3 grams 0.014 pounds
7.3 grams 0.016 pounds
8.3 grams 0.018 pounds
9.3 grams 0.021 pounds
10.3 grams 0.023 pounds
11.3 grams 0.025 pounds
12.3 grams 0.027 pounds
13.3 grams 0.029 pounds
14.3 grams 0.032 pounds
15.3 grams 0.034 pounds
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://convertoctopus.com/5-3-grams-to-pounds", "fetch_time": 1620852383000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-21/segments/1620243989705.28/warc/CC-MAIN-20210512193253-20210512223253-00086.warc.gz", "warc_record_offset": 176180712, "warc_record_length": 7873, "token_count": 505, "char_count": 1709, "score": 4.09375, "int_score": 4, "crawl": "CC-MAIN-2021-21", "snapshot_type": "latest", "language": "en", "language_score": 0.8321442008018494, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00046-of-00064.parquet"}
|
Anonymous
Anonymous asked in Science & MathematicsMathematics · 5 months ago
# Covert 5π /18 into degrees ?
Precalc
Relevance
• 5 months ago
5π/18 = 50 degrees
• 5 months ago
• 5 months ago
Remember a full circle is 2π radians or 360°
Use the conversion that π radians = 180°. That means in your case you want to multiply by 180°/π
5π/18 * (180°/π)
= 5/18 * 180°
= 5 * 180°/18
= 5 * 10°
= 50°
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://ca.answers.yahoo.com/question/index?qid=20200806204842AAf3IlQ", "fetch_time": 1610726205000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00373.warc.gz", "warc_record_offset": 292986873, "warc_record_length": 18819, "token_count": 146, "char_count": 405, "score": 3.8125, "int_score": 4, "crawl": "CC-MAIN-2021-04", "snapshot_type": "latest", "language": "en", "language_score": 0.7938158512115479, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00042-of-00064.parquet"}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.