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A291788 Numbers n whose trajectory under the map k -> (psi(k)+phi(k))/2 (A291784) grows without limit. 4
%I
%S 45,48,50,55,56,60,68,69,70,72,74,75,76,77,78,80,84,85,86,87,88,90,91,
%T 92,93,94,95,96,98,99,102,104,105,108,111,112,115,116,117,118,119,120,
%U 122,123,124,126,133,134,135,136,140,141,142,143,144,145,146,147,152
%N Numbers n whose trajectory under the map k -> (psi(k)+phi(k))/2 (A291784) grows without limit.
%C See A291787 (where A291787(m) = 2*A291787(m-7) for m >= 35) for the trajectory of 45.
%C There is a similar proof that all the terms from 48 though 152 have a trajectory that merges with the trajectory of 45, and so doubles every 7 steps after a certain point. For example, the trajectory of 152 reaches 2^106*33 at step 390, is 2^107*33 at step 397, and thereafter doubles every 7 steps.- _N. J. A. Sloane_, Sep 24 2017
%Y Cf. A291784, A291785, A291786, A291787.
%K nonn,more
%O 1,1
%A _N. J. A. Sloane_, Sep 03 2017, based on data supplied by _Hans Havermann_.
%E Terms 104 to 152 added by _N. J. A. Sloane_, Sep 24 2017
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Last modified April 10 19:18 EDT 2021. Contains 342853 sequences. (Running on oeis4.) | 539 | 1,576 | {"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} | 3.046875 | 3 | CC-MAIN-2021-17 | latest | en | 0.735742 |
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# Mischa Kim
### MathWorks
Last seen: ungefär 2 månader ago Active since 2013
Networker, learner, educator
Professional Interests: Spacecraft dynamics and control, numerical analysis, simulation, animation
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#### Content Feed
View by
Properties of exponentials - symbolic algebra
Hi Luca, use syms x y z = simplify(exp(x)*exp(y))
nästan 2 år ago | 0
how to make matlab functions work not starting at the first element
Hi Naufal, how about: a = [2 5 2 4 2 7 2 2 2 5 2]; pks = findpeaks(a); mypks = pks(2:end); To be able to skip the firs...
ungefär 2 år ago | 0
Hi, how about plot(T0,y,'ro') Since this is only one single data point it is hard to see in the plot, hence I change the color...
ungefär 2 år ago | 0
Problem in displaying or printing final result.
Hi Masoud, use fprintf instead. Check out the examples in the link.
ungefär 2 år ago | 1
| accepted
why didn't the event function detect the events?
In your ode45 call you are not returning any event information. Replace with [tout,yout,te,ye,ie] = ode45(@Tq_Tj,tspan, y0,opti...
ungefär 2 år ago | 0
Basins of attraction and Newtons Method.
One problem was in the governing equations for f. I cleaned up a bit and this works now: % Governing equations f = [-7.716*1e-...
ungefär 2 år ago | 0
Empty windows in EditFields of app.designer?
ErikJon, as a workaround you could use an Edit Field (Text) instead of Edit Field (Numeric). The drawback of this workaround is ...
ungefär 2 år ago | 0
| accepted
Getting extra parameters from ODE45 and the mystery transpose
Hi Simon, to your first question: When you assign values to dx(1), dx(2), and so on you are creating a row vector. However, you ...
ungefär 2 år ago | 0
| accepted
How to solve the error "[t, z]= ode45(@observer, t_temp,z0)" in line 23.
dzdt seems to be a (3x2 ?) matrix. It needs to be a 3x1 vector to be consistent with the initial conditions, z0, which is also a...
ungefär 2 år ago | 1
| accepted
Why am i receiving "Error:Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters."? for S(K)>0
Hi, since you do not know the number of iterations it makes more sense to use a while instead of a for loop: while S(K)>0 %as l...
ungefär 2 år ago | 1
| accepted
How to count the number of different strings in a string list?
Hi, use str = {'AAA','ABB','CDE','ABB','CCR','AAA','FDR'}; numstr = numel(unique(str))
ungefär 2 år ago | 0
| accepted
How to extract value at a given time in ode 45
Hi Shubham, use an events functions, see the example below. With events functions you can identify zero crossings; in your case ...
ungefär 2 år ago | 0
| accepted
Draw a circle on matlab app designer?
Hi ErikJon, if the plot input is a matrix (e.g. your circ) "...the plot function plots the columns of the matrix (circ) versus t...
ungefär 2 år ago | 0
| accepted
Use instead syms y(t) eqn = diff(y,t,2) + 0.5*diff(y,t) + y == 3; Dy = diff(y,t); cond = [y(0) == 0.5, Dy(0) == 0]; w ...
ungefär 2 år ago | 0
confusing ode suits for solving discontinuous odes
Hi, this is because the integrator, ode45, adjusts the integration step size. Broadly speaking where the dynamics is more comple...
ungefär 2 år ago | 0
| accepted
How to Find average value for every 10 sets of a matrix
Hi Magdy, you could use something like: x = 1:1:12; % this would be your matrix A y = reshape(x,3,4) m = mean(y,1)
ungefär 2 år ago | 0
| accepted
How to Use "Ask a Question" Section Correctly?
Hi Ercan, thanks for flagging this. I have seen the same behavior this morning in Chrome, other browsers seemed to work fine. I ...
ungefär 2 år ago | 3
| accepted
ODE Event Location detecting events other than zero
Hi Hariharan, yes it is. You simply define a function that is equal to zero whenever your expression equals a certain value. Fo...
ungefär 2 år ago | 0
| accepted
How do I change the values for x-axis ?
Hi Anne-Lise, do you mean changing the tick labels? See this answer. If this does not address your question, how exactly do you ...
ungefär 2 år ago | 1
Why is the resulting time-domain curve from Euler method and ode45 not stable?
Hi Ni Made Ayu Sinta Dewi, there is nothing you are doing wrong. What you are seeing is expected behavior. This is because you a...
ungefär 2 år ago | 0
| accepted
ode45 fail with pendulum
Hi Bobby, this is the wonderful world of numerical (vs symbolic) computation. ode45 is a numerical integrator that approximates ...
ungefär 2 år ago | 0
| accepted
Changing Length of Pendulum while it is in motion
Dear Yan Koon Ang, this is possible. In the example you are referring to in your question you would have to make the time varia...
ungefär 2 år ago | 1
I am trying to plot x and y with y being the only unknown but I don’t what is wrong in my code please help
Honore, the one thing I notice is that x and y are vectors. Rewrite the y-equation using the dot operator, .^ instead of just ...
ungefär 2 år ago | 0
| accepted
Finding the initial conditions which give a specified event using ode45
Patrick, as Walter pointed out this is a BVP. However, you can and I personally would use ode45 and the so-called shooting meth...
ungefär 2 år ago | 0
Double-klick on the integrator blocks and you'll see that the default initial condition is zero (0). It is also inside the integ...
ungefär 2 år ago | 0
Seif, I believe this is what you are trying to do: %calculating velocity V3 = 10; % velocity of the piston in cm/sec %global...
ungefär 2 år ago | 0
| accepted
How to change a Function by a Variable
Ercan, here you go, for the first term: syms phi x A B eq = A*sin(phi) + B*cos(phi) eq = subs(eq,sin(phi),2*x/(1+x^2))
ungefär 2 år ago | 1
| accepted
How to write Newton method with exact number of iterations and see all digit ? for example if method is required to take exact 5 iterations of newton's method
Add a loop index, e.g. ii. Also, you set your code as a function. This means you can call it, e.g., from the command window usin...
ungefär 2 år ago | 0 | 1,715 | 6,072 | {"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} | 2.8125 | 3 | CC-MAIN-2023-06 | longest | en | 0.72851 |
https://www.excelcalcs.com/calcs/repository/Fluids/Fluid-Mechanics/orderby,1/ | 1,716,177,021,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00344.warc.gz | 686,630,822 | 11,031 | 0 30
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### Air compressor power calculator
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The program calculates air compressor power, for a given flow rate and required pressure. Program takes into account the site elevation a... | 1,454 | 6,080 | {"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} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.773613 |
https://en.m.wikibooks.org/wiki/Physics_Study_Guide/Energy | 1,726,276,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00583.warc.gz | 218,498,940 | 12,428 | # Energy
Kinetic energy is simply the capacity to do work by virtue of motion.
(Translational) kinetic energy is equal to one-half of mass times the square of velocity.
${\displaystyle \mathrm {KE} _{T}={\frac {1}{2}}m\|{\vec {v}}\|^{2}}$
(Rotational) kinetic energy is equal to one-half of moment of inertia times the square of angular velocity.
${\displaystyle \mathrm {KE} _{R}={\frac {1}{2}}I\omega ^{2}}$
Total kinetic energy is simply the sum of the translational and rotational kinetic energies. In most cases, these energies are separately dealt with. It is easy to remember the rotational kinetic energy if you think of the moment of inertia I as the rotational mass. However, you should note that this substitution is not universal but rather a rule of thumb.
Potential energy is simply the capacity to do work by virtue of position (or arrangement) relative to some zero-energy reference position (or arrangement).
Potential energy due to gravity is equal to the product of mass, acceleration due to gravity, and height (elevation) of the object.
${\displaystyle \mathrm {PE} _{g}=-m{\vec {g}}\cdot {\vec {x}}=mgy}$
Note that this is simply the vertical displacement multiplied by the weight of the object. The reference position is usually the level ground but the initial position like the rooftop or treetop can also be used. Potential energy due to spring deformation is equal to one-half the product of the spring constant times the square of the change in length of the spring.
${\displaystyle \mathrm {PE} _{e}={\frac {1}{2}}k\|{\vec {x}}-{\vec {x}}_{e}\|^{2}={\frac {1}{2}}k\|\Delta {\vec {x}}\|^{2}}$
The reference point of spring deformation is normally when the spring is "relaxed," i.e. the net force exerted by the spring is zero. It will be easy to remember that the one-half factor is inserted to compensate for finite '"change in length" since one would want to think of the product of force and change in length ${\displaystyle (k\Delta {\vec {x}})\cdot \Delta {\vec {x}}}$ directly. Since the force actually varies with ${\displaystyle \Delta {\vec {x}}}$ , it is instructive to need a "correction factor" during integration.
## Variables
K: Kinetic energy (J) m: mass (kg) v: velocity (m/s) I: moment of inertia, (kg·m2) ω: ("omega") angular momentum (rad/s) Ug: Potential energy (J) g: local acceleration due to gravity (on the earth’s surface, 9.8 m/s2) h: height of elevation (m) Ue: Potential energy (J) k: spring constant (N/m) Δx: change in length of spring (m)
## Definition of terms
Energy: a theoretically indefinable quantity that describes potential to do work. SI unit for energy is the joule (J). Also common is the calorie (cal). The joule: defined as the energy needed to push with the force of one newton over the distance of one meter. Equivalent to one newton-meter (N·m) or one watt-second (W·s). 1 joule = 1 J = 1 newton • 1 meter = 1 watt • 1 second Energy comes in many varieties, including Kinetic energy, Potential energy, and Heat energy. Kinetic energy (K): The energy that an object has due to its motion. Half of velocity squared times mass. Units: joules (J) Potential energy due to gravity (UG): The energy that an object has stored in it by elevation from a mass, such as raised above the surface of the earth. This energy is released when the object becomes free to move. Mass times height time acceleration due to gravity. Units: joules (J) Potential energy due to spring compression (UE): Energy stored in spring when it is compressed. Units: joules (J) Heat energy (Q): Units: joules (J) Spring compression (Dx): The difference in length between the spring at rest and the spring when stretched or compressed. Units: meters (m) Spring constant (k): a constant specific to each spring, which describes its “springiness”, or how much work is needed to compress the spring. Units: newtons per meter (N/m) Change in spring length (Δx): The distance between the at-rest length of the spring minus the compressed or extended length of the spring. Units: meters (m) Moment of inertia (I): Describes mass and its distribution. (kg•m2) Angular momentum (ω): Angular velocity times mass (inertia). (rad/s) | 1,044 | 4,183 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "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} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.883336 |
https://kipdf.com/statistical-inference-in-the-classical-linear-regression-model_5ab269311723dd349c811948.html | 1,618,484,929,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038084765.46/warc/CC-MAIN-20210415095505-20210415125505-00405.warc.gz | 457,016,993 | 11,251 | ## Statistical Inference in the Classical Linear Regression Model
1 September 2004 Statistical Inference in the Classical Linear Regression Model A. Introduction In this section, we will summarize the properties of e...
Author: Jody Thomas
1 September 2004 Statistical Inference in the Classical Linear Regression Model A. Introduction In this section, we will summarize the properties of estimators in the classical linear regression model previously developed, make additional distributional assumptions, and develop further properties associated with the added assumptions. Before presenting the results, it will be useful to summarize the structure of the model, and some of the algebraic and statistical results presented elsewhere. B. Statement of the classical linear regression model The classical linear regression model can be written in a variety of forms. Using summation notation we write it as yt = \$1 + \$2 xt2 + \$3xt3 + ... + gt t
(linear model)
(1)
E(gt xt1, xt2, ...xtk) = 0 t
(zero mean)
(2)
Var(gt xt1,...,xtk) = F2 t
(homoskedasticity)
(3)
E(gt gs) = 0 t
s
(no autocorrelation)
(4)
xti is a known constant
(x's nonstochastic)
(5a)
No xi is a linear combination of the other x's gt - N(0,F2)
(normality)
(5b) (6)
We can also write it in matrix notation as follows
(1)
The ordinary least squares estimator of \$ in the model is given by
(2)
The fitted value of y
and the estimated vectors of residuals ( e ) in the model are defined by
(3)
2 The variance of , (F2) is usually estimated using the estimated residuals as (4)
C. The fundamental matrices of linear regression 1. M - the residual creation matrix The residuals from the least squares regression can be expressed as
(5)
a. b. c. d. e.
The matrix MX is symmetric and idempotent. MX X = 0. e = MX , eNe = yNMX y. eNe = ,NMX ,.
2. P - the projection matrix Consider a representation of the predicted value of y
(6)
a. P is symmetric and idempotent. b. PX X = X c. PX MX = 0 3. An – The deviation transformation matrix Consider the matrix An below which transforms a vector or matrix to deviations from the mean.
3
(7)
a.
An is symmetric and idempotent.
b. c.
An MX = MX (first column of X is a column of ones) Proof: First write An in a different fashion noting that the vector of ones we called j, is the same as the first column of the X matrix in a regression with a constant term.
4
(8)
Now consider the product of An and MX (9)
From previous results, MX X = 0n×k, which implies that X`MX = 0k×n. This then implies that . Given that this product is a row of zeroes, it is clear that the entire second term vanishes. This then implies (10) D. Some results on traces of matrices The trace of a square matrix is the sum of the diagonal elements and is denoted tr A or tr (A). We will state without proof some properties of the trace operator. a. trace (In ) = n b. tr(kA) = k tr(A) c. trace (A + B) = trace (A) + trace (B)
d. tr(AB) = tr(BA) if both AB and BA are defined e. trace (ABC) = trace (CAB) = trace (BCA) The results in part e hold as along as the matrices involved are conformable, though the products may be different dimensions. We will also use Theorem 17 from the lecture on characteristic roots and vectors. A proof of this theorem is given there. Theorem 17: Let A be a square symmetric idempotent matrix of order n and rank r. Then the trace of A is equal to the rank of A, i.e., tr(A) = r(A). E. Some theorems on quadratic forms and normal variables (stated without proof) 1. Quadratic Form Theorem 1: If y ~N(:y, Ey ), then z = Cy ~ N( :z = C:y; Ez = C Ey CN ) where C is a matrix of constants. 2. Quadratic Form Theorem 2 Let the nx1 vector y~N(0, I ), then yNy - P2(n). 3. Quadratic Form Theorem 3: If y-N(0,F2I) and M is a symmetric idempotent matrix of rank m then (11) Corollary: If the nx1 vector y~N(0,I) and the nxn matrix A is idempotent and of rank m, then y'Ay - P2(m). 4. Quadratic Form Theorem 4: If y~N(0,F2I), M is a symmetric idempotent matrix of order n, and L is a kxn matrix, then Ly and yNMy are independently distributed if LM = 0. 5. Quadratic Form Theorem 5: Let the nx1 vector y~N(0,I), let A be an nxn idempotent matrix of rank m, let B be an nxn idempotent matrix of rank s, and suppose BA = 0. Then yNAy and yNBy are independently distributed P2 variables. 6. Quadratic Form Theorem 6 (Craig’s Theorem) If y~N(:, S) where S is positive definite, then q1 = yNAy and q2 = yNBy are independently distributed iff ASB = 0. 7. Quadratic Form Theorem 7 If y is a nx1 random variable and y-N(:,G) then (y - :)NG-1(y - :) P2(n). 8. Quadratic Form Theorem 8: Let y - N(0, I). Let M be a nonrandom idempotent matrix of dimension nxn (rank(M)=r # n). Let A be a nonrandom matrix such that AM = 0. Let t1 = My and let t2 = Ay. Then t1 and t2 are independent random vectors. F. Some finite sample properties of the ordinary least squares estimator in the classical linear regression model can be derived without specific assumptions about the exact distribution of the error term 1. Unbiasedness of Given the properties of the model, we can show that
is unbiased as follows if X is a nonstochastic
6 matrix of full rank
(12)
2. Variance of y We know that yt depends on the constants xt and \$, and on the stochastic error, ,t. We write this as (13) This implies that (14) Furthermore with E(gt gs) = 0 t
s, i.e., the covariance between yt and yt+s is zero, implying that (15) 3. Variance of We can determine the variance of by writing it out and then using the information we have on the variance of y and the formula for the variance of any quadratic form.
(16)
4.
is the best linear unbiased estimator of We can show that is the best linear unbiased estimator of by showing that any other linear unbiased estimator has a variance which is larger that the variance of by a positive definite matrix. The least squares estimator is given by (17) Consider another linear unbiased estimator given by form of . We can determine the restrictions on G for
. Linearity is imposed by the linear to be unbiased by writing it out as
7 follows.
(18)
The variance of
is similar to the variance of (19)
Now let D = G - C = G - (XNX)-1XN, so that G = D +C. Now rewrite the variance of
as
(20)
Now substitute in equation 20 for GX = Ik and XNGN = Ik.
and
noting that
(21)
The variance of
is thus the variance of
plus a matrix that can be shown to be positive definite.
5. Unbiasedness of s2 Given the properties of the model, we can show that s2 is an unbiased estimator of F2. First write eNe as a function of ,.
8
(22)
Now take the expected value of eNe, use the property of the trace operator that tr (ABC) = tr (BCA), and then simplify
(23)
We find the trace of M using the properties on sums, products, and identity matrices.
(24)
9 6. Covariance of
and e
Given the properties of the model, we can show that the covariance of both and e as functions of , from equations 2 and 5.
and e is zero. First write
(25)
Remember that has an expected valued of expected valued of zero as follows
because it is unbiased. We can show that e has an
(26)
We then have (27)
Now compute the covariance directly
(28)
10 G. Distribution of
given normally distributed errors
1. introduction Now make the assumption that gt - N(0,F2) or
. Given that
(29)
then y is also distributed normally because we are simply adding a constant vector to the random vector ,. The error vector , is not transformed in forming y. Given E(,) = 0, E(y) = X\$, and Var(y) = F2I, we then have (30) 2. exact distribution of We can write
as a linear function of the normal random variable y from equation 2 as follows (31)
We can find its distribution by applying Quadratic Form Theorem 1. From this theorem and . Substituting we obtain
(32)
Therefore we have (33) We can also show this by viewing
directly as a function of , and then applying the theorem.
11
(34)
H. Distribution of s2 Consider the quantity (35)
This can be written
(36)
The random variable (g/F) is a standard normal variable with mean zero and variance I. The matrix MX is symmetric and idempotent. By Theorem 3 on quadratic forms, this ratio is distributed as a P2 variable with (n-k) degrees of freedom, that is (37)
where we found the trace of MX in equation 24. Given that
, we can use information on the properties of chi-squared random
variables to find the variance of s2. First remember that the variance of a P2 variable is equal to twice its degrees of freedom, i.e., Var (P2 ( | 2,272 | 8,534 | {"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} | 3.5 | 4 | CC-MAIN-2021-17 | latest | en | 0.851306 |
http://mail-archives.apache.org/mod_mbox/mahout-user/201108.mbox/%3CCAK9GPcSPiu0RNUxHgJFgKeBoZaAxvD9mQ3dfi1QPV3VMOcTnwg@mail.gmail.com%3E | 1,524,589,546,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946807.67/warc/CC-MAIN-20180424154911-20180424174911-00091.warc.gz | 203,594,072 | 2,902 | # mahout-user mailing list archives
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From Lance Norskog <goks...@gmail.com>
Subject Re: Singular vectors of a recommendation Item-Item space
Date Thu, 11 Aug 2011 05:22:58 GMT
```A picture that might help explain the problem:
http://www.flickr.com/photos/54866255@N00/6031564308/in/photostream
On 8/10/11, Lance Norskog <goksron@gmail.com> wrote:
> Zeroing in on the topic:
>
> I have:
> 1) a set of raw input vectors of a given length, one for each item.
> Each value in the vectors are geometric, not bag-of-words or other.
> The matrix is [# items , # dimensions].
> 2) An SVD of same:
> left matrix of [ # items, #d features per item] * singular
> vector[# features] * right matrix of [#dimensions features per
> dimension, #dimensions].
> 3) The first few columns of the left matrix are interesting singular
> eigenvectors.
>
> I would like to:
> 1) relate the singular vectors to the item vectors, such that they
> create points in the "hot spots" of the item vectors.
> 2) find the inverses: a singular vector has two endpoints, and both
> represent "hot spots" in the item space.
>
> Given the first 3 singular vectors, there are 6 "hot spots" in the
> item vectors, one for each end of the vector. What transforms are
> needed to get the item vectors and the singular vector endpoints in
> the same space? I'm not finding the exact sequence.
>
> A use case for this is a new user. It gives a quick assessment by
> asking where the user is on the few common axes of items:
> "Transformers 3: The Stupiding" v.s. "Crazy Bride Wedding Love
> Planner"?
>
> On Mon, Jul 11, 2011 at 8:56 PM, Lance Norskog <goksron@gmail.com> wrote:
>> SVDRecommender is intriguing, thanks for the pointer.
>>
>> On Sun, Jul 10, 2011 at 12:15 PM, Ted Dunning <ted.dunning@gmail.com>
>> wrote:
>>> Also, item-item similarity is often (nearly) the result of a matrix
>>> product.
>>> If yours is, then you can decompose the user x item matrix and the
>>> desired
>>> eigenvalues are the singular values squared and the eigen vectors are
>>> the
>>> right singular vectors for the decomposition.
>>>
>>> On Sun, Jul 10, 2011 at 2:51 AM, Sean Owen <srowen@gmail.com> wrote:
>>>
>>>> So it sounds like you want the SVD of the item-item similarity matrix?
>>>> Sure,
>>>> you can use Mahout for that. If you are not in Hadoop land then look at
>>>> SVDRecomnender to crib some related code. It is decomposing the user
>>>> item
>>>> matrix though.
>>>>
>>>> But for this special case of a symmetric matrix your singular vectors
>>>> are
>>>> the eigenvectors which you may find much easier to compute.
>>>>
>>>> I might restate the interpretation.
>>>> The 'size' of these vectors is not what matters to your question. It is
>>>> which elements (items) have the smallest vs largest values .
>>>> On Jul 10, 2011 3:08 AM, "Lance Norskog" <goksron@gmail.com> wrote:
>>>>
>>>
>>
>>
>>
>> --
>> Lance Norskog
>> goksron@gmail.com
>>
>
>
>
> --
> Lance Norskog
> goksron@gmail.com
>
--
Lance Norskog
goksron@gmail.com
```
Mime
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## Is there an electric field in a capacitor?
Capacitors store electric energy when charged. The charges on the capacitor plates produce an electric field inside the capacitor. Moving along electric field lines results in a change of electric potential: DV = EDx.
## Is electric field strength constant in a capacitor?
A capacitor is supposed to have infinite dimensions. And Electric field strength of a charged plane sheet of infinite dimensions is constant over infinity i.e. distance does not matter.
What is V ED?
In a uniform electric field, the equation to calculate the electric potential difference is super easy: V = Ed. In this equation, V is the potential difference in volts, E is the electric field strength (in newtons per coulomb), and d is the distance between the two points (in meters).
What is inside electric field?
The electric field is zero inside a conductor. Just outside a conductor, the electric field lines are perpendicular to its surface, ending or beginning on charges on the surface. Any excess charge resides entirely on the surface or surfaces of a conductor.
### What is electric field inside parallel plate capacitor?
The Electric Field strength in a parallel plate capacitor is obtained as Voltage applied to plates divided by Distance between the plate.
### Why is electric field inside a capacitor zero?
Because the charge on the two plates is the same magnitude, and of opposite sign, so that the net charge is zero. A spherical capacitor is spherically symmetric, so that the electric field must point radially outward everywhere, and be equal in magnitude.
Is electric field inside a capacitor zero?
In other words, they are talking about the electric field created by the capacitor itself. The proper analysis of a capacitor ASSUMES that the field is zero at infinity and then uses different gaussian surfaces to prove that the field remains zero everywhere outside the capacitor. Highly active question.
Where is the electric field strongest in a capacitor?
The capacitor plates are finite and there is the edge effect: The electric field lines extend to the outer space near the edges, so the field strength close to the edges is slightly less than at the middle of the capacitor. The edge effect depends on the separation between the plates.
#### Where is the electric field strongest between the plates of a capacitor?
The electric field is strongest where the lines are closest together. Now move on to a single plate of charge. Each point on the plate emanates some electric field lines, but they get crowded together and forced to move perpendicular the plate.
#### What is the electric field strength in a parallel plate capacitor?
The electric field strength in a parallel plate capacitor is determined by the formula, where Q – charge on the plate ε0 – vacuum permittivity, ε0 = 8.85418781762039 × 10-12 ε – permittivity of dielectric
What is the SI unit of electric field strength of a capacitor?
The SI unit of electric field strength is volt (V). A cylindrical capacitor is a capacitor, the plates of which are two cylinders, the inner one with the radius R1 and the outer one with the radius R2. Between the plates there is a dielectric whose permittivity is ε. π – the number Pi (3.1415926535897932384626433832795…)
How do you use the electric field in a capacitor equation?
Use the electric field in a capacitor equation: Converting to and plug in values: Consider the given diagram. If , each plate of the capacitor has surface area , and the plates at apart, determine the electric field between the plates. The voltage rise through the source must be the same as the drop through the capacitor.
## What is the electric field between the plates of a capacitor?
A parallel-plate capacitor is connected to a constant voltage source. If the distance between the plates of this capacitor is and the capacitor holds a charge of , what is the value of the electric field between the plates of this capacitor? An electric field does not exist between the plates of a parallel-plate capacitor | 820 | 4,116 | {"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} | 3.75 | 4 | CC-MAIN-2023-14 | longest | en | 0.922681 |
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### Show Tags
12 Jul 2010, 00:27
GMAT exam is broken down into 3 section. Analytical Writing, Quantitative and Verbal.
May i know under the :
1) Analytical Writing section >> do we need to write like an essay ?
2) Verbal >> are we tested on how we speak and pronounce ?
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Re: New to GMAT [#permalink]
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12 Jul 2010, 01:12
everything-you-need-to-prepare-for-the-gmat-revised-77983.html
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Re: New to GMAT [#permalink]
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12 Jul 2010, 08:42
You may also find that our free guide, the GMAT Uncovered, is a useful place to start: http://www.manhattangmat.com/gmat-uncovered.cfm
Best,
Caitlin
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Re: New to GMAT [#permalink]
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12 Jul 2010, 17:54
Thank you for the reference
Re: New to GMAT [#permalink] 12 Jul 2010, 17:54
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 839 | 2,895 | {"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} | 3.03125 | 3 | CC-MAIN-2017-04 | latest | en | 0.816965 |
https://mathematica.stackexchange.com/questions/1446/syntax-highlighting-for-your-own-functions/1449 | 1,653,319,769,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00231.warc.gz | 435,753,733 | 71,478 | Syntax highlighting for your own functions
Mathematica has a useful feature that for functions and special constructs getting passed local variables (for example Minimize or Block), the affected variables are shown in a different colour. Now when writing your own functions of this type (like the NOptimize function found in this answer), Mathematica of course doesn't know the meaning of the arguments and doesn't change the colours accordingly.
Is there a way to tell Mathematica that an argument is a set of variables to affect other arguments?
To make this post self-contained, here a small (admittedly silly) example:
SetAttributes[SequenceVars,HoldAll]
SequenceVars[vars_List,expr_]:=Block[vars,vars=Range@Length@vars;expr]
(* add syntax highlighting somehow *)
SequenceVars[{a,b,c},foo[a,b,c,d]]
(* here a, b and c, but not d, should appear in a different colour while typing *)
What you want is SyntaxInformation. With this, you can use every highlighting which already exists for things like Table, Solve, etc. for your own functions. You can specify the pattern of the arguments. With this you get the typical red commas if you use too many parameters. Or you can highlight locally used variables inside the function-arguments:
SetAttributes[SequenceVars, HoldAll]
SyntaxInformation[SequenceVars] = {"ArgumentsPattern" -> {_, _},
"LocalVariables" -> {"Solve", {1, 1}}};
SequenceVars[vars_List, expr_] :=
Block[vars, vars = Range@Length@vars; expr]
It looks like this in the front-end:
The usage of the "LocalVariables" highlighting is as follows. First, you choose the general type of highlighting. For instance "Solve" provides simple x or lists like {a,b} while "Plot" provides highlighting for the first element in e.g. {a,1,2}.
You can use the following settings as templates
{"Table", "Solve", "Integrate", "Limit", "Plot", "Manipulate"}
Additionally, you need to specify at which places of your function call the local variable specifications can appear. Say you want a function where in the first argument is some expression and then can follow arbitrary many lists of local variables. The definition for that would look like
SyntaxInformation[f] = {"ArgumentPattern" -> {_, __},
"LocalVariables" -> {"Solve", {2, Infinity}}}
This looks then like
and should make clear how "LocalVariables" has to be used.
Update
Since this function seems to be of unexpected interest, I should add one more important thing: SyntaxInformation can highlight wrong options. If you use OptionsPattern in your "ArgumentsPattern" option, all non-existent options will be highlighted in red
Options[f] = {a -> 1, b -> True};
f[x_] = Identity;
SyntaxInformation[f] = {"ArgumentsPattern" -> {_, OptionsPattern[]}};
Now using f with right/wrong options gives
• Interesting! I had no idea that SyntaxInformation can handle local variables. Big +1. Feb 7, 2012 at 12:34
• Works great except for one small detail: At least in 8.0, it's "ArgumentsPattern", not "ArgumentPattern". Feb 7, 2012 at 13:09
• Wow, 16 upvotes in 2 hours, I didn't know this information was scarce since it's in the docs. @halirutan, do you know if it can be used to simulate a With/Block/Module-like local with the equal sign lala[{x=8}, ...]?
– Rojo
Feb 7, 2012 at 18:47
• @Rojo none of the given option-values for "LocalVariables" seems to fit for that. I would say it cannot be possible regarding the power of the highlighting in Block/Module/With. Consider the following two examples Table[Table[i, {i, 5}], {i, 5}] and Block[{i = 3}, Block[{i = 3}, i]]. There happens more parsing in Block/Module constructs than simple highlighting would be able to. Feb 8, 2012 at 19:52
• @Mr.Wizard: Sorry for the late reply; I only noticed your question just now. Yes, it fails with "ArgumentPattern" in Version 8 (at least in 8.0.0.0), but works with "ArgumentsPattern". Apr 10, 2012 at 16:35
This answer is focussed at more experienced users, to provide a way for them to find out more information. I do not discuss how anything works.
Information in this answer corresponds to version 10.3
Despite the fact that PrintDefinitions@SyntaxInformation gives nothing, we can still see how the function works, by doing
<< GeneralUtilites
PrintDefinitions@SystemUtilitiesGetSystemSyntaxInformation
SystemUtilitiesGetSystemSyntaxInformation is pretty much equivalent to SyntaxInformation.
From the definition of SystemUtilitiesGetSystemSyntaxInformation, we see that the full list of "options of SyntaxInformation" is
(*sio is short for SyntaxInformation options*)
sio = {"ArgumentsPattern", "OptionNames", "LocalVariables", "ColorEqualSigns"}
Of these options "OptionNames" is undocumented (see this answer for an example of use, credit to jkuczm)
The following procedure gives a list of basic template names
informationFile =
ToFileName[{$InstallationDirectory, "SystemFiles", "Kernel", "TextResources",$Language}, "FunctionInformation.m"];
(*sife is short for System Information File Expression*)
sife = If[FileType[informationFile] === File,
Get[informationFile], {}];
templatesWithNames=
Select[
sife[[1,2]]
,
Length@#>3&&#[[4]]=!= None&
][[All,{1,4}]];
DeleteDuplicates@templatesWithNames[[All,2,1]]
{"Manipulate", "Solve", "Plot", "Table", "D", "Integrate", "Limit",
"SumSign", "IntegralSign"}
Of these, {"D", "IntegralSign", "SumSign"} do not appear in the docs.
I had forgotten about the "Lexical" modifier, but that is not undocumented, see the details section of the docs). Here are some examples of templates that involve it and also one that does not.
manipulateTemplates =
Select[templatesWithNames, #[[2, 1]] == "Manipulate" &]
{"Animate",{"Manipulate",{2,∞},"Lexical"}}
{"ControllerManipulate",{"Manipulate",{2,∞},"Lexical"}}
{"Manipulate",{"Manipulate",{2,∞},"Lexical"}}
{"RepeatingElement",{"Manipulate",{2}}}
Also this is weird
"ArgumentsPattern" /. SyntaxInformation@EmbedCode
{_, Optional["/Volumes/Jenkins/workspace/Documentation.Usage.English.release/scratch"],
_., OptionsPattern[]}
I don't know if this is possible for the Input - style cells, but it is certainly possible for the Program-style cells with the syntax highlighter generator I exposed in a recent answer. The work in that direction is underway, but I don't have a complete package at the moment. In fact, apart from code highlighting for languages like C, Java, etc, a strong motivation for that project for me was to create an extensible syntax highlighter for Mathematica based on it. I hope to have more to say on this soon.
Note that, in general, an implementation of the syntax highlighting of the type you mentioned requires context / semantic analysis of the program - lexical analysis is not enough.
• A nice package, but since it didn't exactly answer my question, I've given an upvote on the linked answer instead. Feb 7, 2012 at 13:20
• @celtschk I think you made the right choice. If I knew that SyntaxInformation handles local variables, I would not bother suggesting this. I expect however that the highlighter I may make based on my package will be more flexible at the end, since I have full control over the way tokens are highlighted. Even now, it does paren-matching, braces and bracket - matching, and highlights them in different colors. I don't know how easy it is to enable this for Input` cells, and how flexible it is. So, I will still keep this answer. Feb 7, 2012 at 13:26
• @LeonidShifrin: Will your syntax highlighter be usable as a replacement for the built-in syntax highlighter from Wolfram for Input cells or is it restricted just to formatting/highlighting other programming languages in non-Input style cells? Also what is the current status of the project? May 19, 2012 at 22:15
• @StackExchanger Yes, I plan to cover Mathematica as well. Potentially it will allow more than the built-in syntax highlighter. The current status of the project is that it is perfectly usable for someone to generate highlighters for any given language. In the attached notebook, I gave an example for C. I also did it for Java. It really takes just a few lines of code to write a spec for a given language, and a package for that language is generated automatically. The project needs some work though - I don't support comments yet, need to define this through styles, etc. Hope to address this soon May 20, 2012 at 12:56
• Leonid's syntax highlighter is incredible. I use it for Java, Scala, C#, Ocaml, and F# code. Just reading the code teaches you a lot about M. Dec 17, 2012 at 19:54 | 2,052 | 8,494 | {"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} | 2.78125 | 3 | CC-MAIN-2022-21 | longest | en | 0.77798 |
https://math.stackexchange.com/questions/3692470/integer-side-length-right-triangles-with-area-perimeter/3692495 | 1,621,178,849,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991224.58/warc/CC-MAIN-20210516140441-20210516170441-00399.warc.gz | 411,971,238 | 40,001 | # Integer side length right triangles with area = perimeter
How many right triangles with integer side lengths (up to congruence) are there with the property that the area of the triangle is the same as the perimeter?
I found that $$5, 12, 13$$ has both an area and perimeter of 30 but I'm not sure if there are any more of them.
• $(6,8,10)$ is another. – Maryam May 26 '20 at 14:19
• thanks, is that the only other one? – qs13 May 26 '20 at 14:29
We are looking for $$a,b\in\mathbb N$$ such that $$ab = 2a+2b+2\sqrt{a^2+b^2}$$. After some manipulation, we have $$4a^2+4b^2=a^2b^2-4ab(a+b)+4a^2+8ab+4b^2$$$$a^2b^2-4ab(a+b-2)=0$$$$a^2b^2=4ab(a+b-2)$$$$ab=4(a+b-2)$$$$ab-4a-4b+16=8$$$$(a-4)(b-4)=8$$
So, there are only two possibilities as there are only two possible pairs of $$(a-4,b-4)$$: $$(1,8)$$ and $$(2,4)$$, which correspond to $$(5,12,13)$$ and $$(6,8,10)$$, respectively.
• There are actually infinetly many solutions to this system, of course if we add the fact that this MUST be pithagorean triplets, then we have only two cases – Samuel A. Morales May 26 '20 at 14:39
• @Samorales Look at the beginning of my answer, and the domain to which $a,b$ belong. – Don Thousand May 26 '20 at 14:41
• I did not pay attention to that, my bad. – Samuel A. Morales May 26 '20 at 14:42
The answer also falls directly out of the canonical formula for Pythagorean triples: $$x=m^2-n^2;\ y=2mn;\ z=m^2+n^2$$.
Accordingly, $$P=m^2-n^2+2mn+m^2+n^2=2m(m+n)$$ and $$A=\frac{1}{2}(m^2-n^2)(2mn)=mn(m+n)(m-n)$$
$$P=A \Rightarrow 2m(m+n)=mn(m+n)(m-n)$$. Removing identical factors, $$2=n(m-n)$$
Since $$2$$ is prime, the only solutions are $$m,n=3,1$$ and $$m,n=3,2$$, which yield the triples identified by the previous responders: $$(6,8,10)\ \text{and}\ (5,12,13)$$
To solve this question we need to go to the equations, notice that the area of a right triangle is denoted as:
$$\frac{a b}{2}$$ Where $$a$$ and $$b$$ are the legs of the triangle. On the other side, the perimeter of this triangle is $$a+b+c$$ Where $$c$$ is the hypotenuse of the same triangle. We can draw a relationship between c, a and b because by the pitagorean theorem we now that $$c=\sqrt{a^2+b^2}$$ and now we can group both equation with this last equality to have the following $$\frac{a b}{2}=a+b+c$$ $$\frac{a b}{2}=a+b+\sqrt{a^2+b^2}$$ After some basic algebraic manipulation we get the following expression
$$a=\frac{4 (b-2)}{b-4}$$ If we let $$b=5$$, then $$a=12$$ and by pithagorean theorem $$c=13$$
• You have to be a bit careful about dividing by expressions like $b-4$. – Don Thousand May 26 '20 at 14:33
• Sure, since this is a fractional expression, we can not use values suech as b=4 – Samuel A. Morales May 26 '20 at 14:37
• a has to be an integer, so you should consider when does b-4 divide 4, and when does it divide b-2. A quick think quickly eliminates all candidates for 4 as it is small, and in fact will give you the 2nd solution. Now can you prove from this that these are the only solutions? – Daniel Buck May 27 '20 at 16:32 | 1,028 | 3,032 | {"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": 35, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-21 | latest | en | 0.869149 |
dosinga.medium.com | 1,726,499,487,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00810.warc.gz | 191,190,876 | 33,717 | Predicting the outcome of the World Cup in 150 lines of Python
--
It seems like Machine learning and Deep Learning are everywhere these days. AI can do anything! From writing essays to cheat tests to generating to having an infinite image generating machine running on your local machine, the possibilities are endless. So it should come as no surprise that you can use deep learning to predict the outcome of major sports tournaments.
I mean, it shouldn’t come as a surprise and I am sure you can, but today we’re going to talk about how to predict the Qatar World Cup without any machine learning toolkits, in just 150 lines of Python. I’m going to use Neptyne, the programmable spreadsheet — partly because that’s the startup I co-founded, but also since it’s a good platform for this sort of thing. The principles should work for any Python platform though.
Our basic plan here is to acquire recently played matches, calculate the relative strengths of each of the countries and then simulate the tournament a thousand times so we have an idea of the likely outcomes.
Training Data
Predictions are hard especially about the future they say, but you have to start with the past. Kaggle has a good data set, so let’s download that. This has all the international matches since 1872 in it, so that’s a bit much. We want to only have recent matches and only of countries that have played more than ten:
`def upload_results(): """Call this function and upload the data from kaggle""" tbl = nt.upload("Kaggle International football results", "*.csv").dropna() cutoff_date = sorted(tbl["date"])[-2500] tbl = tbl[tbl["date"] >= cutoff_date] counts = Counter( itertools.chain.from_iterable( ((rec['home_team'], rec['away_team']) for _, rec in tbl.iterrows()) ) ) thress_hold = {country for country, count in counts.items() if count <= 10} res = [ (DATEVALUE(rec['date']), rec['home_team'], rec['away_team'], rec['home_score'], rec['away_score']) for _, rec in tbl.iterrows() if not rec['home_team'] in thress_hold and not rec['away_team'] in thress_hold ] res.insert(0, ["Date", "Home team", "Away team", "Goals for", "Goals against"]) if "History" in nt.sheets: nt.sheets.delete_sheet("History") nt.sheets.new_sheet("History") History!B1 = res`
This code takes the csv we got, keeps the last 2500 matches, filters out teams that have not played enough and sticks all of it in a new sheet in our spreadsheet called History and then keeps only the columns we’re interested in.
Feats of Strength
Next we want to use this historic data to calculate the relative strength of each team. Here’s where deep learning could come in, but it doesn’t. Instead we do the straightforward thing. We try to assign each team a score such that if two teams play each other, the difference in score will correspond to the expected goal difference between the two teams. We start with assigning each team a score of zero and then adjust those scores to match all the matches. Rinse and repeat for about a thousand times and we get reasonable ratings. Here’s the code:
`def train_model(): scores = defaultdict(float) min_goals = [] home_advantage = 0 for i in range(500): delta = 5 / (i *10 + 50) total_diff = [] ddd = 0 for country1, country2, goals1, goals2 in History!C2:F: if i == 0: min_goals.append(min(goals1, goals2)) diff = goals1 - goals2 - (scores[country1] - scores[country2]) - home_advantage total_diff.append(diff) ddd += abs(diff) scores[country1] = scores[country1] + diff * delta scores[country2] = scores[country2] - diff * delta home_advantage += delta * diff / 20 Model!F2:G[i // 10] = [[i + 1, ddd / len(History!C2:F)]] Model!C2 = sorted(scores.items(), key=lambda rec: rec[1], reverse=True) Model!A2 = "Score Std:", "Av Min Goals:", "Min Goals Std:", "Home advantage" Model!B2 = np.std(total_diff), sum(min_goals) / len(min_goals), np.std(min_goals), home_advantage`
We read the history from the sheet (table) we just created and write the results to another sheet called Model. We also keep track of the standard deviation of the outcomes and the minimum number of goals, plus the actual average minimum number of goals. Keeping track of the home advantage makes the model a little more stable. Here are the results:
Brazil’s always a favorite — as is Germany who rank last in our overview of the teams with a decent score.
Predicting Games
We can now simulate individual games using some simple code:
`def predict(team1, team2): score1 = VLOOKUP(team1, Model!C2:D200, 2, False) score2 = VLOOKUP(team2, Model!C2:D200, 2, False) gf = np.random.normal(score1 - score2, Model!B2) ga = 0 if gf < 0: ga = -gf gf = 0 mg = max(0, np.random.normal(Model!B3, Model!B4)) gf = max(int(round(mg + gf)), 0) ga = max(int(round(mg + ga)), 0) return gf, ga`
This depends a bit more on the spreadsheet than the other code, but it should be easy to follow. We lookup the scores for each team and then pick a random value for the goal difference, normally distributed around the predicted goal difference. Add the minimum number of goals again normally distributed and make sure we don’t return values below zero.
Simulate the tournament
Simulating the entire tournament is now a matter of running this over all the matches. In a Neptyne spreadsheet we can easily model the structure of the tournament using some custom functions to model the outcomes of the groups for example:
`def standings(games): all_teams = {} for team1, _flag, team2, result in games: all_teams[team1] = 8 * [0] all_teams[team2] = 8 * [0] for team1, _flag, team2, result in games: if result: goals1, goals2 = split_score(result) for team, gf, ga in zip((team1, team2), (goals1, goals2), (goals2, goals1)): entry = all_teams[team] entry[0] += 1 entry[1] += 1 if gf > ga else 0 entry[2] += 1 if gf == ga else 0 entry[3] += 1 if gf < ga else 0 entry[4] += gf entry[5] += ga entry[6] += gf - ga entry[7] += (3 if gf > ga else (1 if gf == ga else 0)) return sorted(([k, *v] for k, v in all_teams.items()), key=lambda r:r[8], reverse=True)`
This returns the group results given a list of games played. For group A this predicts an outcome like:
Looks reasonable to me. Netherlands taking the lead!
We can take this all the way and simulate all the games until the final. Here’s one predicted finals:
Simulating multiple tournaments
So now all we have to do is run the tournament a thousand times and we get the following prediction:
So, Brazil’s the favorite! Who would have thought. 4.1% for the Netherlands though.
That’s it for today. If you want to have a look at the code, have a look at the Neptyne Spreadsheet. If you want to get your own account, please join our waitlist and we’ll get you coding up a spreadsheet before you know it. | 1,764 | 7,209 | {"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} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.923426 |
http://oeis.org/A014199 | 1,553,023,911,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202125.41/warc/CC-MAIN-20190319183735-20190319205735-00199.warc.gz | 152,626,403 | 3,335 | This site is supported by donations to The OEIS Foundation.
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A014199 Inverse of 190th cyclotomic polynomial. 2
1, -1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Periodic with period length 190. - Ray Chandler, Apr 03 2017 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (-1, 0, 0, 0, 1, 1, 0, 0, 0, -1, -1, 0, 0, 0, 1, 1, 0, 0, 1, 0, -1, 0, 0, -1, 0, 1, 0, 0, 1, 0, -1, 0, 0, -1, 0, 1, 0, -1, 0, 0, -1, 0, 1, 0, 0, 1, 0, -1, 0, 0, -1, 0, 1, 0, 0, 1, 1, 0, 0, 0, -1, -1, 0, 0, 0, 1, 1, 0, 0, 0, -1, -1). MAPLE with(numtheory, cyclotomic); c := n->series(1/cyclotomic(n, x), x, 80); MATHEMATICA CoefficientList[Series[1/Cyclotomic[190, x], {x, 0, 200}], x] (* Vincenzo Librandi, Apr 06 2014 *) CROSSREFS Cf. similar sequences listed in A240328, A240467. Sequence in context: A014124 A014239 A014104 * A014094 A014179 A014074 Adjacent sequences: A014196 A014197 A014198 * A014200 A014201 A014202 KEYWORD sign,easy AUTHOR STATUS approved
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Last modified March 19 15:02 EDT 2019. Contains 321330 sequences. (Running on oeis4.) | 817 | 1,700 | {"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} | 3.328125 | 3 | CC-MAIN-2019-13 | latest | en | 0.570082 |
https://www.coursehero.com/file/6443979/-4/ | 1,487,709,336,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170823.55/warc/CC-MAIN-20170219104610-00032-ip-10-171-10-108.ec2.internal.warc.gz | 799,760,450 | 21,388 | paper 3 2001
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Unformatted text preview: ................................................... (c) According to A, how fast is B moving in order to give this time dilation factor? [2] ......................................................................... ......................................................................... ......................................................................... (d) According to B, how fast is A moving? [1] ......................................................................... (This question continues on the following page) 881-181 – 27 – N01/430/S(3) (Question G1 continued) (e) The second event is at rest with respect to observer A. Observer B measures 6 seconds for this event. What time interval does A measure? [3] ......................................................................... ......................................................................... ......................................................................... ......................................................................... (f) Which version of time is ‘correct’? Explain your answer. [2] ......................................................................... ......................................................................... ......................................................................... 881-181 Turn over – 28 – N01/430/S(3) G2. The mass of a particle called a muon as a function of its speed is shown in the graph below. ( mass / MeV / c 2 ) 1000 800 600 400 200 0 0.0 0.2 0.4 0.6 0.8 1.0 velocity / c Use the information in the graph to answer the following questions. (a) What is the rest mass of the muon? [1] ......................................................................... (b) At what speed would the muon need to be moving in order to have a mass of twice its rest mass? [2] ......................................................................... ......................................................................... ......................................................................... (c) If a constant force is applied to the muon, could this force ever accelerate it to a speed greater than the speed of light? Explain. ......................................................................... ......................................................................... ......................................................................... 881-181 [2] – 29 – N01/430/S(3) G3. One prediction of the General Theory of Relativity is gravitational redshift. Explain what is meant by the term gravitational redshift and outline an experiment that demonstrates the effect. [5] .............................................................................. .............................................................................. .............................................................................. .............................................................................. .............................................................................. .............................................................................. .............................................................................. 881-181 Turn over – 30 – N01/430/S(3) OPTION H — OPTICS H1. A student uses a single converging lens of focal length 12 cm to produce a magnified virtual image. (a) Show the approximate arrangement of object, lens and eye in order...
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## This note was uploaded on 09/30/2011 for the course CHEM 102 taught by Professor Tina during the Spring '11 term at Global.
Ask a homework question - tutors are online | 584 | 3,788 | {"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} | 2.828125 | 3 | CC-MAIN-2017-09 | longest | en | 0.642629 |
https://convert-online.org/how-many-square-feet-in-an-acre | 1,717,070,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00801.warc.gz | 145,525,814 | 4,641 | # Acres to Square Feet
Acres to square feet is a tool to convert acres to square feet. To calculate how many square feet in an acre, multiply by 43,560.
## Convert Acre to sq ft
Acres Square Feet
Acres to Square Feet Conversion Acres: Square Feet: Square Inches: Square Miles: Square Meters: Square Yards: Square Kilometers: Hectares:
## How many square feet in an acre?
1 acre equals 43,560 square feet.
Acres Square feet
1 ac 43,560 sq ft
2 ac 87,120 sq ft
3 ac 130,680 sq ft
4 ac 174,240 sq ft
5 ac 217,800 sq ft
6 ac 261,360 sq ft
7 ac 304,920 sq ft
8 ac 348,480 sq ft
9 ac 392,040 sq ft
10 ac 435,600 sq ft
11 ac 479,160 sq ft
12 ac 522,720 sq ft
13 ac 566,280 sq ft
14 ac 609,840 sq ft
15 ac 653,400 sq ft
16 ac 696,960 sq ft
17 ac 740,520 sq ft
18 ac 784,080 sq ft
19 ac 827,640 sq ft
20 ac 871,200 sq ft
21 ac 914,760 sq ft
22 ac 958,320 sq ft
23 ac 1,001,880 sq ft
24 ac 1,045,440 sq ft
25 ac 1,089,000 sq ft
26 ac 1,132,560 sq ft
27 ac 1,176,120 sq ft
28 ac 1,219,680 sq ft
29 ac 1,263,240 sq ft
30 ac 1,306,800 sq ft
31 ac 1,350,360 sq ft
32 ac 1,393,920 sq ft
33 ac 1,437,480 sq ft
34 ac 1,481,040 sq ft
35 ac 1,524,600 sq ft
36 ac 1,568,160 sq ft
37 ac 1,611,720 sq ft
38 ac 1,655,280 sq ft
39 ac 1,698,840 sq ft
40 ac 1,742,400 sq ft
41 ac 1,785,960 sq ft
42 ac 1,829,520 sq ft
43 ac 1,873,080 sq ft
44 ac 1,916,640 sq ft
45 ac 1,960,200 sq ft
46 ac 2,003,760 sq ft
47 ac 2,047,320 sq ft
48 ac 2,090,880 sq ft
49 ac 2,134,440 sq ft
50 ac 2,178,000 sq ft
51 ac 2,221,560 sq ft
52 ac 2,265,120 sq ft
53 ac 2,308,680 sq ft
54 ac 2,352,240 sq ft
55 ac 2,395,800 sq ft
56 ac 2,439,360 sq ft
57 ac 2,482,920 sq ft
58 ac 2,526,480 sq ft
59 ac 2,570,040 sq ft
60 ac 2,613,600 sq ft
61 ac 2,657,160 sq ft
62 ac 2,700,720 sq ft
63 ac 2,744,280 sq ft
64 ac 2,787,840 sq ft
65 ac 2,831,400 sq ft
66 ac 2,874,960 sq ft
67 ac 2,918,520 sq ft
68 ac 2,962,080 sq ft
69 ac 3,005,640 sq ft
70 ac 3,049,200 sq ft
71 ac 3,092,760 sq ft
72 ac 3,136,320 sq ft
73 ac 3,179,880 sq ft
74 ac 3,223,440 sq ft
75 ac 3,267,000 sq ft
76 ac 3,310,560 sq ft
77 ac 3,354,120 sq ft
78 ac 3,397,680 sq ft
79 ac 3,441,240 sq ft
80 ac 3,484,800 sq ft
81 ac 3,528,360 sq ft
82 ac 3,571,920 sq ft
83 ac 3,615,480 sq ft
84 ac 3,659,040 sq ft
85 ac 3,702,600 sq ft
86 ac 3,746,160 sq ft
87 ac 3,789,720 sq ft
88 ac 3,833,280 sq ft
89 ac 3,876,840 sq ft
90 ac 3,920,400 sq ft
91 ac 3,963,960 sq ft
92 ac 4,007,520 sq ft
93 ac 4,051,080 sq ft
94 ac 4,094,640 sq ft
95 ac 4,138,200 sq ft
96 ac 4,181,760 sq ft
97 ac 4,225,320 sq ft
98 ac 4,268,880 sq ft
99 ac 4,312,440 sq ft
100 ac 4,356,000 sq ft
105 ac 4,573,800 sq ft
110 ac 4,791,600 sq ft
115 ac 5,009,400 sq ft
120 ac 5,227,200 sq ft
125 ac 5,445,000 sq ft
130 ac 5,662,800 sq ft
135 ac 5,880,600 sq ft
140 ac 6,098,400 sq ft
145 ac 6,316,200 sq ft
150 ac 6,534,000 sq ft
155 ac 6,751,800 sq ft
160 ac 6,969,600 sq ft
165 ac 7,187,400 sq ft
170 ac 7,405,200 sq ft
175 ac 7,623,000 sq ft
180 ac 7,840,800 sq ft
185 ac 8,058,600 sq ft
190 ac 8,276,400 sq ft
195 ac 8,494,200 sq ft
200 ac 8,712,000 sq ft
250 ac 10,890,000 sq ft
300 ac 13,068,000 sq ft
350 ac 15,246,000 sq ft
400 ac 17,424,000 sq ft
450 ac 19,602,000 sq ft
500 ac 21,780,000 sq ft
550 ac 23,958,000 sq ft
600 ac 26,136,000 sq ft
650 ac 28,314,000 sq ft
700 ac 30,492,000 sq ft
750 ac 32,670,000 sq ft
800 ac 34,848,000 sq ft
850 ac 37,026,000 sq ft
900 ac 39,204,000 sq ft
950 ac 41,382,000 sq ft
1,000 ac 43,560,000 sq ft
1,500 ac 65,340,000 sq ft
2,000 ac 87,120,000 sq ft
2,500 ac 108,900,000 sq ft
3,000 ac 130,680,000 sq ft
3,500 ac 152,460,000 sq ft
4,000 ac 174,240,000 sq ft
4,500 ac 196,020,000 sq ft
5,000 ac 217,800,000 sq ft
5,500 ac 239,580,000 sq ft
6,000 ac 261,360,000 sq ft
6,500 ac 283,140,000 sq ft
7,000 ac 304,920,000 sq ft
7,500 ac 326,700,000 sq ft
8,000 ac 348,480,000 sq ft
8,500 ac 370,260,000 sq ft
9,000 ac 392,040,000 sq ft
9,500 ac 413,820,000 sq ft
10,000 ac 435,600,000 sq ft
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Convert Online | Terms | Privacy | Contact © 2024 Convert Online | 1,983 | 4,595 | {"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} | 3 | 3 | CC-MAIN-2024-22 | latest | en | 0.417617 |
https://ask.sagemath.org/answers/13690/revisions/ | 1,713,162,833,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816942.33/warc/CC-MAIN-20240415045222-20240415075222-00717.warc.gz | 99,433,579 | 5,574 | Revision history [back]
You can make sure solve returns only positive square roots by making sure that the argument of square root is positive. For eg.
solve(x*x==1,x)
gives x=1,x=-1
however,
assume(x>0)
solve(x*x==1,x)
returns x=1
I might be able to give a better answer if you could give me a simplified example of the problem. | 87 | 338 | {"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} | 2.828125 | 3 | CC-MAIN-2024-18 | latest | en | 0.878848 |
http://www.britannica.com/print/topic/478486 | 1,418,868,026,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802765093.40/warc/CC-MAIN-20141217075245-00114-ip-10-231-17-201.ec2.internal.warc.gz | 494,776,410 | 8,011 | # projective geometry
projective geometry, branch of mathematics that deals with the relationships between geometric figures and the images, or mappings, that result from projecting them onto another surface. Common examples of projections are the shadows cast by opaque objects and motion pictures displayed on a screen.
Projective geometry has its origins in the early Italian Renaissance, particularly in the architectural drawings of Filippo Brunelleschi (1377–1446) and Leon Battista Alberti (1404–72), who invented the method of perspective drawing. By this method, as shown in the The sight lines drawn from the image in the reality plane (RP) to the artist’s eye intersect the picture plane (PP) to form a projective, or perspective, drawing. The horizontal line drawn parallel to PP corresponds to the horizon. Early perspective experimenters sometimes used translucent paper or glass for the picture plane, which they drew on while looking through a small hole to keep their focus steady.Encyclopædia Britannica, Inc., the eye of the painter is connected to points on the landscape (the horizontal reality plane, RP) by so-called sight lines. The intersection of these sight lines with the vertical picture plane (PP) generates the drawing. Thus, the reality plane is projected onto the picture plane, hence the name projective geometry. See also geometry: Linear perspective.
Although some isolated properties concerning projections were known in antiquity, particularly in the study of optics, it was not until the 17th century that mathematicians returned to the subject. The French mathematicians Girard Desargues (1591–1661) and Blaise Pascal (1623–62) took the first significant steps by examining what properties of figures were preserved (or invariant) under perspective mappings. The subject’s real importance, however, became clear only after 1800 in the works of several other French mathematicians, notably Jean-Victor Poncelet (1788–1867). In general, by ignoring geometric measurements such as distances and angles, projective geometry enables a clearer understanding of some more generic properties of geometric objects. Such insights have since been incorporated in many more advanced areas of mathematics.
## Parallel lines and the projection of infinity
A theorem from Euclid’s Elements (c. 300 bc) states that if a line is drawn through a triangle such that it is parallel to one side (see the The formula in the figure reads k is to l as m is to n if and only if line DE is parallel to line AB. This theorem then enables one to show that the small and large triangles are similar.Encyclopædia Britannica, Inc.), then the line will divide the other two sides proportionately; that is, the ratio of segments on each side will be equal. This is known as the proportional segments theorem, or the fundamental theorem of similarity, and for triangle ABC, with line segment DE parallel to side AB, the theorem corresponds to the mathematical expression CD/DA = CE/EB.
Now consider the effect produced by projecting these line segments onto another plane as shown in the In RP, Euclid’s fundamental theorem of similarity states that CD/DA = CE/EB. By introducing a scaling factor, the theorem can be saved in RP as CD′/DA′ = CE′/EB′ ∙ ΩB′/ΩA′. Note that while lines AB and DE are parallel in RP, their projections onto PP intersect at the infinitely distant horizon (Ω).Encyclopædia Britannica, Inc.. The first thing to note is that the projected line segments AB′ and DE′ are not parallel; i.e., angles are not preserved. From the point of view of the projection, the parallel lines AB and DE appear to converge at the horizon, or at infinity, whose projection in the picture plane is labeled Ω. (It was Desargues who first introduced a single point at infinity to represent the projected intersection of parallel lines. Furthermore, he collected all the points along the horizon in one line at infinity.) With the introduction of Ω, the projected figure corresponds to a theorem discovered by Menelaus of Alexandria in the 1st century ad:CD′/DA′ = CE′/EB′ ∙ ΩB′/ΩA′.Since the factor ΩB′/ΩA′ corrects for the projective distortion in lengths, Menelaus’s theorem can be seen as a projective variant of the proportional segments theorem.
## Projective invariants
With Desargues’s provision of infinitely distant points for parallels, the reality plane and the projective plane are essentially interchangeable—that is, ignoring distances and directions (angles), which are not preserved in the projection. Other properties are preserved, however. For instance, two different points have a unique connecting line, and two different lines have a unique point of intersection. Although almost nothing else seems to be invariant under projective mappings, one should note that lines are mapped onto lines. This means that if three points are collinear (share a common line), then the same will be true for their projections. Thus, collinearity is another invariant property. Similarly, if three lines meet in a common point, so will their projections.
The following theorem is of fundamental importance for projective geometry. In its first variant, by Pappus of Alexandria (fl. ad 320) as shown in the Pappus of Alexandria (fl. ad 320) proved that the three points (x, y, z) formed by intersecting the six lines that connect two sets of three collinear points (A, B, C; and D, E, F) are also collinear.Encyclopædia Britannica, Inc., it only uses collinearity:
Let the distinct points A, B, C and D, E, F be on two different lines. Then the three intersection points—x of AE and BD, y of AF and CD, and z of BF and CE—are collinear.
The second variant, by Pascal, as shown in the The 17th-century French mathematician Blaise Pascal proved that the three points (x, y, z) formed by intersecting the six lines that connect any six distinct points (A, B, C, D, E, F) on a circle are collinear.Encyclopædia Britannica, Inc., uses certain properties of circles:
If the distinct points A, B, C, D, E, and F are on one circle, then the three intersection points x, y, and z (defined as above) are collinear.
There is one more important invariant under projective mappings, known as the cross ratio (see the Although distances and ratios of distances are not preserved under projection, the cross ratio, defined as AC/BCBD/AD, is preserved. That is, AC/BCBD/AD = AC′/BC′ ∙ BD′/AD′.Encyclopædia Britannica, Inc.). Given four distinct collinear points A, B, C, and D, the cross ratio is defined asCRat(A, B, C, D) = AC/BC ∙ BD/AD.It may also be written as the quotient of two ratios:CRat(A, B, C, D) = AC/BC : AD/BD.
The latter formulation reveals the cross ratio as a ratio of ratios of distances. And while neither distance nor the ratio of distance is preserved under projection, Pappus first proved the startling fact that the cross ratio was invariant—that is,CRat(A, B, C, D) = CRat(A′, B′, C′, D′).However, this result remained a mere curiosity until its real significance became gradually clear in the 19th century as mappings became more and more important for transforming problems from one mathematical domain to another.
## Projective conic sections
Conic sections can be regarded as plane sections of a right circular cone (see the The conic sections result from intersecting a plane with a double cone, as shown in the figure. There are three distinct families of conic sections: the ellipse (including the circle); the parabola (with one branch); and the hyperbola (with two branches).Encyclopædia Britannica, Inc.). By regarding a plane perpendicular to the cone’s axis as the reality plane (RP), a “cutting” plane as the picture plane (PP), and the cone’s apex as the projective “eye,” each conic section can be seen to correspond to a projective image of a circle (see the The conic sections (ellipse, parabola, and hyperbola) can be generated by projecting the circle formed by the intersection of a cone with a plane (the reality plane, or RP) perpendicular to the cone’s central axis. The image of the circle is projected onto a plane (the projective plane, or PP) that is oriented at the same angle as the cutting plane (Ω) passing through the apex (“eye”) of the double cone. In this example, the orientation of Ω produces an ellipse in PP.Encyclopædia Britannica, Inc.). Depending on the orientation of the cutting plane, the image of the circle will be a circle, an ellipse, a parabola, or a hyperbola.
A plane Ω passing through the apex and parallel to PP defines the line at infinity in the projective plane PP. The situation of Ω relative to RP determines the conic section in PP: If Ω intersects RP outside the base circle (the circle formed by the intersection of the cone and RP), the image of the circle will be an ellipse. If Ω is tangent to the base circle (in effect, tangent to the cone), the image will be a parabola. If Ω intersects the base circle (thus, cutting the circle in two), a hyperbola will result.
Pascal’s theorem, quoted above, also follows easily for any conic section from its special case for the circle. Start by selecting six points on a conic section and project them back onto the base circle. As given earlier, the three relevant intersection points for six points on the circle will be collinear. Now project all nine points back to the conic section. Since collinear points (the three intersection points from the circle) are mapped onto collinear points, the theorem holds for any conic section. In this way the projective point of view unites the three different types of conics.
Similarly, more complicated curves and surfaces in higher-dimensional spaces can be unified through projections. For example, Isaac Newton (1643–1727) showed that all plane curves defined by polynomials in x and y of degree 3 (the highest power of the variables is 3) can be obtained as projective images of just five types of polynomials. | 2,228 | 9,908 | {"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} | 3.40625 | 3 | CC-MAIN-2014-52 | longest | en | 0.943108 |
https://www.greenbuildingadvisor.com/article/what-is-heat | 1,713,570,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00463.warc.gz | 696,978,616 | 32,461 | # What Is Heat?
## An exploration into the deeper meaning of a word we use all the time
While you slept last night, Santa Claus was putting his knowledge of physics to work. No, not with that silly anti-gravity stuff. Everyone knows the whole reindeer thing is just a cover for the way he really gets to all those houses in just one night. He uses one of the original Time Turners. In fact, Professor McGonagall got her first Time Turner from Santa himself.
No, the physics of Santa’s job is dealing with all those fireplaces and the fires that he encounters in so many of them. It’s understanding heat that saves his hide — and all those presents!
Back in grad school, I was taking a class in surface physics, and my friend Steve asked, “What is temperature?” On first hearing that question, you might think it’s so obvious that it doesn’t even bear asking, and anyone who does give voice to it must be an idiot. But have you really thought about it?
Perhaps the most important lesson I learned in my seven years of grad school was to ask questions, even about things that seem so basic as not to need questioning. There a lot of insights and deeper understanding available when you start questioning your basic beliefs.
### Temperature, thermal energy, and heat
So, let’s look at a question related to Steve’s: What is heat? All of us who work or play in the world of building science talk about it, so it’s a good idea to know a little about this basic concept.
Actually, the two questions are related. Heat and temperature go together, but they’re not the same thing. Before we can understand heat and temperature, though, we’ve got to talk about thermal energy. It sounds like thermal energy and heat might be the same thing, too, but they’re not.
Thermal energy is the combined energy of all the jiggling going on in the molecules of an object.
Temperature is a measure of the average kinetic energy (energy of motion) of the molecules.
Heat is a measure of how much thermal energy transfers from one object to another because of a temperature difference.
Of the three quantities, temperature is the one we have direct access to. It’s the one we learn about at an early age, when we touch something hot and get burned. It’s also the key to understanding heat because heat flows only when there’s a temperature difference. You don’t learn any life lessons by touching something that’s the same temperature as you because no temperature difference means no heat flow.
The term “heat” should be thought of only in conjunction with “flow” or “transfer” because it’s what happens when there’s a temperature difference, and one body (at a higher temperature) transfers thermal energy to another (at a lower temperature). We don’t really care about thermal energy in building science because heat flow is what matters.
### Calculating heat flow
We can easily calculate the amount of heat that we lose through the building enclosure (walls, floors, and ceilings that separate conditioned from unconditioned space) by conduction on a cold night. The equation is Q = U x A x ΔT.
Q is the amount of heat flow (in BTU per hour), U is the conductance, A is the area, and ΔT is the temperature difference. (For more on the details of this equation, see my articles U R A ΔT, and Other Building Science Blandishments and Flat or Lumpy.)
Now, where this really comes in handy is when you want to do some energy modeling or heating and cooling load calculations for the purpose of sizing the heating and cooling equipment. You use that formula for all parts of the building enclosure, add in your other loads (ducts, people, appliances…), and you can find out how much energy a home uses or needs for heating and cooling. It’s all about heat flow, which is all about temperature difference, which is all about thermal energy.
Let me emphasize one point here before I end: Heat flows only when there’s a temperature difference, and it naturally flows from a warmer body to a cooler body. You can pump it the other way, and that’s exactly what refrigerators, air conditioners, and heat pumps do. But the Second Law of Thermodynamics says heat wants to flow from warmer to cooler bodies.
That law is one of the most important and interesting in all of physics, and we could spend many hours engrossed in a discussion of the Second Law topics of entropy and the arrow of time, the heat death of the universe, why some processes are irreversible, and how it’s possible, according to statistical mechanics, for all the air in the room to be suddenly in one little corner, leaving you gasping in your chair as you read this blog. Of course, we’d need plenty of adult beverages on hand. Even better would be to have that discussion at the end of a long day of skiing.
And speaking of skiing, Santa has put his Time Turner back in its box and is now heading out to the slopes himself. That’s how he keeps in shape during the off-season, you know.
Allison Bailes of Decatur, Georgia, is a speaker, writer, energy consultant, RESNET-certified trainer, and the author of the Energy Vanguard Blog. You can follow him on Twitter at @EnergyVanguard.
1. | | #1
I'm not sure...
I appreciate your articles, and your efforts to explain building science concepts to us. I'm not convinced that your new definition of heat, which contrasts with all the textbooks and references that I have seen, is really moving comprehension forward.
2. | | #2
Response to Derek Roff
What, specifically, is the contrast between Allison's admittedly compact definition and those of the textbooks you refer to?
3. | | #3
Question
Allison,
I too am a bit confused by this definition. You say that heat is a measure of how much thermal energy transfers from one object to another because of a temperature difference.
Then you emphasize that heat flows only when there's a temperature difference, and it naturally flows from a warmer body to a cooler body.
When you say that “heat flows only when there's a temperature difference,” it follows that heat does not flow if there is no temperature difference. The distinction of heat that is not flowing shows that heat can exist without flowing. Yet how can heat exist without flowing if flowing is what defines heat?
4. | | #4
variations on heat
Responding to Bruce, the term "heat" shows up in lots of fairly basic equations, many of which are relevant to building design at one point or another. For example, the BTU is only a meaningful measurement unit, because we have a definition for the [specific] heat of water. The definition of BTU is incompatible with Allison's definition here. So are the classical laws of thermodynamics.
I can understand what Allison means in his new definition, and I can guess why he chose it. He wants to communicate to readers a more active and functional experience of how heat moves in a house. But I think equating heat and heat flow adds confusion, rather than reducing it. Will it not be confusing to readers, to suggest that adding insulation to a house will reduce its heat? That a really well-insulated electric water heater, that is up to temperature and idling, has virtually no heat?
According to quantum theory, we can't determine whether Schroedinger's beer is cold or not, until we open the cooler. I don't think we need similar confusion when discussing building science concepts.
5. GBA Editor
| | #5
It's basic physics, not a new definition
What I've stated above is just basic physics that you can find in just about any introductory physics book. I haven't redefined anything. Heat flow is analogous to work. It's a transfer of energy, just as work is.
And speaking of work, there is an exception to what I said above about heat flowing only when there's a temperature difference. When work is done on or by a system or when there's a phase change, you can have heat flow without a temperature difference. Building enclosures, however, usually don't do work or have work done on them, and few have phase change materials to simulate the effect of thermal mass.
6. GBA Editor
| | #6
Derek wrote:
For example, the BTU is only a meaningful measurement unit, because we have a definition for the [specific] heat of water. The definition of BTU is incompatible with Allison's definition here. So are the classical laws of thermodynamics.
Actually, you're putting the cart before the horse here. Heat capacity comes from the definition of the BTU, not the other way around. It's the proportionality constant that determines the amount of temperature change you get for a given heat flow in a material.
• ## How Does a Heat Pump Get Heat From Cold Air?
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• ## What is Comfort?
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• ## Why Doesn’t Heat Flow Backwards?
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• ## How Heat Moves Through Homes — Building Science Podcast
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• |
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• | | 1,966 | 9,209 | {"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} | 3.3125 | 3 | CC-MAIN-2024-18 | latest | en | 0.966653 |
http://www.weegy.com/?ConversationId=0AA6B1B1 | 1,529,624,844,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864303.32/warc/CC-MAIN-20180621231116-20180622011116-00575.warc.gz | 538,495,315 | 8,389 | You have new items in your feed. Click to view.
Q: multiply and simplify. (2p)^7 * (2p)^5
A: answer is 4p^12 you just need to add the exponent for multiplication but always remember that you can only do that if they have the same variable
Original conversation
User: multiply and simplify. (2p)^7 * (2p)^5
Weegy: answer is 4p^12 you just need to add the exponent for multiplication but always remember that you can only do that if they have the same variable
jameshand|Points 287|
User: multiply. (4-4x^2)^2
Weegy: (4-4^2)(4-4^2) 16 -16X^2-16X^2+16X^4 then 16-32x^2+16x^4 rearranging 16x^4-32x^2+16
Question
Updated 7/13/2014 1:34:25 PM
Rating
3
(2p)^7 * (2p)^5
= 128p^7 * 32p^5
= 4,096p^12
Questions asked by the same visitor
multiply (4-4x^2)^2
Weegy: (4-4x^2)^2 = 16x^4 - 32x^2 + 16 (More)
Question
Updated 7/16/2014 10:45:26 AM
*
Get answers from Weegy and a team of really smart lives experts.
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 657 | 1,763 | {"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} | 3.609375 | 4 | CC-MAIN-2018-26 | latest | en | 0.827548 |
https://tutswiki.com/data-structures-algorithms/heap-sort/ | 1,695,828,518,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00125.warc.gz | 648,583,307 | 8,320 | # Heap Sort
Sorting as you might already know is basically an algorithm that is used for arranging elements of a list in a certain order. (Usually ascending or descending). Sorting is one of the most important categories of algorithms, it can significantly reduce the complexity of problems, and is generally used for efficient searching.
There are an ample number of sorting algorithms available like the Bubble sort, Selection sort, Insertion sort, Merge sort, Quick sort, Heap sort, Counting sort and more. The type of algorithm to choose depends on the type of problem. (Generally Merge Sort and Quick sort are used.)
Today we will be discussing Heap Sort. Now before directly jumping to Heap Sort, we must be aware of a few terminologies.
## 1. Complete Binary Tree
We can define a complete binary tree as a tree in which every level is completely filled except possibly the last, and it is as left as possible.
## 2. Binary Heap
A binary heap is a complete binary tree in which the value of parent is greater or lesser than its children.
If the value of parent is greater than its children, then it’s called max-heap else we call it min-heap.
We can represent the heap as a binary tree or an array.
## Array Representation of Heap:
As we have already discussed that heap is a type of complete binary tree, therefore it is easy to represent it as an array.
Let’s suppose that the parent node is at index `i`
Then the left child will be at `(2 * i) + 1`
And the right child will be at `(2 * i) + 2`
## Examples
`Parent index= 0`
`Left child index= 2*0 + 1 = 1`
`Right child index= 2*0 +2=2`
`Parent index= 1`
`Left child index= 2*1 + 1 = 3`
`Right child index= 2*1 +2=4`
`Parent index= 2`
`Left child index= 2*2 + 1 = 5`
`Right child index= 2*2 +2=6`
## HeapSort
Heap sort algorithm is a comparison based sorting technique, it’s basic working is similar to that of insertion sort. It is an in-place sorting algorithm but is not stable, that is, the original order of keys is not maintained.
### Understanding the algorithm
In the heap sort algorithm, we insert all the elements from the unsorted list or array into a heap. We then create max-heap which brings the largest element at the root of the heap, we exchange this value with the last value and then decrement size of the array. Then, we heapify the first element. This process is continued until there is only one element left in the array.
### Java Code for Heap Sort
``````import java.util.Arrays;
public class Sorting {
public void heapSort(int arr[]) {
int lengthOfArray = arr.length;
// creating heap
for (int i = (lengthOfArray - 1) / 2; i >= 0; i--) {
heapify(arr, lengthOfArray, i);
}
// Sorting
for (int i = lengthOfArray - 1; i >= 0; i--) {
// Swap the root node with last node
int temp = arr[i];
arr[i] = arr[0];
arr[0] = temp;
heapify(arr, i, 0);
}
}
public void heapify(int[] arr, int index, int i) {
// Initializing parent and children
int parentIndex = i;
int leftChild = (2 * i) + 1;
int rightChild = (2 * i) + 2;
// comparing the left child value
if (leftChild < index && arr[leftChild] > arr[parentIndex]) {
parentIndex = leftChild;
}
// comparing the right child value
if (rightChild < index && arr[rightChild] > arr[parentIndex]) {
parentIndex = rightChild;
}
if (parentIndex != i) {
int temp = arr[parentIndex];
arr[parentIndex] = arr[i];
arr[i] = temp;
heapify(arr, index, parentIndex); // recursive call
}
}
// Driver Code
public static void main(String args[]) {
Sorting sort = new Sorting();
int[] arr = {
46,
76,
24,
1,
9,
7,
11,
38,
79,
13
};
sort.heapSort(arr);
System.out.println("Array after applying heap sort is " + Arrays.toString(arr));
}
}``````
The above program will generate the following output:
``Array after applying heap sort is [1, 7, 9, 11, 13, 24, 38, 46, 76, 79]``
## Performance of Heap Sort
• Worst case time complexity: `O(nlogn)`
• Best case time complexity: `O(nlogn)`
• Average case time performance: `O(nlogn)`
• Worst case space complexity: `O(n)`
• Auxiliary space complexity: `O(1)` | 1,058 | 4,026 | {"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} | 4.25 | 4 | CC-MAIN-2023-40 | latest | en | 0.887203 |
https://mathoverflow.net/questions/373551/scottish-book-problem-172 | 1,627,250,191,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151866.98/warc/CC-MAIN-20210725205752-20210725235752-00083.warc.gz | 389,923,659 | 31,054 | # Scottish Book Problem 172
The problem is formulated using old terminology and I want to understand what it actually says.
The problem reads: "A space $$E$$ of type (B) has the property (a) if the weak closure of an arbitrary set of linear functionals is weakly closed. [A sequence of linear functionals $$f_n(x)$$ converges weakly to $$f(x)$$ if $$f_n(x) \to f(x)$$ for every $$x$$.] The space $$E$$ of type (B) has the property (b) if every sequence of linear functionals weakly convergent converges weakly as a sequence of elements in the conjugate space $$\bar{E}$$. Question: Does every separable space of type (B) which has property (a) also possess property (b)?"
I know that "space of type (B)" means "Banach space" in modern terminology.
Q1: Does "linear functional" include the assumption of continuity?
Q2: What is the conjugate space? Does it mean the dual space of continuous linear functionals on $$E$$?
If the answers to Q1 and Q2 are yes, then this would mean Banach spaces with property (b) are exactly what nowadays are called Grothendieck spaces. [A Grothendieck space is a Banach space $$E$$ such that every sequence in the dual space $$E^*$$ that converges weak* - that is, with respect to $$\sigma(E^*,E)$$ - also converges weakly - that is, with respect to $$\sigma(E^*,E^{**})$$.]
I am also unsure how to understand the condition "the weak closure of an arbitrary set of linear functionals is weakly closed", as this seems like a tautology. Is the point here that we consider a sequential closure?
• Thank you. So the problem is: Assume that $E$ is a Banach space such that the weak* sequential closure of every subset of $E^*$ is weak* sequentially closed. Is $E$ a Grothendieck space? – Hannes Thiel Oct 8 '20 at 7:47
• The terminology "weak* sequential closure" is counterintuitive. Tthe weak* sequential closure of a set $X\subset E^*$, lets denote it by $c(X)$, is the set of all limits of weak* convergent sequences in $X$. It is not necessarily true that $c(c(X))=c(X)$. Thus, the weak* sequential closure need not be weak* sequentially closed. I found a paper of Godun (Weak* derivatives of sets of linear functionals), which shows that a separable Banach space $E$ is quasireflexive iff every subspace $F\subseteq E^*$ satisfies $c(c(F))=c(F)$. Thus, separable Banach spaces with (a) are quasireflexive. – Hannes Thiel Oct 8 '20 at 12:50
• In the same paper, it is also shown that a Banach space $E$ is Grothendieck if and only if $c(F)=F$ for every norm-closed subspace $F\subset E^*$. Further, for every (not necessarily norm-closed) subspace $F\subset E^*$ with norm-closure $\overline{F}$, we have $c(F)=c(\overline{F})$. – Hannes Thiel Oct 8 '20 at 13:03 | 753 | 2,701 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-31 | longest | en | 0.942159 |
https://fr.scribd.com/doc/294323701/Autocad-3D-and-2D-Practice-Activities | 1,591,319,932,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492295.88/warc/CC-MAIN-20200604223445-20200605013445-00074.warc.gz | 357,102,987 | 70,610 | Vous êtes sur la page 1sur 19
Figure 1. To make this object, notice the absolute coordinate location of the smaller circle (4,4.5) which is where you are going to start with. And the rest of the object follows with it.
Figure 2. This object is made with the advantage of the direct distance entry command to first make the perimeter (first without the curves). The curves in the sides can be made by using the fillet or circle-ttr command then trim.
Figure 3. Again notice the absolute coordinate location of the line in 1, 3 and the point in 5,7 which is where the rest of the drawing follows.
Figure 4. This object again has absolute coordinates to follow. This is made through circles-2P, lines, and ars (start-center-radius).
Figure 5. To make this object, start with the line, then the arcs. The center of the arc in 45º is the endpoint of the location 1.50.
Figure 6. This drawing again has absolute coordinate locations to follow. The triangle is made by following those commands given in numbered order. To circle inside the triangle is made through cicle-ttr command.
Figure 7. Though this object can be made through lines and arcs, the objective is to make this one using polylines wherein the lines and arcs are connected.
Figure 8. This object starts with the smaller circles with the diameter or 0.50 and the rest follows. A circle-ttr command is more useful than using arc to make the curves.
Figure 9. This object is almost similar to figure 9 which also starts with the smaller circles. Again a cicle-ttr command is used here than arcs to make the curves. To copy the arc inside or outside at a different radius, the offset command could be useful.
Figure 10. In this objects the direct distance entry could be useful to make the perimeter. For the curves on the edge, fillet command could also be used. Relative rectangular entry with snap-from command could locate the circles’ center. To make the dash-dot center line you need to load the line in the linetype box.
Figure 11. The objective of making this object is to learn the array (rectangular and polar) command. There are a total of 176 smaller circles. There are 40 circles around each inner circles and 96 circles inside.
Figure 12. This exercise will help you use the chamfer command which will help you make the cut on the edge. The mirror command is also used to reflect an object to its opposite side.
. | 541 | 2,396 | {"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} | 3.28125 | 3 | CC-MAIN-2020-24 | latest | en | 0.904918 |
www.eigenmath.org | 1,571,470,548,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692126.27/warc/CC-MAIN-20191019063516-20191019091016-00027.warc.gz | 246,451,245 | 8,140 | Returns the absolute value or vector length of x.
```X = (x,y,z)
abs(X)
```
``` 1/2
2 2 2
(x + y + z )
```
Returns the adjunct of matrix m. Adjunct is equal to determinant times inverse.
```A = ((a,b),(c,d))
```
```1
```
Returns 1 if all arguments are true (nonzero). Returns 0 otherwise.
```and(1=1,2=2)
```
```1
```
Returns the arc cosine of x.
```arccos(1/2)
```
``` 1
--- pi
3
```
Returns the arc hyperbolic cosine of x.
Returns the arc sine of x.
```arcsin(1/2)
```
``` 1
--- pi
6
```
Returns the arc hyperbolic sine of x.
Returns the arc tangent of y over x. If x is omitted then 1 is used.
```arctan(1,0)
```
``` 1
--- pi
2
```
Returns the arc hyperbolic tangent of x.
Returns the angle of complex z.
```arg(2 - 3i)
```
```arctan(-3,2)
```
Returns a solution to the Bessel differential equation.
```besselj(x,1/2)
```
``` 1/2
2 sin(x)
-------------
1/2 1/2
pi x
```
The result of evaluating a symbol can differ from the symbol's binding. For example, the result may be expanded. The binding function returns the actual binding of a symbol.
```p = quote((x + 1)^2)
p
```
``` 2
p = x + 2 x + 1
```
```binding(p)
```
``` 2
(x + 1)
```
Returns the coefficient of xk yn-k in (x + y)n. Binomial and choose are the same function.
```binomial(52,5)
```
```2598960
```
Returns the smallest integer greater than or equal to x.
```ceiling(1/2)
```
```1
```
If x is true (nonzero) then continue in a script, else stop. Use A=B or A==B to test for A equals B.
```A = 1
B = 1
check(A=B) -- script stops here if A not equal to B
```
Returns the number of combinations of n items taken k at a time. The following example computes the number of poker hands.
```choose(52,5)
```
```2598960
```
Returns expression x with circular and hyperbolic functions converted to exponentials.
```circexp(cos(x) + i sin(x))
```
```exp(i x)
```
Clears all symbol definitions.
Returns complex z in polar form with base of negative 1 instead of e.
```clock(2 - 3i)
```
``` 1/2 arctan(-3,2)/pi
13 (-1)
```
Returns the coefficient of xn in polynomial p.
```p = x^3 + 6x^2 + 12x + 8
coeff(p,x,2)
```
```6
```
Returns a cofactor of matrix m. The cofactor matrix is the transpose of the adjunct of m. This function returns the cofactor component at row i and column j.
```A = ((a,b),(c,d))
```
```1
```
Returns the complex conjugate of z.
```conj(2 - 3i)
```
```2 + 3 i
```
Returns tensor a summed over indices i and j. If i and j are omitted then 1 and 2 are used. The expression contract(m) computes the trace of matrix m.
```A = ((a,b),(c,d))
contract(A)
```
```a + d
```
Returns the cosine of x.
```cos(pi/4)
```
``` 1
------
1/2
2
```
```cos(arctan(y,x))
```
``` x
--------------
1/2
2 2
(x + y )
```
Returns the hyperbolic cosine of x.
```circexp(cosh(x))
```
``` 1 1
--- exp(-x) + --- exp(x)
2 2
```
Returns the cross product of vectors u and v. It is OK to redefine cross. This is the default definition.
```cross(u,v) = (u[2] v[3] - u[3] v[2],
u[3] v[1] - u[1] v[3],
u[1] v[2] - u[2] v[1])
```
Returns the curl of vector u. It is OK to redefine curl. This is the default definition.
```curl(u) = (d(u[3],y) - d(u[2],z),
d(u[1],z) - d(u[3],x),
d(u[2],x) - d(u[1],y))
```
Returns the partial derivative of f with respect to x.
```d(x^2,x)
```
```2 x
```
Note: Argument f can be a tensor of any rank. Argument x can be a vector. When x is a vector the result is the gradient of f.
```F = (f(),g(),h())
X = (x,y,z)
d(F,X)
```
```d(f(),x) d(f(),y) d(f(),z)
d(g(),x) d(g(),y) d(g(),z)
d(h(),x) d(h(),y) d(h(),z)
```
Returns the definite integral of f with respect to x evaluated from a to b. The argument list can be extended for multiple integrals as shown in the following example.
```f = (1 + cos(theta)^2) sin(theta)
defint(f, theta, 0, pi, phi, 0, 2pi) -- integrate over theta then over phi
```
``` 16
---- pi
3
```
Returns the degree of polynomial p(x).
```deg(x^3,x)
```
```3
```
Returns the denominator of expression x.
```denominator(a/b)
```
```b
```
Returns the determinant of matrix m.
```A = ((a,b),(c,d))
det(A)
```
```a d - b c
```
Returns the dimension of the nth index of tensor a. Index numbering starts with 1.
```A = ((1,2),(3,4),(5,6))
dim(A,1)
```
```3
```
Returns the divergence of vector u. It is OK to redefine div. This is the default definition.
```div(u) = d(u[1],x) + d(u[2],y) + d(u[3],z)
```
Evaluates each argument from left to right. Returns the result of the last argument.
```do(A=1,B=2,A+B)
```
```3
```
Returns the dot or matrix product of vectors, matrices, and tensors.
```-- solve for X in AX=B
A = ((1,2),(3,4))
B = (5,6)
X = dot(inv(A),B)
X
```
``` -4
X =
9
---
2
```
Draws a graph of f(x). Drawing ranges can be set with xrange and yrange.
```xrange = (0,1)
yrange = (0,1)
draw(x^2,x)
```
Symbol e is initialized to the natural number e. It is OK to clear or redefine e and use the symbol for something else.
```e^x
```
```exp(x)
```
Computes eigenvalues and eigenvectors numerically. Matrix m is required to be both numerical and symmetric. Eigenvectors are returned in Q and eigenvalues are returned in D. Each row of Q is an eigenvector. Each diagonal element of D is an eigenvalue.
```A = ((1,2),(2,1))
eigen(A)
dot(transpose(Q),D,Q)
```
```1.0 2.0
2.0 1.0
```
Error function of x.
Complementary error function of x.
Returns f evaluated at x equals a.
```eval(x^2 + 3,x,0)
```
```3
```
Returns the exponential of x.
```exp(i pi)
```
```-1
```
Returns the partial fraction expansion of the ratio of polynomials r in x.
```p = (x + 1)^2
q = (x + 2)^2
expand(p/q,x)
```
``` 2 1
- ------- + -------------- + 1
x + 2 2
x + 4 x + 4
```
Returns the cosine of z in exponential form.
```expcos(z)
```
``` 1 1
--- exp(i z) + --- exp(-i z)
2 2
```
Returns the hyperbolic cosine of z in exponential form.
```expcosh(z)
```
``` 1 1
--- exp(-z) + --- exp(z)
2 2
```
Returns the sine of z in exponential form.
```expsin(z)
```
``` 1 1
- --- i exp(i z) + --- i exp(-i z)
2 2
```
Returns the hyperbolic sine of z in exponential form.
```expsinh(z)
```
``` 1 1
- --- exp(-z) + --- exp(z)
2 2
```
Returns the tangent of z in exponential form.
```exptan(z)
```
``` i i exp(2 i z)
---------------- - ----------------
exp(2 i z) + 1 exp(2 i z) + 1
```
Returns the hyperbolic tangent of z in exponential form.
```exptanh(z)
```
``` 1 exp(2 z)
- -------------- + --------------
exp(2 z) + 1 exp(2 z) + 1
```
Factors integer n.
```factor(10!)
```
``` 8 4 2 1
2 3 5 7
```
factor(p,x)
Factors polynomial p(x). The polynomial should be factorable over integers. The argument list can be extended for multivariate polynomials.
```p = 2x + x y + y + 2
factor(p,x,y)
```
```(x + 1) (y + 2)
```
Note: Factor returns an unexpanded expression. If the result is assigned to a symbol, evaluating the symbol will expand the result. Use binding to retrieve the unexpanded expression.
```q = factor(p,x)
binding(q)
```
```(x + 1) (y + 2)
```
Returns the factorial of n. The expression n! can also be used.
```100!
```
```93326215443944152681699238856266700490715968264381621468592963895217599993229915
608941463976156518286253697920827223758251185210916864000000000000000000000000
```
Returns f excluding any terms containing a, b, etc.
```p = x^2 + 3x + 2
filter(p,x^2)
```
```3 x + 2
```
Returns expression x with rational numbers and integers converted to floating point values. The symbol pi and the natural number are also converted.
```float(212^17)
```
``` 39
3.52947 10
```
Returns the largest integer less than or equal to x.
```floor(1/2)
```
```0
```
For i equals j through k evaluate a, b, etc. The previous value of symbol i is restored after for completes.
```for(k,1,3,A=k,print(A))
```
```A = 1
A = 2
A = 3
```
Returns the greatest common divisor of expressions.
```gcd(x,x y)
```
```x
```
Returns the nth Hermite polynomial in x.
```hermite(x,3)
```
``` 3
8 x - 12 x
```
Returns an n by n Hilbert matrix.
```hilbert(3)
```
``` 1 1
1 --- ---
2 3
1 1 1
--- --- ---
2 3 4
1 1 1
--- --- ---
3 4 5
```
Symbol i is initialized to the imaginary unit (−1)1/2. It is OK to clear or redefine i and use the symbol for something else.
```exp(i pi)
```
```-1
```
Returns the imaginary part of complex z.
```imag(2 - 3i)
```
```-3
```
Returns the inner product of tensors. Inner and dot are the same function.
```A = ((a,b),(c,d))
B = (x,y)
inner(A,B)
```
```a x + b y
c x + d y
```
Note: Inner product is equivalent to an outer product followed by contraction.
```contract(outer(A,B),2,3)
```
```a x + b y
c x + d y
```
Returns the integral of f with respect to x.
```integral(x^2,x)
```
``` 1 3
--- x
3
```
Returns the inverse of matrix m.
```A = ((1,2),(3,4))
inv(A)
```
``` -2 1
3 1
--- - ---
2 2
```
Returns 1 if n is a prime number. Returns zero otherwise.
```isprime(2^31 - 1)
```
```1
```
Set j=sqrt(-1) to use j for the imaginary unit instead of i.
```j = sqrt(-1)
1/sqrt(-1)
```
```-j
```
Returns the nth Laguerre polynomial in x. If argument a is omitted then zero is used.
```laguerre(x,3)
```
``` 1 3 3 2
- --- x + --- x - 3 x + 1
6 2
```
The result of the previous calculation is stored in last.
```212^17
```
```3529471145760275132301897342055866171392
```
```last^(1/17)
```
```212
```
Note: Symbol last is an implied argument when a function has no argument list.
```212^17
```
```3529471145760275132301897342055866171392
```
```float
```
``` 39
3.52947 10
```
Returns the least common multiple of expressions.
```lcm(x,x y)
```
```x y
```
Returns the leading coefficient of polynomial p(x).
```leading(3x^2 + 1,x)
```
```3
```
Returns the nth Legendre polynomial in x. If m is omitted then zero is used.
```legendre(x,3)
```
``` 5 3 3
--- x - --- x
2 2
```
Evaluates expression x and returns the result as a string in prefix notation. Useful for debugging scripts.
```lisp(x^2 + 1)
```
```(+ (^ x 2) 1)
```
Returns the natural logarithm of x.
```log(x^y)
```
```y log(x)
```
Returns the magnitude of complex z. Mag treats undefined symbols as real while abs does not.
```mag(x + i y)
```
``` 1/2
2 2
(x + y )
```
Returns the remainder of integer a divided by integer b.
```mod(10,7)
```
```3
```
Returns 0 if x is true (nonzero). Returns 1 otherwise.
```not(1=1)
```
```0
```
Returns all roots, both real and complex, of polynomial p(x). The roots are computed numerically. The coefficients of p can be real or complex.
Returns 1 if x is a rational or floating point number. Returns 0 otherwise.
```number(1/2)
```
```1
```
Returns the numerator of expression x.
```numerator(a/b)
```
```a
```
Returns 1 if at least one argument is true (nonzero). Returns 0 otherwise.
```or(1=1,2=2)
```
```1
```
Returns the outer product of tensors. Also known as the tensor product.
```A = (a,b,c)
B = (x,y,z)
outer(A,B)
```
```a x a y a z
b x b y b z
c x c y c z
```
Symbol for π.
```exp(i pi)
```
```-1
```
Returns complex z in polar form.
```polar(x - i y)
```
``` 1/2
2 2
(x + y ) exp(i arctan(-y,x))
```
Use ^ to raise something to a power. Use parentheses for negative powers.
```x^(-2)
```
``` 1
----
2
x
```
Returns the nth prime number. The domain of n is 1 to 10000.
```prime(100)
```
```541
```
Evaluate expressions and print the results. Useful for printing from inside a for loop.
```for(j,1,3,print(j))
```
```j = 1
j = 2
j = 3
```
For i equals j through k evaluate f. Returns the product of all f.
```product(j,1,3,x + j)
```
``` 3 2
x + 6 x + 11 x + 6
```
Returns expression x without evaluating it first.
```p = quote((x + 1)^2)
binding(p)
```
``` 2
(x + 1)
```
```p = quote(p) -- clear symbol p
binding(p)
```
```p
```
Returns the quotient of polynomial p(x) over q(x).
```p = x^2 + 1
q = x + 3
quotient(p,q,x)
```
```x - 3
```
```p - q quotient(p,q,x) -- remainder of p/q
```
```10
```
Returns the number of indices that tensor a has.
```A = ((1,0),(0,1))
rank(outer(A,A,A))
```
```6
```
Returns expression x with everything over a common denominator.
```rationalize(1/a + 1/b + 1/2)
```
``` 2 a + a b + 2 b
-----------------
2 a b
```
Note: Rationalize returns an unexpanded expression. If the result is assigned to a symbol, evaluating the symbol will expand the result. Use binding to retrieve the unexpanded expression.
```f = rationalize(1/a + 1/b + 1/2)
binding(f)
```
``` 2 a + a b + 2 b
-----------------
2 a b
```
Returns the real part of complex z.
```real(2 - 3i)
```
```2
```
Returns complex z in rectangular form.
```rect(exp(i x))
```
```cos(x) + i sin(x)
```
Returns the values of x such that polynomial p(x) equals zero. The polynomial should be factorable over integers. Returns a vector for multiple roots.
```roots(x^2 + 3x + 2,x)
```
```-2
-1
```
Returns expression x in a simpler form.
```simplify(sin(x)^2 + cos(x)^2)
```
```1
```
Returns the sine of x.
```sin(pi/4)
```
``` 1
------
1/2
2
```
```sin(arctan(y,x))
```
``` y
--------------
1/2
2 2
(x + y )
```
Returns the hyperbolic sine of x.
```circexp(sinh(x))
```
``` 1 1
- --- exp(-x) + --- exp(x)
2 2
```
Returns the square root of x.
```sqrt(10!)
```
``` 1/2
720 7
```
Prints memory statistics.
```status
```
```block_count 1
free_count 99258
gc_count 1
bignum_count 370
string_count 0
tensor_count 5
```
In a script, it does what it says.
Evaluates expression x and returns the result as a string. Useful for testing scripts.
```string((x + 1)^2) == "x^2 + 2 x + 1"
```
```1
```
Substitutes a for b in c and returns the result.
```subst(x,y,y^2)
```
``` 2
x
```
For i equals j through k evaluate f. Returns the sum of all f.
```sum(j,1,5,x^j)
```
``` 5 4 3 2
x + x + x + x + x
```
Returns the tangent of x.
```simplify(tan(x) - sin(x)/cos(x))
```
```0
```
Returns the hyperbolic tangent of x.
```circexp(tanh(x))
```
``` 1 exp(2 x)
- -------------- + --------------
exp(2 x) + 1 exp(2 x) + 1
```
Returns the Taylor expansion of f(x) near x equals a. If argument a is omitted then zero is used. Argument n is the degree of the expansion.
```taylor(sin(x),x,5)
```
``` 1 5 1 3
----- x - --- x + x
120 6
```
If argument a is true (nonzero) then b is returned, else if c is true then d is returned, etc. If the number of arguments is odd then the last argument is returned if all else fails. Use A=B or A==B to test for A equals B.
```A = 1
B = 1
test(A=B,"yes","no")
```
```yes
```
Set trace=1 in a script to print the script as it is evaluated. Useful for debugging.
```trace = 1
```
Note: The contract function is used to obtain the trace of a matrix.
Returns the transpose of tensor a with respect to indices i and j. If i and j are omitted then 1 and 2 are used. Hence a matrix can be transposed with a single argument.
```A = ((a,b),(c,d))
transpose(A)
```
```a c
b d
```
Set tty=1 to print results in a flat format.
```tty = 1
(x + 1/2)^2
```
```x^2 + x + 1/4
```
Returns an n by n identity matrix.
```unit(3)
```
```1 0 0
0 1 0
0 0 1
```
Returns a null tensor with dimensions i, j, etc. Useful for creating a tensor and then setting the component values.
```A = zero(3,3)
for(k,1,3,A[k,k]=k)
A
```
``` 1 0 0
A = 0 2 0
0 0 3
``` | 5,203 | 15,656 | {"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} | 3.796875 | 4 | CC-MAIN-2019-43 | longest | en | 0.697725 |
https://doc.flix.dev/enums.html | 1,717,029,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00408.warc.gz | 185,319,629 | 6,333 | ## Enums
### Enumerated Types
Enumerated types are used to define a type that has a finite (enumerated) set of values. Enumerated types are useful for things such as modeling compass directions, the cards in a deck, and the days in a week.
For example, here is an enumeration of the days in a week:
enum Weekday {
case Monday,
case Tuesday,
case Wednesday,
case Thursday,
case Friday,
case Saturday,
case Sunday
}
Here Monday, Tuesday and so on are referred to as the constructors of the enum.
We can refer to a weekday as Monday or Weekday.Monday. The latter is required if we have multiple enums in scope with similarly named constructors.
We can use pattern matching to destruct an enum value. For example:
enum Animal {
case Cat,
case Dog,
case Giraffe
}
def isTall(a: Animal): Bool = match a {
case Animal.Cat => false
case Animal.Dog => false
case Animal.Giraffe => true
}
The function isTall takes a value of type Animal and performs a pattern match on it. If the value is Giraffe the function returns true. Otherwise it returns false.
Flix guarantees that pattern matches are exhaustive, i.e. that all cases have been covered. It is a compile-time error if a pattern match is non-exhaustive. A pattern match can always be made exhaustive by adding a default case as the last case. A default case is written with an underscore case _ => ???.
### Recursive Types
Recursive types are used to define types that are self-referential.
For example, we can define a binary tree of integers as follows:
enum Tree {
case Leaf(Int32),
case Node(Tree, Tree)
}
A tree is either a Leaf with an Int32 value or an internal Node with a left and a right sub-tree. Note that the definition of Tree refers to itself.
We can write a function, using pattern matching, to compute the sum of all integers in such a tree:
def sum(t: Tree): Int32 = match t {
case Tree.Leaf(x) => x
case Tree.Node(l, r) => sum(l) + sum(r)
}
The sum function pattern matches on a tree value. If the tree is a leaf its value is simply returned. Otherwise the function recurses on both subtrees and adds their results.
### Polymorphic Types
Polymorphic types are types parameterized by other types. For example, we can write:
enum Bottle[a] {
case Empty,
case Full(a)
}
def isEmpty(b: Bottle[a]): Bool = match b {
case Bottle.Empty => true
case Bottle.Full(_) => false
}
Here the Bottle type is parameterized by the type parameter a. In Flix, type parameters, like ordinary parameters are always written in lowercase. The Bottle type has two cases: either the bottle is empty (and contains no value) or it is full (and contains one value of type a). The isEmpty function takes a bottle, type parameterized by a, and determines if the bottle is empty.
The careful reader might have noticed that Bottle is equivalent to the more well-known Option type.
In general, polymorphic types can have more than one type argument. For example, the standard library implement of the Result has two type parameters:
enum Result[e, t] {
case Ok(t),
case Err(e)
}
### Shorthand Enum Syntax
A typical enum may look like:
enum Weekday {
case Monday,
case Tuesday,
case Wednesday,
case Thursday,
case Friday,
case Saturday,
case Sunday
}
The same enum can also be declared as:
enum Weekday {
case Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday
}
This shorthand syntax is always available, but should only be used for simple enums.
### Singleton Enum Syntax
An enum with a single case:
enum USD {
case USD(Int32)
}
can be shortened to:
enum USD(Int32) | 814 | 3,569 | {"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} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.862149 |
http://www.space-awareness.org/es/activities/1610/lets-map-the-earth/ | 1,721,226,595,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00348.warc.gz | 54,981,400 | 9,970 | # Let's Map the Earth
Discover how to read and draw maps.
## Breve descripción
In this activity, students familiarise themselves with the concept of a map by observing and describing maps, and drawing a map from an aerial photograph. They understand that any location on Earth is described by two numbers, latitude and longitude. The notion of scale and ratio is also explored.
## Objetivos
• to introduce the concept of maps to students,
• to teach students to recognise different elements like oceans, mountains, rivers, main roads, etc.;
• to introduce the concepts of scale and cardinal points, and their meanings when reading a map;
• to talk about the importance of latitude and longitude to identify any location on Earth;
• to make students aware of how modern maps are made.
## Objetivos de aprendizaje
After implementing the activity, students will be able:
• to identify a map and classify different elements on it, like mountains, oceans, cities, rivers, main roads, etc.;
• to identify the map scale and the cardinal points, and to understand the significance of these concepts when reading a map;
• to sketch a map from an aerial photograph of their country or city;
• to identify some elements related to the geography of the country.
## Evaluación
Teachers can check if students have correctly labelled features on their maps, as listed in the learning objectives, and if students have identified the important features to include in their own maps, including cardinal points and a scale. Check if students understand how distance is represented on a map. Ask them to calculate the distance between two objects using the scale. Students can present their maps in small groups or to the whole class to show the features they included. Invite each group to present their work through a short video or a presentation.
## Materiales
Per student:
Per group of students:
• 10 blank stickers or labels
• World map
• Pencils
Optional:
• Different maps with different scales or purposes
## Información de referencia
### Cardinal points
There are four cardinal directions or cardinal points: north, south, east, and west, denoted by the initials N, S, E, and W, respectively. On a map, it is common to indicate only north. Cartographers defined the cardinal points and draw maps with the north at the top by convention. Cardinal points provide a systematic way to tell someone where places are. In cultures that do not descend from European traditions, the cardinal directions could be shown with colours or numbers.
There are many ways to find the cardinal directions: some use the sun or stars, others the magnetic properties of the Earth (e.g., a compass). Today, satellite navigation (GPS) is widely used.
### Cartography
Cartography is the study and the creation of maps. Humans have been creating and using maps since ancient times but the oldest map that has survived over time is a map of the Babylonian city of Nippur. The technologies used to create maps have evolved enormously since ancient times.
The first maps were made by hand. Gradually, new techniques and tools like the compass, telescope, and sextant started being used, increasing the accuracy of maps. In the 20th century, aerial photography and satellite imagery tremendously improved the mapping of physical elements such as buildings, roads, rivers, etc. More recently, GPS has been used to perform direct mapping.
### Scales
A scale is a ratio that compares a measurement on a map (or figure) to the actual distance between locations identified on the map (or figure). It's the mathematical relationship between the length or figurative distance on the map and the actual length of the represented distance. Scales have no units. A scale is represented by an ordinary fraction (such as 1 / 1 000 000) or a mathematical ratio (1: 1 000 000). The number 1 means the unit on the map (1 cm) and number 1 000 000 means the actual size (1 000 000 cm is 10 km).
### Maps
A map is a simplified representation of populated areas, geographical features, roads, rivers, continents, and oceans. In addition to maps of the Earth and its countries, city plans or ordnance survey charts are also types of map. In the past, maps were drawn using a compass, an astrolabe, and a tape measure. Nowadays, they are charted using aerial photography and GPS. Maps are used for several purposes. The world map shows us countries, large rivers and mountains, the position of oceans and deserts, etc. A country map shows us larger cities and towns, roads, etc. Ordnance survey maps show contours, roads, houses, forests, etc. A city plan shows us streets, monuments, parks, etc. All maps show the cardinal points as well as a scale.
## Descripción completa de la actividad
Image: a world map showing the continents and the sea beds in relief. The blue represents the sea. Dark blue represents the deeper ocean beds and light blue the shallower beds. The continents are shown in yellow and the darker areas are the contours. The positions of the Equator, the Greenwich meridian, the Tropics of Cancer and Capricorn, the Arctic and the Antarctic Circles are shown on the map.
## What do you see on a map?
Image: Archipelago of the Azores. Credit: Varp/Wikimedia Commons
### Step 1
Ask students how we could find out where all the countries in the world are, or how we know the way if we want to travel somewhere new. We use a map: mobile applications and GPS are all based on this concept.
### Step 2
Ask students to look at the map or the globe. Ask them to describe what they see and encourage them to explain what they think different colours represent on the map.
### Step 3
Divide students into small groups. Distribute one map per group and several stickers or labels. Ask the group to write down “mountain”, “ocean”, “[your country]” and to stick the label to the corresponding feature on the map.
### Step 4
Identify the the colour on your map that refers to mountains on the sea bed (often a lighter blue). Ask students if they see differences in the colour blue. Do they know what these difference refer to? Explain that they show the mountains on the sea bed. As an example find the Azores archipelago and notice how they all form part of a chain which rises out of the sea bed and crosses the Earth from north to south.
## Cardinal points, latitude, and longitude
### Step 1
Find the cardinal points on their world map. Explain that cardinal points are used to describe where a place is.
### Step 2
Using the stickers on the maps, ask the students to localise the different elements relative to each other using different cardinal points. For example, is the Atlantic Ocean west of Europe?
### Step 3
Ask students to consider how we might define a particular location on a map. What are the issues involved? The cardinal points help us to describe an object's location relative to another but to be more general and precise, we need a system of reference. We can define a grid where the direction of the grid lines is defined using the cardinal points, and with grid lines at specific locations. This allows us to define a location on Earth using longitude (giving the east to west position from the Prime Meridian) and latitude (giving the north to south position from the Equator).
## How to build a map? Aerial photography.
Image: An aerial photograph of Lisbon from Sentinel-1A’s radar. Credit: Copernicus data/ESA (2014)
### Step 1
Distribute the aerial photographs of your city. Ask students to recognize different elements in the photo like buildings, sea, rivers, etc. You can ask students if they can locate some known places on the map like your school. Identify the scale and relevant features.
### Step 2
Discuss with students what the scale represents. Ask them to measure the distance between two objects on the map, either chosen by the educator or by free choice of the students. Students then use the scale to calculate the distance between the objects in real life.
### Step 3
Distribute tracing paper to students and ask them to copy the picture. Students draw a map of this area and identify what they think are the relevant features. For example, rivers, bridges, roads, etc. Students lay out their own map.
### Step 4
Explain that this is how modern maps are made. Aerial photographs are taken with balloons, planes or other flying objects. The different elements in the photos are then identified.
### Step 5
You can ask students if they know any other device that allows us to see aerial photographs. Ask students if they have seen a GPS. Explain that some objects fly even higher than airplanes around the Earth. These objects are called satellites; they are so high that they need to fly on a rocket to space. Satellites watch the Earth and send us very precise information about locations on the planet, in terms of an object's longitude and latitude.
## Plan de estudio
### Space Awareness curricula topics (EU and South Africa)
Navigation through the ages, latitude and longitude, instruments
### National Curricula UK
KS2, Geography: Locational knowledge: Identify the position and significance of latitude, longitude, Equator, Northern Hemisphere, Southern Hemisphere, the Tropics of Cancer and Capricorn, Arctic and Antarctic Circle, the Prime/Greenwich Meridian and time zones (including day and night). Geographical skills and fieldwork: Use maps, atlases, globes and digital/computer mapping to locate countries and describe features studied Use the eight points of a compass, four and six-figure grid references, symbols and key (including the use of Ordnance Survey maps) to build knowledge of the United Kingdom and the wider world.
KS2: Year 6, Maths: Ratio and proportion: Solve problems involving the relative sizes of two quantities where missing values can be found by using integer multiplication and division facts.
### National Curricula Wales
KS2, Geography: Skills: Locating places, environments, and patterns.
## Más información
Nowadays, maps are charted using aerial photography and the Global Positioning System (GPS). This may be linked to talking about the importance of satellites and about its uses in our everyday life.
Land and offshore surveying were some of the first sectors to exploit this new technology. Ordnance Survey uses GPS for the majority of its surveys to provide national mapping infrastructure.
Since the beginning of time, human beings have looked to the sky to determine their whereabouts. Traditionally, the Sun and the pattern of fixed stars have been their guides. But in the second millennium, constellations of man-made satellites have taken over as beacons to guide the way. Using the ‘fixed’ stars, you can determine your position anywhere on the Earth’s surface to within a few hundred metres, provided the weather is fine. But navigation satellites can tell you where you are to the nearest few metres or better, whatever the weather. Enhanced instruments can even pinpoint the position of a stationary object to within a few centimetres by measuring the object’s position many thousands of times over several hours and then working out the average of the measurements. The European Space Agency's (ESA’s) Galileo mission is a new generation of European navigation satellites, launching in 2016.
## Conclusión
In this activity, students familiarise themselves with the concept of a map. They observe, discuss, and draw a map. They learn how to identify different elements on a map like mountains and oceans. They learn that each map has a scale and cardinal points. They learn how modern maps are made using aerial photographs or GPS.
This resource was developed by Ciencia Viva, peer-reviewed by astroEDU and revised by Space Awareness.
Descargar
Tema del plan de estudio
latitude and longitude, instruments
La gran idea de la ciencia
Palabras clave
Maps, Earth, Latitude, Longitude, Geography, GPS
8 - 12
Nivel del sistema educativo
Primary, Middle School
Tiempo
45min
Tamaño del grupo
Group
No
Gasto
Low Cost
Ubicación
Small Indoor Setting (e.g. classroom)
Competencias básicas
Asking questions, Analysing and interpreting data, Communicating information | 2,570 | 12,182 | {"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} | 2.890625 | 3 | CC-MAIN-2024-30 | latest | en | 0.900017 |
https://www.scribd.com/document/198407793/Grid-fortran | 1,545,235,512,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376832559.95/warc/CC-MAIN-20181219151124-20181219173124-00382.warc.gz | 1,032,333,749 | 48,900 | You are on page 1of 11
# M1 International - Arts et M´ etiers PariTech ”Programing and Numerical Methods” Fortran Project
The second Fortran project is an homework due November 23rd 2012. It represents 25% of the total evaluation. You can send a pdf or doc file including your comments, results, figures, listings of routines to xavier.gloerfelt@paris.ensam.fr. To be sure that your email is well delivered, please verify that you have a confirmation by e-mail return specifying that your work is received.
1
In the first Fortran project, we have considered 2-D heat transfer problems in which the tem-
peratures are independent of time. In many applications, however the temperatures are varying with time, and we require the understanding of the complete time history of the temperature variation. For example, in metallurgy, the heat treating process can be controlled to directly affect the characteristics of the processed materials. Annealing (slow cool) can soften metals and improve ductility. On the other hand, quenching (rapid cool) can harden the strain boundary and increase strength. In order to characterize this transient behavior, the unsteady equation is needed. The onedimensional version is ∂φ ∂2φ = α 2 , 0 ≤ x ≤ L, t ≥ 0 (1) ∂t ∂x where φ = φ(x, t) is the dependent variable, and α is a constant coefficient. Equation (1) is a model of transient heat conduction in a slab of material with thickness L. The domain of solution is a semi-infinite strip of width L that continues indefinitely in time. The material property α is the thermal diffusivity. In a practical computation, the solution is obtained only for a finite time tmax . Solution to Equation (1) requires specification of boundary conditions at x = 0 and x = L, and initital conditions at t = 0. Simple boundary and initital conditions are : φ(0, t) = φ0 , φ(L, t) = φL , φ(x, 0) = f0 (x) (2)
2
Discretization
A finite-difference method is used for obtaining numerical solutions to Equation (1). Applying
finite differences, a discrete difference equation is obtained from the original PDE for φ(x, t). Then
1
m = 1.M where M is the total number oftime steps and ∆t is the size of a time step ∆t = tmax M −1 Both the time and space derivatives are replaced by finite differences. t) = φ(L... t) = 0. . Given L and N . The first one is an explicit scheme. t). The second method is implicit and will require the inversion of a system (with a tridiagonal matrix).the solution method. the discrete x are uniformly spaced in the interval 0 ≤ x ≤ L such that : xj = (j − 1)∆x . and the time step is limited by Fourier criterion. 4 Fortran program The proposed structure of the program could be – a module part.N where N is the total number of spatial nodes. and initial condition f0 = sin(πx/L). The exact solution for this problem is derived in the Appendix. . 2.. 3 Test problem The finite difference codes are tested by solving the heat equation with boundary condition φ(0. will provide an approximation φm j to φ(x. you will use a second-order central difference (as in the first Fortran project). including those on the boundary. For the second-order spatial operator. where specification of variables which are used in several subroutines are defined : specif. 2. coded in Fortran. j = 1. the discrete time t are uniformly spaced in 0 ≤ t ≤ tmax : tm = (m − 1)∆t . the spacing between the xj is computed with ∆x = L N −1 Similarly..f90 2 . Two different time integration will be used : – Forward Euler scheme – Crank-Nicolson scheme This is respectively Adams-Bashforth of order 1 and Adams-Moulton of order 2. For the codes you have to develop.
forward_euler.f90 including the forward and backward substitution steps for solving a system given its LU decomposition (here provided by tridiag.f90 forward_euler. and output in a file of the solution and error to be plotted with Matlab. You then have to change consequently the makefile.o main.f90 tridiag.o tridiag.f90 OBJS = specif.f90 where the Thomas algorithm is used to obtain directly the LU decomposition. computation of the final error.o LIBS = # for CYGWIN with gfortran F90 = gfortran F90FLAGS = -fdefault-real-8 # for LINUX PC with intel fortran #F90 = ifort #F90FLAGS = -O3 -r8 -fpe0 LDFLAGS = all: \$(PROG) \$(PROG): \$(OBJS) \$(F90) -o \$@ \$(OBJS) \$(LIBS) \$(LDFLAGS) clean: rm -f \$(PROG) \$(OBJS) *.o solve_LU.f90. crank_nicolson.o forward_euler.f90 forward_euler.o main. where the following operations are realized : initialization of the variables. A sample makefile is provided using these names for the f90-files.f90 main. calls to solving algorithm (Forward Euler or Crank-Nicolson). – a subroutine to solve a tridiagonal system tridiag.o forward_euler.f90 #OBJS = specif.– a main program main.o crank_nicolson. – a subroutine solve_LU. – a subroutine for the Forward Euler/Centered differences method.f90). – a subroutine for the Crank-Nicolson/Centered differences method.f90 main.o SRCS = specif. Example makefile PROG = prog #SRCS = specif.f90 crank_nicolson.f90 solve_LU.mod *.out 3 . You are free to change routine names or the number or scope of the subroutines.f90.f90.
o Example specif.out .f90 .o: specif.o forward_euler.o: specif. dimension(:). allocatable :: x ! location of finite difference nodes ! time discretization integer :: nt ! number of steps real :: tmax ! maximum time for the simulation real :: dt ! time step ! constants real :: alpha ! diffusivity real :: pi ! constant pi end module variables 4 real. allocatable :: u ! spatial boundary conditions at x=0 and x=L real :: u0.:). dimension(:).o crank_nicolson.o: specif.f90 !============================================================================ ! Common variables !============================================================================ module variables implicit none ! unknown variable (first dimension nx for space and second nt for time) real.debug: \$(F90) -g \$(SRCS) \$(LIBS) dxladebug a. dimension(:. allocatable :: t ! values of time at which solution is obtained (time nod .f90.SUFFIXES: \$(SUFFIXES) .uL ! space discretization integer :: nx ! number of mesh points in x direction real :: L ! length of the domain real :: dx ! mesh size real.o: \$(F90) \$(F90FLAGS) -c \$< main.
nt print*.5 ires=1 !!\$ !!\$ !!\$ !!\$ !!\$ !!\$ !!\$ !!\$ !!\$ !!\$ 5 print*.Example main.) ! Values of parameters nt = 10 nx = 20 alpha = 0.f90 ! ================================= ! main program ! ================================= program unsteady_heat ! USE ZONE: common variables use variables ! put implicit none to avoid declaration problem implicit none ! Local variables integer:: i. allocatable :: ue ! analytical solution (for each spatial node nx) real :: err ! error at last time step nt ! Constant Pi pi=acos(-1.t_cpu integer:: done real.’tmax’ read*.L . dimension(:).ires real :: t_final.tmax print*.t_initial.’L’ read*.m.1 L = 1 tmax = 0.’Size of the problem?’ print*.’nt’ read*.nx print*.’nx’ read*.
/(nt-1) ! uniform mesh t do i=1.!!\$ !!\$ print *. ! u is initialized at initial time (f0) do i=1.ires call cpu_time(t_initial) ! Memory allocations for dynamic variables ! ======================================== allocate(x(nx)) allocate(t(nt)) allocate(u(nx./(nx-1) ! uniform mesh x do i=1.nt t(i)=(i-1)*dt end do ! Initialization stage ! ==================== u=0.nx u(i.nx x(i)=(i-1)*dx end do ! Time step dt=1.nt)) allocate(ue(nx)) ! Compute mesh spacing and time step ! ================================== ! Mesh size dx=1.1)=sin(pi*x(i)/L) end do ! Initialization of the boundary values ! needed to apply BC inside time step loop u0 = 0 6 .’choice of the method ->1: FE/CD . 2: CN/CD’ read *.
do i=1.*) err close(10) ! The final solution is written in a file 7 .nx err=err+abs(u(i.uL = 0 ! Exact solution (used to compute the error) ! ============== ! at end of simulation ue = sin(pi*x/L)*exp(-t(nt)*alpha*(pi/L)**2) ! Choice of the method ! ==================== select case (ires) case (1) call forward_euler case (2) call crank_nicolson case default print*.file=’error.nt)-ue(i)) end do err=err/nx ! Opening of output files to store the errors ! =========================================== open(10.’the option is not defined’ stop end select ! Error computation ! ================= err=0.dat’) rewind(10) write(10.
m=1.ue) end program unsteady_heat Hereafter an example of visualization file reading a binary file is proposed.t_cpu ! The dynamic arrays are deallocated ! ================================== deallocate(x.nt) ! write the computed solution write(40) ((u(i. 8 .m).nx print *.err print *.i=1.’Error L2 : ’.nt) ! write the analytical solution write(40) (ue(i).m=1.’CPU time’.! ======================================= ! MATLAB file (particular BINARY file) open(40.nx) close(40) ! Note that no format is used for a binary file ! That’s why the ASCII file is also named FORMATTED file ! by default.’Mesh size: ’.form=’unformatted’. You can of course improve the visualization and add other plot such as error.nx).status=’unknown’) write(40) nx write(40) nt ! write the meshgrid and time nodes write(40) (x(i).nx) write(40) (t(m).i=1.log10(err) call cpu_time(t_final) t_cpu=t_final-t_initial print *.t.nt print *.’Log(Error L2 norm) : ’.u. the "format" attribute is "formatted" ! Some informations are written to the screen ! =========================================== print *.i=1.file=’solution.bin’.’Number of iterations: ’. | 2,526 | 9,370 | {"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} | 3.296875 | 3 | CC-MAIN-2018-51 | latest | en | 0.897121 |
http://www.aaaknow.com/xlate/lessonFull.php?slug=mult6d3d&menu=Multiplication | 1,548,219,385,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583897417.81/warc/CC-MAIN-20190123044447-20190123070447-00218.warc.gz | 230,196,152 | 6,127 | ## Multiplying 6-Digit by 3-Digit Numbers
### Multiply 6-Digit by 3-Digit Numbers: Learn
Multiplying a six digit number by a three digit number (for example 524639 * 687) is illustrated below.
5 2 4 6 3 9 * 6 8 7 3 6 7 2 4 7 3 4 1 9 7 1 1 2 + 3 1 4 7 8 3 4 3 6 0 4 2 6 9 9 3
### Multiplication: Practice
#### What is the product of the two numbers?
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Extra time is awarded for each correct answer.
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How fast can you get 20 more correct answers than wrong answers? 999 | 238 | 712 | {"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} | 2.84375 | 3 | CC-MAIN-2019-04 | latest | en | 0.853052 |
https://www.hackmath.net/en/example/1117 | 1,560,666,769,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997801.20/warc/CC-MAIN-20190616062650-20190616084650-00305.warc.gz | 791,561,364 | 6,766 | # Gasholder
The gasholder has spherical shape with a diameter 20 m. How many m3 can hold in?
Result
V = 4188.8 m3
#### Solution:
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#### To solve this example are needed these knowledge from mathematics:
Tip: Our volume units converter will help you with converion of volume units.
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Solve equations (use adding and subtracting of linear equations): -4x+11y=5 6x-11y=-5 | 714 | 2,714 | {"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} | 3.71875 | 4 | CC-MAIN-2019-26 | latest | en | 0.928233 |
https://www.mathworksheets4kids.com/counting.php | 1,726,204,288,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00380.warc.gz | 821,247,241 | 8,541 | # Counting Worksheets
Counting blocks or counting coins, counting cookies or candies, our printable counting worksheets teeming with adequate practice should be your pick if understanding cardinalities of small sets of objects is on your mind. Get preschool through grade 5 kids counting a few objects to help them know that the last number counted is equal to the quantity of the set. Add to their subitizing skills using manipulatives, relate counting to real world with quantitative word problems, count and produce sets of given sizes, and juggle between estimation and exact count. Try our free counting worksheets for a sneak-peek into what lies in store.
## Explore the Counting Worksheets in Detail
Counting to 5 Worksheets
Hardwired into the brain, basic counting comes naturally. Preschool and kindergarten kids try their hands at counting objects with our pdfs, and understand that the last number name tells the number of objects.
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Making a connection between "how many there are" and the number, and building skills in recognizing the number of objects without counting are the objectives of our printable counting worksheets.
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Sort and Count Worksheets
Doubling concepts equates to doubling the fun. Categorizing is an important skill, and when it comes along with counting, it adds a feather to the cap. Kids get to sort objects, classify, and then count them.
Word Problems on Counting Worksheets
Counting is ubiquitous! Counting candies and cookies, or trees and butterflies, or be it counting money at the supermarket, you do it everywhere. Kids in grades up to 5 prove their mettle in counting with our real-life scenarios.
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This stock of printable counting worksheets acts as a bridge and propels kids toward subtraction. With a variety of engaging exercises to reverse skip count, these pdfs are a perfect blend of fun and learning.
Counting with Base-10 Blocks Worksheets
Break away from the humdrum and add an extra-ordinary charm to your 1st grade, 2nd grade, and 3rd grade kids' counting practice, as they visualize counting with base-10 manipulatives: units, rods, flats, and cubes.
Counting Using Tally Marks Worksheets
Let's get back in time, and count like our ancestors did! Pique the minds of kids in grade 3 as they learn to draw tally marks to represent objects and count them. They also get to read and interpret tally marks.
One More or One Less Worksheets
Do your students need extra practice with counting one more or one less than a given number? Our ready-to-print one more or one less worksheets include finding one more than with pictures, choosing one less than, doing a cut and glue activity, and lots more.
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Counting by Grouping | Bundles of Tens and Ones Worksheets
Making groups of tens and ones to count objects, quickens the process. Kids in kindergarten, grade 1, and grade 2 count the sticks in bundles and the individual sticks to bolster their place value skills.
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Add a spark of joy with these pdfs that are sure to strike a chord with the kids. Color the pictures according to the color key to reveal the hidden image, and create amazing posters to display their number skills.
Dot-to-Dot Worksheets
Up the fun quotient with this set of worksheets and watch your counting skills improve. Connect the numbers by counting forward, and watch the image take shape. Attempting to color the pictures will double the fun!
Sample Worksheets | 1,100 | 5,447 | {"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} | 3.0625 | 3 | CC-MAIN-2024-38 | latest | en | 0.938189 |
https://www.jiskha.com/display.cgi?id=1172024814 | 1,503,360,890,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109682.23/warc/CC-MAIN-20170821232346-20170822012346-00379.warc.gz | 922,383,102 | 4,292 | # Algebra 1
posted by .
Simplifying Radicals, Use the quotient property to simplify the expression. 3 times the square root of 6/3
3*sqrt(6/3) is the same thing as saying 3*sqrt(2)
3√(6/3) is the same as 3√6/√3
To rationalize the denominator, multiply both the top and bottom by √3/√3 which is the same as multiplying by 1.
Now we have:
3√6/√3 * √3/√3 = 3√18/√9
√18 can be simplified to √(9*2) = 3√2
√9 = 3
Putting it together:
3*3√2/ 3
Cancel out 3 in both the top and bottom.
You are left with 3√2
I hope this will help.
is that not what i just said without all the words?
Yes, you are correct! Since the question asked to use the quotient property, I just expanded on the problem in a different way.
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More Similar Questions | 994 | 2,636 | {"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} | 3.984375 | 4 | CC-MAIN-2017-34 | latest | en | 0.861343 |
https://www.mrexcel.com/board/tags/golf/ | 1,721,294,428,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00389.warc.gz | 780,688,381 | 17,306 | # golf
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Hi all - thanks in advance. I think this is fairly straightforward with an array formula but can't get it to function. If I have a column that has multiple occurences of a word, and the second column contains either a "n" or a "y", if any of the the occurences contain a "Y" I want the formuala...
12. ### Conditional Formating in Calendar
I want to highlight the days on my calendar that match my Event list, but I want the different events to have different colors. example: Event List 1. 1/3/2020 - Baseball 2. 1/4/2020 - Football 3. 1/5/2020 - Golf 4. 1/6/2020 - Baseball 5. 1/7/2020 - Baseball 6. 1/8/2020 - Golf I want the...
13. ### Golf majors; retreive data from Wikipedia articles
Can someone please help me? I am trying to use PowerQuery to create a table in Excel of all the golf majors. I wish to extract their dates from the Wikipedia articles for each tournament. For example, from this web address https://en.wikipedia.org/wiki/1895_U.S._Open_(golf) I want the...
14. ### 8 team Golf Schedule - need formula to choose teams
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16. ### Countif Table
Hello, I have a reception Im going to be keeping track of those people who arrive and I need 2 formulas to keep track of how many group of people have arrived, and of those how many of them are getting an award. I'm not sure how to attach my spreadsheet anymore since I've gone to Excel 2016 so...
17. ### Golf Handicaps
I have a list of 100 names with associated golf scores/handicaps. I need to develop 5 Flights (A, B, C, D, E) into which players will be placed based on their score/handicap, e.g., players who score par to 10 over par (this is a variable) would be placed into Flight A; Players 11 to ? would be...
18. ### Looking for an Hlookup and Vlookup criss cross combined formula...
Hello, I have data in my Sheet that look like this... <colgroup><col width="64" span="8" style="width:48pt"> </colgroup><tbody> 1 2 3 4 5 6 7 alpha 100 1 1 43 22 1 1 bravo 2 2 2 2 2 2 2 charlie 3 3 3 3 3 3 35...
19. ### Average lowest 5 of last 6 golf scores in a column
I'm having difficulty with the formula to find the lowest 5 of the last 6 golf scores listed in a column. The scores are auto-populated from another sheet so each cell with a score already has a formula in it. I will also have weeks when someone doesn't play so blank cells are common. My...
20. ### Working Calculate Golf Skins Spreadsheet by entering scores
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Go back | 1,871 | 7,426 | {"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} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.924567 |
https://lutz-heilmann.info/which-would-you-use-to-estimate-the-demand-for-a-product-at-various-prices/ | 1,643,176,524,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00681.warc.gz | 425,700,223 | 8,927 | Both the demand and supply curve show the relationship in between price and also the number of systems demanded or gave. Price elasticity is the proportion in between the portion adjust in the amount demanded (Qd) or supplied (Qs) and the equivalent percent adjust in price. The price elasticity of demand is the percentage change in the amount demanded of an excellent or business split by the percentage change in the price. The price elasticity of supply is the portion change in quantity supplied separated by the percentage change in price.
You are watching: Which would you use to estimate the demand for a product at various prices?
Elasticities can be usetotally divided right into three wide categories: elastic, inelastic, and also unitary. An elastic demand or elastic supply is one in which the elasticity is better than one, indicating a high responsiveness to changes in price. Elasticities that are less than one suggest low responsiveness to price transforms and correspond to inelastic demand or inelastic supply. Unitary elasticities suggest proportional responsiveness of either demand also or supply, as summarized in Table 1.
If . . .Then . . .And It Is Called . . .
\%;change;in;amount > \%;change;in;pricefrac\%;change;in;quantity\%;change;in;price) > 1Elastic
\%;change;in;amount = \%;change;in;pricefrac\%;change;in;quantity\%;change;in;price) = 1Unitary
\%;change;in;amount
Before we get into the nitty gritty of elasticity, reap this article on elasticity and ticket prices at the Super Bowl.
To calculate elasticity, rather of using straightforward percent transforms in amount and also price, economists use the average percent adjust in both amount and also price. This is dubbed the Midsuggest Method for Elasticity, and is represented in the complying with equations:
= l}\%;change;in;amount & frac Q _ 2 - Q _ 1 ( Q _ 2 + Q _ 1 )/2 imes 100 \<1em> \%;change;in;price & frac P _ 2 - P _ 1 ( P _ 2 + P _ 1 )/2 imes 100 endarray
The advantage of the is Midsuggest Method is that one obtains the exact same elasticity in between two price points whether tbelow is a price rise or decrease. This is because the formula offers the exact same base for both situations.
Calculating Price Elasticity of Demand
Let’s calculate the elasticity in between points A and also B and also between points G and also H presented in Figure 1.
Figure 1. Calculating the Price Elasticity of Demand. The price elasticity of demand is calculated as the percent change in amount split by the percent readjust in price.
First, use the formula to calculate the elasticity as price decreases from \$70 at allude B to \$60 at point A:
= l}\%;change;in;quantity & frac 3,000 - 2,800 ( 3,000 + 2,800 )/2 imes 100 \<1em> & frac 200 2,900 imes 100 \<1em> & = 6.9 \<1em> \%;change;in;price & frac 60 - 70 ( 60 + 70 )/2 imes 100 \<1em> & frac -10 65 imes 100 \<1em> & -15.4 \<1em> Price;Elasticity;of;Demand & frac 6.9\% -15.4\% \<1em> & 0.45 endarray
Thus, the elasticity of demand in between these two points is frac 6.9\% -15.4\% which is 0.45, an amount smaller sized than one, mirroring that the demand also is inelastic in this interval. Price elasticities of demand are always negative given that price and also quantity demanded always relocate in oppowebsite directions (on the demand curve). By convention, we always talk about elasticities as positive numbers. So mathematically, we take the absolute value of the result. We will certainly neglect this information from currently on, while remembering to interpret elasticities as positive numbers.
This suggests that, alengthy the demand also curve between allude B and A, if the price changes by 1%, the quantity demanded will adjust by 0.45%. A change in the price will certainly lead to a smaller percent adjust in the quantity demanded. For example, a 10% increase in the price will certainly cause only a 4.5% decrease in amount demanded. A 10% decrease in the price will lead to only a 4.5% increase in the quantity demanded. Price elasticities of demand also are negative numbers indicating that the demand curve is downward sloping, yet are check out as absolute values. The adhering to Work It Out function will certainly walk you through calculating the price elasticity of demand.
### Finding the Price Elasticity of Demand
Calculate the price elasticity of demand utilizing the data in Figure 1 for a rise in price from G to H. Has the elasticity raised or decreased?
Step 1. We recognize that:
Price;Elasticity;of;Demand also = frac \%;change;in;amount \%;change;in;price
= l}\%;change;in;amount & frac Q _ 2 - Q _ 1 ( Q _ 2 + Q _ 1 )/2 imes 100 \<1em> \%;change;in;price & frac P _ 2 - P _ 1 ( P _ 2 + P _ 1 )/2 imes 100 endarray
= l}\%;change;in;quantity & frac 1,600 - 1,800 ( 1,600 + 1,800 )/2 imes 100 \<1em> & frac -200 1,700 imes 100 \<1em> & -11.76 \<1em> \%;change;in;price & frac 130 - 120 ( 130 + 120 )/2 imes 100 \<1em> & frac 10 125 imes 100 \<1em> & 8.0 endarray
= l}Price;Elasticity;of;Demand also & frac \%;change;in;amount \%;change;in;price \<1em> & frac -11.76 8 \<1em> & 1.47 endarray
Thus, the elasticity of demand from G to H 1.47. The magnitude of the elasticity has actually increased (in absolute value) as we moved up alengthy the demand also curve from points A to B. Respeak to that the elasticity between these 2 points was 0.45. Demand also was inelastic in between points A and B and also elastic in between points G and H. This shows us that price elasticity of demand also changes at various points along a straight-line demand curve.
Calculating the Price Elasticity of Supply
Assume that an apartment leas for \$650 per month and at that price 10,000 devices are rented as displayed in Figure 2. When the price rises to \$700 per month, 13,000 devices are provided into the industry. By what percentage does apartment supply increase? What is the price sensitivity?
Figure 2. Price Elasticity of Supply. The price elasticity of supply is calculated as the percent readjust in quantity divided by the percentage readjust in price.
Using the Midsuggest Method,
= l}\%;change;in;quantity & frac 13,000 - 10,000 ( 13,000 + 10,000 )/2 imes 100 \<1em> & frac 3,000 11,500 imes 100 \<1em> & 26.1 \<1em> \%;change;in;price & frac \$700 - \$650 ( \$700 + \$650 )/2 imes 100 \<1em> & frac 50 675 imes 100 \<1em> & 7.4 \<1em> Price;Elasticity;of;Demand & frac 26.1\% 7.4\% \<1em> & 3.53 endarray
Again, as with the elasticity of demand also, the elasticity of supply is not complied with by any systems. Elasticity is a proportion of one percentage readjust to an additional portion change—nothing more—and also is review as an absolute value. In this case, a 1% rise in price causes a boost in quantity supplied of 3.5%. The better than one elasticity of supply implies that the percent readjust in quantity supplied will be higher than a one percent price adjust. If you"re founding to wonder if the idea of slope fits into this calculation, review the complying with Clear It Up box.
### Is the elasticity the slope?
It is a widespread misrequire to confuse the slope of either the supply or demand curve with its elasticity. The slope is the rate of adjust in devices alengthy the curve, or the rise/run (readjust in y over the readjust in x). For instance, in Figure 1, each point presented on the demand also curve, price drops by \$10 and the variety of systems demanded increases by 200. So the slope is –10/200 alengthy the whole demand also curve and also does not change. The price elasticity, but, changes alengthy the curve. Elasticity in between points A and B was 0.45 and also raised to 1.47 in between points G and H. Elasticity is the percentage readjust, which is a various calculation from the slope and has a various interpretation.
When we are at the top finish of a demand curve, wright here price is high and also the quantity demanded is low, a little adjust in the quantity demanded, also in, say, one unit, is pretty massive in percent terms. A readjust in price of, say, a dollar, is going to be much much less vital in percentage terms than it would certainly have gone to the bottom of the demand curve. Likewise, at the bottom of the demand also curve, that one unit readjust once the amount demanded is high will be little as a percentage.
So, at one finish of the demand curve, wbelow we have actually a huge percentage readjust in quantity demanded over a tiny portion readjust in price, the elasticity value would certainly be high, or demand also would be reasonably elastic. Even via the same change in the price and also the very same change in the quantity demanded, at the other finish of the demand curve the quantity is a lot better, and also the price is much lower, so the portion change in quantity demanded is smaller sized and also the percent readjust in price is a lot better. That indicates at the bottom of the curve we"d have a tiny numerator over a big denominator, so the elasticity measure would certainly be a lot reduced, or inelastic.
As we relocate alengthy the demand curve, the values for amount and price go up or dvery own, depending upon which means we are moving, so the percentages for, say, a \$1 difference in price or a one unit difference in quantity, will change also, which means the ratios of those percentperiods will adjust.
Key Concepts and Summary
Price elasticity measures the responsiveness of the quantity demanded or provided of a good to a readjust in its price. It is computed as the percentage adjust in quantity demanded (or supplied) divided by the portion change in price. Elasticity can be described as elastic (or incredibly responsive), unit elastic, or inelastic (not incredibly responsive). Elastic demand or supply curves show that amount demanded or gave respond to price transforms in a better than proportional manner. An inelastic demand or supply curve is one wbelow a given percent readjust in price will cause a smaller percent readjust in quantity demanded or provided. A unitary elasticity implies that a provided percent readjust in price leads to an equal percentage readjust in amount demanded or offered.
### Recheck out Questions
What is the formula for calculating elasticity?What is the price elasticity of demand? Can you define it in your own words?What is the price elasticity of supply? Can you describe it in your own words?
### Critical Thinking Questions
Transatlantic air travel in company class has an approximated elasticity of demand of 0.40 much less than transatlantic air travel in economy class, through an approximated price elasticity of 0.62. Why carry out you think this is the case?What is the relationship in between price elasticity and also position on the demand curve? For instance, as you relocate up the demand also curve to higher prices and also reduced quantities, what happens to the measured elasticity? How would certainly you describe that?
### Problems
The equation for a demand curve is P = 48 – 3Q. What is the elasticity in relocating from a quantity of 5 to a quantity of 6?The equation for a demand also curve is P = 2/Q. What is the elasticity of demand as price drops from 5 to 4? What is the elasticity of demand as the price drops from 9 to 8? Would you intend these answers to be the same?The equation for a supply curve is 4P = Q. What is the elasticity of supply as price rises from 3 to 4? What is the elasticity of supply as the price rises from 7 to 8? Would you suppose these answers to be the same?The equation for a supply curve is P = 3Q – 8. What is the elasticity in relocating from a price of 4 to a price of 7?
## Glossary
elastic demandas soon as the elasticity of demand also is greater than one, indicating a high responsiveness of quantity demanded or supplied to changes in priceelastic supplywhen the elasticity of either supply is better than one, indicating a high responsiveness of quantity demanded or supplied to changes in priceelasticityan business economics concept that measures responsiveness of one variable to transforms in an additional variableinelastic demandwhen the elasticity of demand is less than one, indicating that a 1 percent boost in price passist by the customer leads to much less than a 1 percent change in purchases (and vice versa); this indicates a low responsiveness by consumers to price changesinelastic supplywhen the elasticity of supply is much less than one, indicating that a 1 percent increase in price paid to the firm will result in a much less than 1 percent rise in manufacturing by the firm; this shows a low responsiveness of the firm to price boosts (and also vice versa if prices drop)price elasticitythe partnership between the percent change in price causing a equivalent percentage adjust in the amount demanded or suppliedprice elasticity of demandpercent readjust in the amount demanded of an excellent or organization split the percent adjust in priceprice elasticity of supplyportion change in the quantity supplied divided by the percent readjust in priceunitary elasticitywhen the calculated elasticity is equal to one indicating that a change in the price of the good or service results in a proportional change in the amount demanded or supplied
### Solutions
From allude B to point C, price rises from \$70 to \$80, and Qd decreases from 2,800 to 2,600. So:
= l}\%;change;in;amount & frac 2,600 - 2,800 ( 2,600 + 2,800 )/2 imes 100 \<1em> & frac -200 2,700 imes 100 \<1em> & -7.41 \<1em> \%;change;in;price & frac 80 - 70 ( 80 + 70 )/2 imes 100 \<1em> & frac 10 75 imes 100 \<1em> & 13.33 \<1em> Elasticity;of;Demand also & frac -7.41\% 13.33\% \<1em> & 0.56 endarray
The demand curve is inelastic in this area; that is, its elasticity value is less than one.
Answer from Point D to suggest E:
= l}\%;change;in;quantity & frac 2,200 - 2,400 ( 2,200 + 2,400 )/2 imes 100 \<1em> & frac -200 2,300 imes 100 \<1em> & -8.7 \<1em> \%;change;in;price & frac 100 - 90 ( 100 + 90 )/2 imes 100 \<1em> & frac 10 95 imes 100 \<1em> & 10.53 \<1em> Elasticity;of;Demand also & frac -8.7\% 10.53\% \<1em> & 0.83 endarray
The demand curve is inelastic in this area; that is, its elasticity worth is much less than one.
Answer from Point G to point H:
= l}\%;change;in;quantity & frac 1,600 - 1,800 ( 1,600 + 1,800 )/2 imes 100 \<1em> & frac -200 1,700 imes 100 \<1em> & -11.76 \<1em> \%;change;in;price & frac 130 - 120 ( 130 + 120 )/2 imes 100 \<1em> & frac 10 125 imes 100 \<1em> & 7.81 \<1em> Elasticity;of;Demand also & frac -11.76\% 7.81\% \<1em> & -1.51 endarray
The demand also curve is elastic in this interval.From point J to suggest K, price rises from \$8 to \$9, and also amount rises from 50 to 70. So:
= l}\%;change;in;amount & frac 70 - 50 ( 70 + 50 )/2 imes 100 \<1em> & frac 20 60 imes 100 \<1em> & 33.33 \<1em> \%;change;in;price & frac \$9 - \$8 ( \$9 + \$8 )/2 imes 100 \<1em> & frac 1 8.5 imes 100 \<1em> & 11.76 \<1em> Elasticity;of;Supply & frac 33.33\% 11.76\% \<1em> & 2.83 endarray
The supply curve is elastic in this area; that is, its elasticity value is better than one.
See more: Explain Why The Function Is Discontinuous At The Given Number A.
From point L to point M, the price rises from \$10 to \$11, while the Qs rises from 80 to 88:
= l}\%;change;in;quantity & frac 88 - 80 ( 88 + 80 )/2 imes 100 \<1em> & frac 8 84 imes 100 \<1em> & 9.52 \<1em> \%;change;in;price & frac \$11 - \$10 ( \$11 + \$10 )/2 imes 100 \<1em> & frac 1 10.5 imes 100 \<1em> & 9.52 \<1em> Elasticity;of;Demand also & frac 9.52\% 9.52\% \<1em> & 1.0 endarray
The supply curve has actually unitary elasticity in this location.
From allude N to suggest P, the price rises from \$12 to \$13, and also Qs rises from 95 to 100:
= l}\%;change;in;quantity & frac 100 - 95 ( 100 + 95 )/2 imes 100 \<1em> & frac 5 97.5 imes 100 \<1em> & 5.13 \<1em> \%;change;in;price & frac \$13 - \$12 ( \$13 + \$12 )/2 imes 100 \<1em> & frac 1 12.5 imes 100 \<1em> & 8.0 \<1em> Elasticity;of;Supply & frac 5.13\% 8.0\% \<1em> & 0.64 endarray | 4,295 | 15,997 | {"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} | 3.703125 | 4 | CC-MAIN-2022-05 | latest | en | 0.908331 |
http://metamath.tirix.org/mpeuni/matvsca2 | 1,721,354,400,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00487.warc.gz | 17,322,603 | 2,791 | Metamath Proof Explorer
Theorem matvsca2
Description: Scalar multiplication in the matrix ring is cell-wise. (Contributed by Stefan O'Rear, 5-Sep-2015)
Ref Expression
Hypotheses matvsca2.a 𝐴 = ( 𝑁 Mat 𝑅 )
matvsca2.b 𝐵 = ( Base ‘ 𝐴 )
matvsca2.k 𝐾 = ( Base ‘ 𝑅 )
matvsca2.v · = ( ·𝑠𝐴 )
matvsca2.t × = ( .r𝑅 )
matvsca2.c 𝐶 = ( 𝑁 × 𝑁 )
Assertion matvsca2 ( ( 𝑋𝐾𝑌𝐵 ) → ( 𝑋 · 𝑌 ) = ( ( 𝐶 × { 𝑋 } ) ∘f × 𝑌 ) )
Proof
Step Hyp Ref Expression
1 matvsca2.a 𝐴 = ( 𝑁 Mat 𝑅 )
2 matvsca2.b 𝐵 = ( Base ‘ 𝐴 )
3 matvsca2.k 𝐾 = ( Base ‘ 𝑅 )
4 matvsca2.v · = ( ·𝑠𝐴 )
5 matvsca2.t × = ( .r𝑅 )
6 matvsca2.c 𝐶 = ( 𝑁 × 𝑁 )
7 1 2 matrcl ( 𝑌𝐵 → ( 𝑁 ∈ Fin ∧ 𝑅 ∈ V ) )
8 7 adantl ( ( 𝑋𝐾𝑌𝐵 ) → ( 𝑁 ∈ Fin ∧ 𝑅 ∈ V ) )
9 eqid ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) = ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) )
10 1 9 matvsca ( ( 𝑁 ∈ Fin ∧ 𝑅 ∈ V ) → ( ·𝑠 ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) = ( ·𝑠𝐴 ) )
11 8 10 syl ( ( 𝑋𝐾𝑌𝐵 ) → ( ·𝑠 ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) = ( ·𝑠𝐴 ) )
12 11 4 eqtr4di ( ( 𝑋𝐾𝑌𝐵 ) → ( ·𝑠 ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) = · )
13 12 oveqd ( ( 𝑋𝐾𝑌𝐵 ) → ( 𝑋 ( ·𝑠 ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) 𝑌 ) = ( 𝑋 · 𝑌 ) )
14 eqid ( Base ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) = ( Base ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) )
15 8 simpld ( ( 𝑋𝐾𝑌𝐵 ) → 𝑁 ∈ Fin )
16 xpfi ( ( 𝑁 ∈ Fin ∧ 𝑁 ∈ Fin ) → ( 𝑁 × 𝑁 ) ∈ Fin )
17 15 15 16 syl2anc ( ( 𝑋𝐾𝑌𝐵 ) → ( 𝑁 × 𝑁 ) ∈ Fin )
18 simpl ( ( 𝑋𝐾𝑌𝐵 ) → 𝑋𝐾 )
19 simpr ( ( 𝑋𝐾𝑌𝐵 ) → 𝑌𝐵 )
20 1 9 matbas ( ( 𝑁 ∈ Fin ∧ 𝑅 ∈ V ) → ( Base ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) = ( Base ‘ 𝐴 ) )
21 8 20 syl ( ( 𝑋𝐾𝑌𝐵 ) → ( Base ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) = ( Base ‘ 𝐴 ) )
22 21 2 eqtr4di ( ( 𝑋𝐾𝑌𝐵 ) → ( Base ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) = 𝐵 )
23 19 22 eleqtrrd ( ( 𝑋𝐾𝑌𝐵 ) → 𝑌 ∈ ( Base ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) )
24 eqid ( ·𝑠 ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) = ( ·𝑠 ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) )
25 9 14 3 17 18 23 24 5 frlmvscafval ( ( 𝑋𝐾𝑌𝐵 ) → ( 𝑋 ( ·𝑠 ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) 𝑌 ) = ( ( ( 𝑁 × 𝑁 ) × { 𝑋 } ) ∘f × 𝑌 ) )
26 6 xpeq1i ( 𝐶 × { 𝑋 } ) = ( ( 𝑁 × 𝑁 ) × { 𝑋 } )
27 26 oveq1i ( ( 𝐶 × { 𝑋 } ) ∘f × 𝑌 ) = ( ( ( 𝑁 × 𝑁 ) × { 𝑋 } ) ∘f × 𝑌 )
28 25 27 eqtr4di ( ( 𝑋𝐾𝑌𝐵 ) → ( 𝑋 ( ·𝑠 ‘ ( 𝑅 freeLMod ( 𝑁 × 𝑁 ) ) ) 𝑌 ) = ( ( 𝐶 × { 𝑋 } ) ∘f × 𝑌 ) )
29 13 28 eqtr3d ( ( 𝑋𝐾𝑌𝐵 ) → ( 𝑋 · 𝑌 ) = ( ( 𝐶 × { 𝑋 } ) ∘f × 𝑌 ) ) | 1,422 | 2,142 | {"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} | 3.46875 | 3 | CC-MAIN-2024-30 | latest | en | 0.360242 |
https://www.liuchuo.net/archives/1036 | 1,726,755,975,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00724.warc.gz | 794,055,966 | 11,748 | # LeetCode 112. Path Sum
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
❤ 点击这里 -> 订阅《PAT | 蓝桥 | LeetCode学习路径 & 刷题经验》by 柳婼
❤ 点击这里 -> 订阅《从放弃C语言到使用C++刷算法的简明教程》by 柳婼
❤ 点击这里 -> 订阅PAT甲级乙级、蓝桥杯、GPLT天梯赛、LeetCode题解离线版 | 203 | 482 | {"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} | 2.859375 | 3 | CC-MAIN-2024-38 | latest | en | 0.569135 |
http://www.mathworks.com/help/symbolic/mupad_ref/stats-geometricrandom.html?nocookie=true | 1,405,306,581,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776439852.43/warc/CC-MAIN-20140707234039-00019-ip-10-180-212-248.ec2.internal.warc.gz | 364,117,714 | 10,426 | Accelerating the pace of engineering and science
# Documentation Center
• Trial Software
# stats::geometricRandom
Generate a random number generator for geometric deviates
### Use only in the MuPAD Notebook Interface.
This functionality does not run in MATLAB.
## Syntax
```stats::geometricRandom(p, <Seed = s>)
```
## Description
stats::geometricRandom(p) returns a procedure that produces geometric deviates (random numbers) with `probability parameter' p.
The procedure f:=stats::geometricRandom(p) can be called in the form f(). The return value of f() is a positive integer if p is a real number satisfying 0 ≤ p ≤ 1.
Otherwise, stats::geometricRandom(p)() is returned symbolically.
Numerical values for p are only accepted if they satisfy 0 ≤ p ≤ 1.
The values X = f() are distributed randomly according to the discrete distribution function of the geometric distribution with parameter p, i.e., for 1 ≤ x, the probability of Xx is given by .
Without the option Seed = s, an initial seed is chosen internally. This initial seed is set to a default value when MuPAD® is started. Thus, each time MuPAD is started or re-initialized with the reset function, random generators produce the same sequences of numbers.
Note: With this option, the parameter p must evaluate to a numerical value at the time, when the generator is created.
Note: In contrast to the function random, the generators produced by stats::geometricRandom do not react to the environment variable SEED.
For efficiency, it is recommended to produce sequences of K random numbers via
f := stats::geometricRandom(p): f() \$k = 1..K;
rather than by
stats::geometricRandom(p)() \$k = 1..K;
The latter call produces a sequence of generators each of which is called once. Also note that
stats::geometricRandom(p, Seed = s)() \$k = 1..K;
does not produce a random sequence, because a sequence of freshly initialized generators would be created each of them producing the same number.
## Environment Interactions
The function is sensitive to the environment variable DIGITS which determines the numerical working precision.
## Examples
### Example 1
We generate geometric deviates with :
`f := stats::geometricRandom(1/3): f() \$ k = 1..10`
`delete f:`
### Example 2
With symbolic parameters, no random floating-point numbers can be produced:
`f := stats::geometricRandom(p): f()`
When p evaluates to a real number between 0 and 1, the generator starts to produce random numbers:
`p := 1/sqrt(70): f(), f(), f()`
`delete f, p:`
### Example 3
We use the option Seed = s to reproduce a sequence of random numbers:
`f := stats::geometricRandom(1/10, Seed = 1): f() \$ k = 1..10`
`g := stats::geometricRandom(1/10, Seed = 1): g() \$ k = 1..10`
`f() = g(), f() = g()`
`delete f, g:`
## Parameters
p The "probability parameter": an arithmetical expression representing a real number 0 ≤ p ≤ 1.
## Options
Seed Option, specified as Seed = s Initializes the random generator with the integer seed s. s can also be the option CurrentTime, to make the seed depend on the current time. This option serves for generating generators that return predictable sequences of pseudo-random numbers. The generator is initialized with the seed s which may be an arbitrary integer. Several generators with the same initial seed produce the same sequence of numbers. When this option is used, the parameter p must be convertible to a floating-point number between 0.0 and 1.0 at the time when the random generator is generated. | 832 | 3,521 | {"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} | 3.03125 | 3 | CC-MAIN-2014-23 | longest | en | 0.798911 |
www.exotransinternational.com | 1,679,993,157,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00639.warc.gz | 841,307,270 | 19,134 | # What’s an algorithm in math?
An algorithm is just a group of guidelines. Computers will use a algorithm proceed to the next step of discovering how exactly to compute that answer and then in order to figure out the answer to a issue.
An algorithm is most usually related to the math of calculus. The method that defines an algorithm can be also called as the integral formula.
An algorithm https://blog.themeson.com/go-to-science-town-kansas-city/ can be also utilised in different branches of mathematics. Algorithms are used to address issues in probability, logarithms, complexity theory, optimization, combinatorics, analysis, and algebra. Some of the sites are tree algorithms, matrix formulations, game algorithms, graph algorithms, and fuzzy logic.
What is an algorithm in math? An algorithm is a set of instructions to be followed. Computers will use an algorithm in order to figure out the correct answer to a problem and then proceed to the next step of determining how to compute that answer. useful source The rules of the algorithm are the most important aspects to consider.
An algorithm can be compared to a set of rules, and a set of instructions. The algorithm can be used to solve a problem by assigning a solution, where the instructions can be used by the computer to perform computations. In addition, the instructions can also be used to track progress of the algorithm.
What is an algorithm in math? The way how the algorithm is applied determines how it can be considered to be useful or useless. The instructions of the algorithm are simple to understand and can be applied in a number of different ways. Algorithms are usually found in software.
One case of an algorithm could possibly become a pair of directions which could be utilized to figure the amounts of polynomials. This really is a practical example, as the code applied to figure out the sums might be changed and also used by the software’s users. Still another example is that a script used to figure that a succession of best research paper logarithms.
Algorithms are most often used in computer programs. It can be complicated for humans to work with these kinds of codes, and it is very difficult for computers to understand them. Algorithms are also commonly used in e-commerce websites.
An algorithm is not utilised inside the vast bulk of people’s own lives. Algorithms have been introduced. This usually means it is important to learn regarding the algorithms before it’ll soon be helpful in your life. There are lots of guides and the internet comprises a lot of details on the subject.
Algorithm is also. An algorithm could be described as”a succession of actions which are required to be followed closely to accomplish that task” There are unique kinds of calculations that are employed.
The most common uses of algorithms are used in computer systems. They can be used to find a solution to a mathematical problem or a solution to a business problem. In fact, there are a lot of companies that do not use a computer at all, but use an algorithm to solve their problems.
It is important to understand why type of an algorithmic undertaking Although the algorithm is still much phrase. You ought to not make use of this procedure for a non-technical endeavor. Make sure you understand what it is that you’re doing when you use a tool such as this. | 657 | 3,360 | {"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} | 3.859375 | 4 | CC-MAIN-2023-14 | longest | en | 0.944242 |
http://mathhelpforum.com/pre-calculus/60874-matrix-determinant-complex-numbers-print.html | 1,506,107,378,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689102.37/warc/CC-MAIN-20170922183303-20170922203303-00200.warc.gz | 228,315,429 | 3,152 | # Matrix Determinant / Complex numbers.
• Nov 21st 2008, 01:21 PM
battery2004
Matrix Determinant / Complex numbers.
Hello,
Ehh, i left this homework till the last day thinking its not too hard, but actualy i have no clue how to completely solve any of them -
Determinant:
http://img374.imageshack.us/img374/9...ricaqh7.th.jpghttp://img374.imageshack.us/images/thpix.gif
http://img355.imageshack.us/img355/539/kskva4.th.jpghttp://img355.imageshack.us/images/thpix.gif
If anyone could solve just one of them it would be much appreciated!
• Nov 21st 2008, 01:31 PM
Krizalid
For the first one, try to turn that matrix into a left triangular one, then you'll get easily the determinant.
As for the second question, it's just Euler's formula application.
• Nov 21st 2008, 01:32 PM
Mathstud28
Quote:
Originally Posted by battery2004
Hello,
Ehh, i left this homework till the last day thinking its not too hard, but actualy i have no clue how to completely solve any of them -
http://img355.imageshack.us/img355/539/kskva4.th.jpghttp://img355.imageshack.us/images/thpix.gif
If anyone could solve just one of them it would be much appreciated!
We know by Euler's Formula that $e^{ix}=\cos(x)+i\sin(x)$. So \begin{aligned}\sqrt[4]{7\left(\cos\left(\frac{\pi}{2}\right)+i\sin\left( \frac{\pi}{2}\right)\right)}&=\sqrt[4]{7}e^{\frac{\pi{i}}{8}}\\ | 395 | 1,346 | {"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": 2, "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} | 3.71875 | 4 | CC-MAIN-2017-39 | longest | en | 0.843578 |
https://tipsfolder.com/much-5-ml-grams-340e611a83704528fad9e30820f4307f/ | 1,713,801,089,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818312.80/warc/CC-MAIN-20240422144517-20240422174517-00752.warc.gz | 501,177,243 | 19,816 | # How much is .5 ml in grams?
Is 1 gram equal to 1 gram? 21 ml = 21 grams 70 ml = 70 grams 4 ml = 4 grams 24 ml = 4 grams 100 ml = 100 grams 6 ml = 6 grams 25 ml = 25 grams 120 ml = 120 grams 7 ml = 7 grams 27 ml = 27 grams 130 ml = 130 grams So, is 1 gram equal to 1 gram? How much is 75ml in grams? Simple Conversion Chart for Water-Based Substances Unit equals to 1 gram 1 milliliter or 0.001 liters 1 kilogram 1000 milliliters or 1 milligram 0.001 milliliters or 0.000001 liters
– Because 1 gram equals 1 ml, there are 75 grams in 75 ml, use the 75 ml to grams converter to figure out how many grams is 75 ml. Simply multiply 75 ml by 1 to get grams to convert 75 ml into g. Also, how much is 250ml in grams?
Because 1 milliliter equals 1 gram, 250 grams are present in 250 ml. What is the difference between 5 mg and ML?
## In cups, how much is 250 grams?
250 grams equals 1.06 cups, or 1.06 cups in 250 grams.
## What is the weight of 25 grams in ml?
How many grams is 25 ml? 25 grams equals 25 ml. Because 1 milliliter equals 1 gram, 25 ml equals 25 grams.
Related Articles:
## How much butter is 250 grams?
Tablespoons 52.86 Teaspoons 1.10 Us. Volume of 250 Grams of Butter 250 Grams of Butter = 17.62 Tablespoons 52.86 Teaspoons Cups 0.92 Imperial Cups 250 Grams of Butter Volume 200 Grams of Butter = 17.62 Tablespoons 52.86 Teaspoons 1.10 United States Cups 0.92 Imperial Cups
## In 1 mL, how many grams is there?
The correct way to convert milliliters to grams is determined by the substance you’re measuring. 1 milliliter of water equals 1 gram, while 1 milliliter of flour equals 0.57 grams.
## In grams, what is 200ml?
g to ml conversion table 1 gram = 1 ml 21 grams = 21 ml 70 grams = 70 ml 13 grams = 33 ml 13 grams = 33.5 ml 14 grams = 1 ml 17 grams = 1 ml 18 grams = 1 ml 20 grams = 1 ml 30 grams = 1 ml 30 grams = 1 ml 30 grams = 1 ml 30 grams = 1 ml 30 grams = 1 ml 30 grams = 1
1 gram 1 gram
## In 100g, how many ml is there?
The answer is that the change of 1 100g (100 grams portion) unit in a cake flour measuring equals 236.59 ml (milliliter) as per the same cake flour type and the equivalent measure.
## How many grams of milk does 250ml contain?
Calculate the amount of milk in grams in 250 ml Ingredient: gram kilogram pound Results: 250 milliliters of milk weighs 259 grams, according to the most significant figures. (*) (*) or 259 grams, to be precise.
All of the values are approximate. Calculate the amount of milk in grams in 250 ml Ingredient: gram kilogram pound Results: 250 milliliters of milk weighs 259 grams, according to the most significant figures.
(*) (*) or 259 grams, to be precise. All of the values are approximate.
## What is the best way to measure grams?
Using a scale is the most accurate way to measure grams. Select a digital or mechanical scale with the metric system and then press the tare button to set it aside. Place your item in the center of the scale, wait for the digital display or needle on the scale to come to a halt, and record the grams.
## What exactly is 600 ml in grams?
In 600 ml, how many grams do you have? 600 Grams: 600 Liter: 0.6 US Fluid Ounce/fl oz: 20.28841 How many grams per 600 ml? ML Conversion milliliter: 600 Grams: 600 Liter: 600 Liter: 0.6 US Fluid Ounce/fl oz: 20.28841 ML Conversion milliliter: 600 Grams: 600 Liter: 0.6 US Fluid Ounce/fl oz: 20.28841
## How many grams is there in a single liter?
1000 grams (g) = 1 liter (l). In the Metric system, Liter (l) is a volume unit.
## How many grams of water does 250ml contain?
100 mL 100 grams 150 mL 150 gram 200 mL 200 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL 250 mL
## Is 250 grams the same as 250 milliliters of water?
Is 250 milliliters of water equal to 250 grams of water? Please explain. No, because a gram is a measurement of mass, whereas a milliliter is a measurement of capacity or volume.
## In grams, how much 250ml flour is?
The answer is that the change of 1 cup 250 mL (Metric cup) unit in a teff flour measuring unit equals 121.52 g (gram) as per the equivalent measurement and for the same teff flour type.
## How many grams of cream does 250ml contain?
Heavy cream Conversion Chart Near 70 milliliters Milliliters to grams of Heavy cream 250 milliliters = 254 grams, 260 milliliters = 264 grams, 270 milliliters = 274 grams Heavy cream Conversion Chart Near 70 milliliters Milliliters to grams of Heavy cream 250 milliliters = 254 grams, 260 milliliters = 264 grams, 260 milliliters = 264 grams, 260 milliliters = 274 grams
## How many grams of water does 50ml contain?
How much water is 50 grams? What is the weight of 50 grams of water? How much is 50 grams of water in a cup? 50 Grams of Water = 0.18 Imperial Cups 0.20 Metric Cups 50.00 Milliliters
What is the weight of 50 grams of water? 50 Grams of Water = 0.18 Imperial Cups 0.20 Metric Cups 50.00 Milliliters 50 Grams of Water | 1,510 | 5,002 | {"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} | 3.625 | 4 | CC-MAIN-2024-18 | latest | en | 0.837889 |
https://socratic.org/physics/thermodynamics/first-law-of-thermodynamics | 1,670,408,819,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00806.warc.gz | 557,498,308 | 8,656 | # First Law of Thermodynamics
## Key Questions
It is a law that relates the exchange of energy of a system with its internal energy.
#### Explanation:
If you have a system (say, a gas) it is possible to change its internal energy by exchanging heat or work with the surrounding.
The first law tells us that when heat and/or work is exchanged the internal energy of the system changes.
This is written in mathematical form as:
$\textcolor{red}{\Delta E = Q - W}$
where $E$ is the internal energy. $Q$ is heat and $W$ is work.
The signs are a bit tricky though; I studied this law as I wrote it because there is a convention in it:
When heat enters the System it is considered POSITIVE;
When heat exits the System it is considered NEGATIVE;
When work enters the System it is considered NEGATIVE;
When work exits the System it is considered POSITIVE;
[I suspect that this convention comes from the Industrial Revolution and the birth of Thermodynamics; a machine that gave out work meant GOOD work...that gave money; while a machine that required work to operate meant BAD work or money you had to input. Heat out was also bad because meant loss in performance and so in money gained....I think]
The first law may seem complicated at first but you have to think at it in terms of exchange of energy.
I give you an example:
Imagine you have a gas in a cylinder that has a movable piston; visualize the internal energy of the gas as the vibration and movement of the gas particles.
Now let us give energy to the gas:
1) we heat the gas (we lit a fire under the cylinder): energy is communicated to the gas through heat ${Q}_{\text{in}} > 0$ so the particles of the gas start to vibrate and move more rapidly: the Internal Energy Increased and:
$\Delta E = Q - 0 > 0$ [no work]
ex: you give $100 J$ of heat (heat in so positive) so the internal energy increases of: $\Delta E = 100 J$
2) we compress the gas: work is done on the gas ${W}_{\text{in}} < 0$ and again energy enters the gas and visually we compress the molecules of the gas that start to bump into each other more frequently increasing their energy=internal energy:
$\Delta E = 0 - W > 0$ [no heat and work "negative"]
ex: you give $100 J$ of work compressing the gas so you get: $W = - 100 J$ because is in :
$\Delta E = 0 - \left(- 100\right) = 100 J$
3) Now you heat the cylinder and let the piston expand: the internal energy increases because the heat BUT also decreases because work is done by the system!!!
ex: heat in (positive) ${Q}_{\text{in}} = 100 J$ and expansion of the piston (work out, positive) ${W}_{\text{out}} = 40 J$ we get:
$\Delta E = Q - W = 100 - \left(+ 40\right) = 60 J$ the internal energy increased only $60 J$
What I do it is always let the minus (in red below) of the formula fixed and insert in brackets the numerical values for $Q$ and $W$ with right signs (in or out) and perform the operation to get $\Delta E$:
$\Delta E = \left(\pm Q\right) \textcolor{red}{-} \left(\pm W\right)$
Hope it helps! | 755 | 3,013 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 21, "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} | 3.96875 | 4 | CC-MAIN-2022-49 | latest | en | 0.951053 |
https://community.fabric.microsoft.com/t5/Desktop/Price-development-of-a-basket-of-products/m-p/3152302/highlight/true | 1,716,502,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00540.warc.gz | 144,771,306 | 130,886 | cancel
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Helper I
Price development of a basket of products
Hi all.
Im trying to figure out how to analyze the total cost of a basket of products on each date.
The basket is a sum of all sales in a given periode. Lets say 3 months.
I also have a price table. The prices changes on a monthly basis.
Any idea how to solve this?
Example data
1 ACCEPTED SOLUTION
Super User
Sorry. I did not check the total that is not correct.
After omitting the relationship between your DateTable and StartDate,
You can fix it with :
1-Create a measure:
SalesQuantityNew=
SUM(ItemSales[Quantity])
Then Change your NewSales as below:
NewSales =
SUMX (
DateTable,
SUMX (
FILTER (
ItemPrice,
ItemPrice[StartDate] <= DateTable[Date]
&& ItemPrice[NewEndDate] >= DateTable[Date]
),
[SalesQuantityNew] * 'ItemPrice'[Price]
)
)
If this is your answer, please mark it as an accepted solution so others can find it easily between the long discussions that happened.
Regards,
Loran
17 REPLIES 17
Helper I
@MohammadLoran25 Thank you so much for taking your time on this one.
It seems like i get the amount on every entry-date. But wat i want is total quantity pr item (regardless of entry-date) * price on the spesific entry-date. Should end up something like this:
Totals:
Yellow = total quantity on all entries.
Green = price * total quantity
Final desired output:
Super User
Okay. Then it would be easier.
You do not need any calculated columns.
Just create these 2 measures:
1-
SalesQuantity =
VAR _ITEM =
SELECTEDVALUE ( ItemPrice[ItemNo] )
RETURN
CALCULATE (
SUM ( ItemSales[Quantity] ),
FILTER ( ItemSales, ItemSales[ItemNo] = _ITEM )
)
2-
SalesAmount =
VAR _PRICE = SELECTEDVALUE ( ItemPrice[Price] ) RETURN _PRICE * [SalesQuantity]
If this answer solves your problem, please mark it as an accepted solution so the others would find what they need easier.
Regards,
Loran
Helper I
Thank you!
We are almost there. The SalesAmount only seem to work when filter on each product.
Is it any way to sum the total amount on all 3 items?
As it is in the table above in column named "SUM Amount"
Super User
You're welcome.
Yes, if you want it as a column you can make a calculated column as below:
SUM Quantity=[SalesQuantity]
[SalesQuantity] It is the measure in my previous message.
To get total amount on all 3 items, you can change the previous measure as below:
SalesAmount=
SUMX(ItemPrice, 'ItemPrice'[SUM Quantity]*'ItemPrice'[Price])
Helper I
I want the sales amount for all items on each and every date. The measures seems to give me the amount on the specific StartDate. For example 16.02.2023 gives me the numer 1624. But it should give me (item1: 1624 + item2: 2007+ item 3: 392= ) 4.023
Super User
Oh man,
What a long discussion. LOLL
Finally I think I got what you need.
Assume your date table name is DateTable and Check the measure below:
NewSalesMeasure =
SUMX (
DateTable,
SUMX (
FILTER (
ItemPrice,
ItemPrice[StartDate] <= DateTable[Date]
&& ItemPrice[NewEndDate] >= DateTable[Date]
),
SUMX ( ItemPrice, [SalesQuantity] * 'ItemPrice'[Price] )
)
)
Then by putting date from your DateTable and ItemNo and This measure you can get what you need.
Helper I
Thank you.
I have set a relationship between Date and StartDate.
As you could see there is no data on for example 17.02.2023. And the 16.02.2023 only has 1624, witch is the StartDate for item 1. It should be amount on the other two items, and the total sum.
Top table has "Show items with no data" on. (Date), the bottom table is off.
Super User
Because you want the measure per each date, so you should write the measure on a separate datetable and only filter the datetable like what I did in my newmeasure.
You can make a new datetable with no relationahip or use CROSSFILTER in the measure to deactivate the relationship in the context of measure.
Super User
In my measure, you should delete the relationship betweem startdate and your datetable. Then it works. Check it.
Helper I
This is what i get without any relationships:
Super User
Sorry. I did not check the total that is not correct.
After omitting the relationship between your DateTable and StartDate,
You can fix it with :
1-Create a measure:
SalesQuantityNew=
SUM(ItemSales[Quantity])
Then Change your NewSales as below:
NewSales =
SUMX (
DateTable,
SUMX (
FILTER (
ItemPrice,
ItemPrice[StartDate] <= DateTable[Date]
&& ItemPrice[NewEndDate] >= DateTable[Date]
),
[SalesQuantityNew] * 'ItemPrice'[Price]
)
)
If this is your answer, please mark it as an accepted solution so others can find it easily between the long discussions that happened.
Regards,
Loran
Helper I
I changed the starting date on price for item #3, the earliest date is now 01.02.23.
My visual looks now like this:
As there is no price from 01.01.23 to 31.01.23 there is nothing to show. Even though there is sales data.
I want to use the price starting from 01.02.23 for sales in the dateperiode 01.01.23 -> 31.01.23. (see green line)
Something like: If SalesDate < earliest StartDate (pr item), use earliest StartDate. Else < startDate && >EndDate
Super User
Hi @Wikolin ,
What a good question !
1-Create a measure as below:
MinStartDateMeasure =
MINX (
FILTER ( ALL ( ItemPrice ), ItemPrice[ItemNo] = MIN ( ItemPrice[ItemNo] ) ),
ItemPrice[StartDate]
)
2-Then Change your NewSales Measure to the this one:
NewSales =
SWITCH (
TRUE (),
SUMX (
FILTER (
DateTable,
DateTable[Date]
<= CALCULATE ( MAX ( ItemPrice[NewEndDate] )ALL ( ItemPrice ) )
),
SUMX (
FILTER (
ItemPrice,
ItemPrice[StartDate] <= DateTable[Date]
&& ItemPrice[NewEndDate] >= DateTable[Date]
),
1
)
) > 0,
SUMX (
FILTER (
DateTable,
DateTable[Date]
<= CALCULATE ( MAX ( ItemPrice[NewEndDate] )ALL ( ItemPrice ) )
),
SUMX (
FILTER (
ItemPrice,
ItemPrice[StartDate] <= DateTable[Date]
&& ItemPrice[NewEndDate] >= DateTable[Date]
),
[SalesQuantity] * 'ItemPrice'[Price]
)
),
SUMX (
FILTER (
DateTable,
DateTable[Date]
<= CALCULATE ( MAX ( ItemPrice[NewEndDate] )ALL ( ItemPrice ) )
),
SUMX (
FILTER (
ItemPrice,
ItemPrice[ItemNo] = MIN ( ItemPrice[ItemNo] )
&& ItemPrice[StartDate] = [MinStartDatemEASURE]
),
[SalesQuantity] * 'ItemPrice'[Price]
)
)
)
If this answer solves your problem, please give it a thumbs up and mark it as an accepted solution so the others would find what they need easier.
Regards,
Loran
Helper I
It worked in one case, but the total does not change.
As you see here, the calculations on each item is the same on each date in this periode.
But the total changes from 01.02.23. (because the real price starts then on item #3)
Would like the total to be (1392+2007+392 = ) 3791.
Super User
Just need to create a measure based on the last measure we created:
NewMeasureWithCorrectTotal=
SUMX(VALUES(ItemPrice[ItemNo]),[NewSales])
**Please keep in mind that because of using SUMX in several measures, the measure might be slow. So based on your data size, if there is any performance problem, of course you need to revise the measures accordingly. Now the logic is clear so changing it would be easier.
Regards,
Helper I
This works perfectly. A minor adjustment. I used [SalesQuantity] instead of [SalesQuantityNew]. Thank you so much @MohammadLoran25
Super User
Hi @Wikolin ,
Assume the name of your first table is ItemPrice
the name of your second table is ItemSales.
1-In your ItemPrice table, Create a calculated column as below:
NewEndDate = IF(ISBLANK(ItemPrice[EndDate]),TODAY(),ItemPrice[EndDate])
2-In your ItemSales table, Create a calculated column as :
PriceBasedOnDate =
CALCULATE (
MIN ( ItemPrice[Price] ),
FILTER (
ItemPrice,
ItemPrice[ItemNo] = ItemSales[ItemNo]
&& ItemSales[Date] >= ItemPrice[StartDate]
&& ItemSales[Date] <= ItemPrice[NewENDDate]
)
)
3-Your sales would be a measure like:
Amount=
SUMX('ItemSales',
'ItemSales'[Quantity]*'ItemSales'[PriceBasedOnDate]
)
If this answer solves your problem, please mark it as an accepted solution so the others would find what they need easier.
Regards,
Loran
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One of the more unintuitive facts about sailing is that it is possible to harness the power of the wind to sail in a direction against that of the wind or to sail with a speed faster than the wind itself, even when the water itself is calm. It is somewhat less known, but nevertheless true, that one can (in principle) do both at the same time – sail against the wind (even directly against the wind!) at speeds faster than the wind. This does not contradict any laws of physics, such as conservation of momentum or energy (basically because the reservoir of momentum and energy in the wind far outweighs the portion that will be transmitted to the sailboat), but it is certainly not obvious at first sight how it is to be done.
The key is to exploit all three dimensions of space when sailing. The most obvious dimension to exploit is the windward/leeward dimension – the direction that the wind velocity $v_0$ is oriented in. But if this is the only dimension one exploits, one can only sail up to the wind speed $|v_0|$ and no faster, and it is not possible to sail in the direction opposite to the wind.
Things get more interesting when one also exploits the crosswind dimension perpendicular to the wind velocity, in particular by tacking the sail. If one does this, then (in principle) it becomes possible to travel up to double the speed $|v_0|$ of wind, as we shall see below.
However, one still cannot sail against to the wind purely by tacking the sail. To do this, one needs to not just harness the power of the wind, but also that of the water beneath the sailboat, thus exploiting (barely) the third available dimension. By combining the use of a sail in the air with the use of sails in the water – better known as keels, rudders, and hydrofoils – one can now sail in certain directions against the wind, and at certain speeds. In most sailboats, one relies primarily on the keel, which lets one sail against the wind but not directly opposite it. But if one tacks the rudder or other hydrofoils as well as the sail, then in fact one can (in principle) sail in arbitrary directions (including those directly opposite to $v_0$), and in arbitrary speeds (even those much larger than $|v_0|$), although it is quite difficult to actually achieve this in practice. It may seem odd that the water, which we are assuming to be calm (i.e. traveling at zero velocity) can be used to increase the range of available velocities and speeds for the sailboat, but we shall see shortly why this is the case.
If one makes several simplifying and idealised (and, admittedly, rather unrealistic in practice) assumptions in the underlying physics, then sailing can in fact be analysed by a simple two-dimensional geometric model which explains all of the above statements. In this post, I would like to describe this mathematical model and how it gives the conclusions stated above. | 653 | 2,962 | {"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": 5, "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} | 2.90625 | 3 | CC-MAIN-2014-52 | latest | en | 0.943155 |
https://www.emedicalprep.com/study-material/chemistry/states-of-matter/gaseous-state/molecular-collisions/ | 1,503,563,571,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133447.78/warc/CC-MAIN-20170824082227-20170824102227-00151.warc.gz | 888,258,543 | 177,439 | Molecular Collisions
# Molecular Collisions
(1) The closest distance between the centres of two molecules taking part in a collision is called molecular or collision diameter (s). The molecular diameter of all the gases is nearly same lying in the order of 10-8m
(2) The number of collisions taking place in unit time per unit volume, called collision frequency (z).
(i) The number of collision made by a single molecule with other molecules per unit time are given by, ZA = √2πσ2uav.n
Where n is the number of molecules per unit molar volume,
(ii) The total number of bimolecular collision per unit time are given by,
(iii) If the collisions involve two unlike molecules, the number of bimolecular collision are given by,
MA, MB are molecular weights (M = mN0)
(iv) (a) At particular temperature; Z∝P2
(b) At particular pressure; Z∝ T-3/2
(c) At particular volume; Z∝ T1/2
(3) During molecular collisions a molecule covers a small distance before it gets deflected. The average distance travelled by the gas molecules between two successive collision is called mean free path (λ).
(4) Based on kinetic theory of gases mean free path, λ∝(T/P). Thus,
(i) Larger the size of the molecules, smaller the mean free path, i.e., λ∝ (1/(radius)2)
(ii) Greater the number of molecules per unit volume, smaller the mean free path.
(iii) Larger the temperature, larger the mean free path.
(iv) Larger the pressure, smaller the mean free path.
(5) Relation between collision frequency (Z) and mean free path (l) is given by, Z = Urms/ λ
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# Graphing exponential functions on the SAT
## Studying graphing exponential functions
On the SAT test graphing exponential functions topic is part of passport to advanced mathematics subscore that includes 9 advanced topics (see the full topics list on the top menu).
Graphing exponential functions topic is the fifth topic of passport to advances mathematics subscore. It is recommended to start learning passport to advances mathematics subscore with its first four topics (the first topic is quadratic equations and quadratic functions).
Graphing exponential functions topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.
Finish studying heart of algebra subscore topics before you study this topic or any other passport to advances mathematics subscore topic. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding passport to advanced mathematics subscore topics). Particularly before learning this topic learn the exponential expressions topic
## Graphing exponential functions- summary
Exponential function is a function with a positive constant other than 1 raised to an exponent that includes a variable.
The basic form of the exponential function is f(x)=bx (b is the base and x is the exponent).
The base b is always positive (b>0) and not equal to one (b≠1).
For example: the function f(x)=3x is an exponential function where the base is a constant b=3 and the exponent is the variable x.
The y axis intercept of the basic exponential function graph f(x)=bx is equal to 1 for all values of b.
An exponential function slope is always increasing or always decreasing. The slope form depends on the value of the base b. All graphs of exponential functions are curved.
The ends of the exponential function graph: The graph of a basic exponential function f(x)=bx has a horizontal asymptote on one of its ends (positive x axis or negative x axis). The other end of the function approaches infinity. The end behavior depends on the value of the base b.
We can shift an exponential function graph by adding a constant to the function or by multiplying the exponential term by a coefficient, so that the basic function f(x)=bx will become f(x)=a*bx+d.
### Two types of exponential functions
The form of the graph of the basic exponential function f(x)=bx depends on the size of the base b:
If b>1 then the slope of the graph is always positive (the graph is increasing), as x increases the graph approaches infinity and as x decreases the graph approaches zero (horizontal asymptote at x=0).
If 0<b<1 then the slope of the graph is always negative (the graph is decreasing), as x increases the graph approaches zero (horizontal asymptote at x=0) and as x decreases the graph approaches infinity.
### The features of exponential functions
#### Y intercept of an exponential function
Y intercept this is the point where the function crosses the y axis.
To find the y intercept, we set x=0 and evaluate the function.
Note that any number raised to a power of 0 is equal to 1, therefore in the basic function f(x)=bx the intercept is equal to 1 for all values of b.
Consider the following example:
Find the y intercepts of a functions f(x)=3x+1 and f(x)=3x-5.
f(x)=3x+1
f(0)=30+1=1+1=2
f(x)=3x-5
f(0)=3x-5=1-5=-4
#### The slope of an exponential function
An exponential function slope is always increasing or always decreasing, therefore, to find the slope of the function we only need to evaluate 2 points from the function.
Let’s mark point 1 (x1,y1) and point 2 (x2, y2).
If the y value increases when the x value increases (exponential grows), then the slope of the function is positive, and the function graph is always increasing (meaning that x2>x1 and y2>y1).
If the y value decreases when the x value increases (exponential decay), then the slope of the function is negative, and the function graph is always decreasing (meaning that x2>x1 and y2<y1).
Consider the following example:
Determine if the slope of the function f(x)=3x+1 is negative or positive.
f(x)=3x+1
f(0)=30+1=1+1=2
f(1)=31+1=3+1=4
The points are (0,2) and (1,4). We got a bigger y for a bigger x, therefore the slope of the function is positive.
Consider the following example:
Determine if the slope of the function f(x)=0.5x+1 is negative or positive.
f(x)=0.5x+1
f(0)=0.50+1=1+1=2
f(1)=0.51+1=1.5
The points are (0,2) and (1,1.5). We got a smaller y for a bigger x, therefore the slope of the function is negative.
The graphs below show the functions f(x)=0.5x+1 and f(x)=3x+1.
The points x+0 and x=1 that were used to find the slopes are marked in red on the graphs.
#### The horizontal asymptote- the end behavior
Asymptote is a line that the graph approaches but never touches.
Horizontal asymptote is a horizontal line that the graph approaches when x gets very large or very small.
The graph of a basic exponential function f(x)=bx has a horizontal asymptote on one of its ends (positive x axis or negative x axis):
If b>1 then the function f(x)=bx has a horizontal asymptote at the negative end of the x axis.
When we raise a base that is bigger than 1 to a high negative exponent the output is a very small number, therefore the output of the function f(x)=bx is very close to zero.
For example:
Given the function f(x)=bx
if x=-4 and b=3 we get f(x)=3-4=1/34=0.012
if x=-10 and b=3 we get f(x)=3-10=1/310=0.00002
If 0<b<1 then the function f(x)=bx has a horizontal asymptote at the positive end of the x axis.
When we raise a base that is smaller than 1 to a high positive exponent the output is a very small number, therefore the output of the function f(x)=bx is very close to zero.
For example:
Given the function f(x)=bx
if x=4 and b=0.5 we get f(x)=0.54=0.5*0.5*0.5*0.5=0.063
if x=10 and b=0.5 we get f(x)=0.510=0.5*0.5*0.5*0.5…*0.5=0.00098
Consider the following example:
Determine the asymptotes of the functions f(x)=3x and f(x)=0.5x.
The graphs below show the functions f(x)=0.5x and f(x)=3x.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.
The horizontal asymptote x=0 is marked in red on the graph and inside the table.
For the graph f(x)=3x small x values will get an output that is close to zero, therefore the function has a horizontal asymptote at x=0.
For the graph f(x)=0.5x big x values will get an output that is close to zero, therefore the function has a horizontal asymptote at x=0.
#### Horizontal asymptote of a function with a constant term f(x)=bx+d
In this type of functions, a constant term d is added to the basic function f(x)=bx so we get a function form of f(x)=bx+d.
When the function has a constant term, the asymptote will approach the constant instead of zero.
For example: b values of 0<b<1:
if x=4, b=0.5 and c=5 we get f(x)=0.54+5=5.063
if x=10, b=0.5 and c=-4 we get f(x)=0.510-4=-4.00098
For example: b values of b>1:
if x=-4, b=3 and c=5 we get f(x)=3-4+5=1/34+5=5.012
if x=-10, b=3 and c=-4 we get f(x)=3-10-4=1/310-4=-4.00002
Consider the following example:
Determine the asymptotes of the functions f(x)=3x+1 and f(x)=0.5x+1.
The graphs below show the functions f(x)=0.5x+1 and f(x)=3x+1.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.
The horizontal asymptote x=1 is marked in red on the graph and inside the table.
For the graph f(x)=3x+1 small x values will get an output that is close to zero plus 1, therefore the function has a horizontal asymptote at x=1.
For the graph f(x)=0.5x+1 big x values will get an output that is close to zero plus 1, therefore the function has a horizontal asymptote at x=1.
#### Graphing an exponential function steps
An exponential function f(x)=bx behavior is divided into 2 areas by the y axis: a positive x area and a negative x area. Therefore, to graph an exponential function, we need to include a point from each area.
The steps for graphing an exponential function:
Step 1: Plotting points:
Evaluate and plot 3 points- its y intercept, a point with a positive x value (like x=1) and a point with a negative x value (like x=-1).
Step 2: Sketching a curve:
Sketch a curve between the 3 points and extend it on both sides. One end will approach a horizontal asymptote of zero along the x axis (if the graph is f(x)=bx+d the asymptote will be x=k instead of x=0). The other and will approach infinity along the y axis.
The graphs below show the functions f(x)=0.5x and f(x)=3x.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.
The 3 points from the different areas that were needed to plot the graphs (x=0, x=3 and x=-3) are marked in red and blue in the table and on the graphs.
Both functions f(x)=3x and f(x)=0.5x have a y intercept at a point (1,0).
The green function f(x)=3x where b>1 has big y outputs at positive x values (approaching to infinity) and small y outputs at negative x values (approaching to zero).
The orange function f(x)=0.5x where 0<b<1 has big y outputs at negative x values (approaching to infinity) and small y outputs at positive x values (approaching to zero).
### Shifting an exponential function
We can shift an exponential function by adding a constant to the function or by multiplying the exponential term by a coefficient.
#### Shifting an exponential function- adding a constant term
To shift an exponential function up we need to add a constant term to the basic function f(x)=bx getting f(x)=bx+d.
To shift an exponential function down we need to subtract a constant term from the basic function f(x)=bx getting f(x)=bx-d.
The shifting results:
The y axis intercept shifts up or down d units.
The asymptote shifts up or down d units.
The other end of the function that approaches infinity remains unchanged.
In the basic function f(x)=bx the horizontal asymptote is x=0 and the y intercept is y=1.
In the function f(x)=bx+d the horizontal asymptote will be x=d and the y intercept will be y=1+d.
In the function f(x)=bx-d the horizontal asymptote will be x=-d and the y intercept will be y=1-d.
The graphs below show the functions f(x)=0.5x , f(x)=0.5x+10, f(x)=3x and f(x)=3x+10.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.
The y axis intersection points are marked in red in the table and on the graph. The y axis intersection point before shifting was y=1 and after the shifting it became y=d+1=10+1=11.
Every point on both functions shifted up by the size of the constant d (d=10), you can see the change comparing the columns of the tables. Two points from each function are marked in blue in the tables and on the graphs as an example.
The asymptotes of both functions shifted up by the size of the constant d (d=10), they are marked in red in the table and on the graphs.
#### Shifting an exponential function- multiplying by a coefficient
We can shift an exponential function by multiplying the exponential term by a coefficient so that the function f(x)=bx will become f(x)=a*bx.
The shifting results:
The y axis intercept will become a*1=a.
The asymptote remains unchanged (x=0).
The other end of the function that approaches infinity remains unchanged.
The graphs below show the functions f(x)=0.5x , f(x)=5*0.5x, f(x)=3x and f(x)=5*3x.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.
The y axis intersection points are marked in red in the table and on the graph. The y axis intersection point before shifting was y=1 and after the shifting it became y=a*1=5.
Note that the shifting size of each point is different, since bigger x values result in a bigger output and therefore a bigger shifting, you can see the change comparing the columns of the tables.
The asymptotes of both functions remained unchanged, they are marked in black in the tables and on the graphs.
#### Shifting an exponential function- adding a constant term and multiplying by a coefficient
The basic function f(x)=bx will become f(x)=a*bx+d.
The shifting results:
The y axis intercept will become a*1+d=a+d.
The asymptote will be x=d.
The other end of the function that approaches infinity remains unchanged.
The graphs below show the functions f(x)=0.5x , f(x)=5*0.5x+10, f(x)=3x and f(x)=5*3x+10.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.
The y axis intersection points are marked in red in the table and on the graphs. The y axis intersection point before shifting was y=1 and after the shifting it became y=a*1+d=5*1+10=15.
Note that the shifting size of each point is different, since bigger x values result in a bigger output and therefore a bigger shifting, you can see the change comparing the columns of the tables.
The asymptotes of both functions shifted up by the size of the constant d (d=10), they are marked in black in the table and on the graphs.
You just finished studying graphing exponential functions topic, the fifth topic of passport to advanced mathematics subscore!
Continue studying the next passport to advanced mathematics subscore topic- polynomial functions and graphs. | 3,370 | 13,320 | {"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} | 4.59375 | 5 | CC-MAIN-2022-27 | longest | en | 0.875115 |
https://artofproblemsolving.com/wiki/index.php/2007_iTest_Problems/Problem_11 | 1,701,778,683,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100551.17/warc/CC-MAIN-20231205105136-20231205135136-00402.warc.gz | 134,870,988 | 11,111 | # 2007 iTest Problems/Problem 11
## Problem 11
Consider the "tower of power" $2^{2^{2^{\cdot^{\cdot^{\cdot^{2}}}}}}$, where there are 2007 twos including the base. What is the last (units digit) of this number?
$\text{(A) }0\qquad \text{(B) }1\qquad \text{(C) }2\qquad \text{(D) }3\qquad \text{(E) }4\qquad \text{(F) }5\qquad \text{(G) }6\qquad \text{(H) }7\qquad \text{(I) }8\qquad \text{(J) }9\qquad \text{(K) }2007\qquad$
## Solution
Note that $2^1 = 2$, $2^2 = 4$, $2^3 = 8$, $2^4 = 16$, and $2^5 = 32$. The units digit of $2^n$ cycle every time $n$ is increased by $4$.
Since $2^{2^{2^{\cdot^{\cdot^{\cdot^{2}}}}}}$ with $2006$ twos is a multiple of four, the units digit is $\boxed{\textbf{(G) }6}$. | 300 | 711 | {"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": 13, "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} | 4.3125 | 4 | CC-MAIN-2023-50 | latest | en | 0.620723 |
https://math.stackexchange.com/questions/2579432/lipschitz-property-of-fx-y-fracxy-sqrtx2y2-for-x-y-neq-0-0/2579461 | 1,563,943,553,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195530385.82/warc/CC-MAIN-20190724041048-20190724063048-00382.warc.gz | 464,304,937 | 30,480 | # Lipschitz property of $f(x,y)= \frac{xy}{\sqrt{x^2+y^2}}$ for $(x,y)\neq (0,0)$ and $f(0,0)=0.$
Prove that $$f(x,y)=\left \{\begin {array}{lll} \displaystyle \frac{xy}{\sqrt{x^2+y^2}}, & ~(x,y)\neq (0,0)\\ 0, & ~\textrm{διαφορετικά}\\ \end{array} \right.$$ is Lipschitz, with constant $M=1$, that is: $$|f(x_1,y_1)-f(x_2,y_2)|\leq \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$ for all $(x_1,y_1),~(x_2,y_2).$
Attempt. By using MVT for one variable, we get for some $y'$ and some $x'$ $$f(x_1,y_1)-f(x_2,y_2)=\big(f(x_1,y_1)-f(x_1,y_2)\big)+\big(f(x_1,y_2)-f(x_2,y_2)\big)=f_y(x_1,y')(y_1-y_2)+f_x(x',y_2)(x_1-x_2).$$ By triangle inequality and since $|f_x|,~|f_y|$ are bounded above by $1$: $$|f(x_1,y_1)-f(x_2,y_2)|\leq |f_y(x_1,y')|\,|y_1-y_2|+|f_x(x',y_2)| \,|x_1-x_2|\leq |y_1-y_2|+|x_1-x_2|\leq \sqrt{2}\,\sqrt{(x_1-x_2)^2+(y_1-y_2)^2},$$ so we get the desired property for $M=\sqrt{2}$. Is $M=1$, as stated above, incorrect or am I missing something?
It looks like treating each variable separately makes you overestimate the bound. If you use MVT for $f$ as a function of two variables, you have $$f(x_1,y_1)-f(x_2,y_2)=\nabla f(\vec{p}) \cdot \big( (x_1,y_1)-(x_2,y_2)\big)$$ for some $\vec p \in \mathbb R^2$ lying in the segment that joins both points. So by Cauchy-Schwartz inequality you can say $$|f(x_1,y_1)-f(x_2,y_2)|\le||\nabla f(\vec{p})|| \cdot ||(x_1,y_1)-(x_2,y_2)||$$ and so it's enough to show that the gradient has norm not greater than one, which is true.
• There is a little more argument to give if the segment passes through $(0,0)$ where the gradient is not defined, but this is easily solved by using an intermediate point tending to $(0,0)$. – Gribouillis Dec 25 '17 at 8:08
• Indeed, we have $\|\nabla f(x,y)\|^2=\Big(\frac{y^2-x^2}{y^2+x^2}\Big)^2\leq 1$ for $(x,y)\neq (0,0)$. – Nikolaos Skout Dec 25 '17 at 8:16
• Yes. Actually it's a little trickier: both partial derivatives exist at $(0,0)$ and equal $0$, so the gradient exists at that point (in fact $\nabla f (0,0)= (0,0)$). But the MVT is not applicable at that point because (as it can be proven) $f$ is not differentiable there. – Alejandro Nasif Salum Dec 25 '17 at 8:23 | 893 | 2,157 | {"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} | 4.125 | 4 | CC-MAIN-2019-30 | latest | en | 0.733056 |
https://issuu.com/stephanieguttmann/docs/dd.m1.stephanieguttmannjournal | 1,553,549,150,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204300.90/warc/CC-MAIN-20190325194225-20190325220225-00517.warc.gz | 516,425,684 | 20,999 | Digital Design - Module 01 Semester 1, 2019 Stephanie Guttmann 992350 Dan Parker Studio 12
Week One
Reading: Zeara Polo, A. 2010. Between Ideas and Matters.
According to Zeara-Polo, the diagram does not play a representational role in the design process but provides an organisational and can have a performative quality depending on how it is deployed. Explain how Diagram is different from Signs and Symbols? (100 words Maximum)
Signs and symbols both represent the role of, as Zeara-Polo put, the “dynamic object.� They express and represent the object in an organisational manner, while a diagram’s purpose is of defining. Rather than representing the object, diagrams represent the information from the object. A diagram collects the information, interprets it and then presents it.
2
Week One
Precedent Analysis
(Top Left): Plan of Pavilion, showing the traced curves from a scaled plan. (Top right): Perspective, showing the initial modeling stage where I was able to use simple extrusions. (Bottom): Perspective, this shows the process of creating the complicated curves. As they varied in height, each point needed to be moved vertically according to the designed height. The first step for modeling the pavilion was tracing the curves in rhino from the plan. The lower half of the pavilion (walls and benches) allowed for simple extrusions from these curves. However, the more complicated multidirectional curves the pavilions canopy required more modeling skills, and individually moving points up and down vertically to ensure the heights are correct at every angle of the moduled pavilion.
3
Week Two
Reading: Hertzberger H. 2005. The in-between and The Habitable Space Between Things, from Lessons for Students in Architecture. Herzberger discusses how design should not be extreme in its functionality. Use your precedent study to explain how the pavilion allows for an appropriation of use. (100 words Maximum)
Barkow Leibinger’s Serpentine Summer House initially appears inflexible due to the obvious benches which define a seating area. Yet, upon further analysis the pavilion does allow for an appropriation of use through its curved structure. The Pavilion has no clear entry and exit points, nevertheless the open curved walls invite people in from all directions and creates a fluid space where people can gather, lean on the walls and children can play, while also creating some more private spaces. Additionally, the canopy roof while appearing to be a decorative extension to the gallery acts as a shelter and provides protection from the sun.
4
Week Two
Isometric
South East Isometric of Barkow Leibinger’s Serpentine Summer House I chose to show the roof canopy as the major focal point of the isometric. The canopy while apearing to be decorative, defines the space of the pavilion and create interesting shadows below. This view also shows the part of the canopy where one curve tilts outwards, this creates visual interest but also a wider shade area. The bench structure is also visible in this view along with its various large and small openings. The key learning in terms of circulation from modeling this pavilion would be its lack of defined entry and exit points and round shape. People regardless if they are or are not engaging with the pavilion must walk around the structure. Yet, the curved walls also provide points that draw people in from any angle creating a public gathering place. The same walls also deifne a smaller space that is more secluded and private. As well as the walls the canopy acts as a threshold. Gathering spaces will vary throughout the day from the shadow casted, if its a hot day people are more likely to sit in the shade and if its cold the sun. As this pavillion was located in England, most of the time people would move away from the shade into the suns warmth.
5
Week Two Diagrams
Circulation Diagram The diagram defines the primary gathering areas and circulation paths. Crowd gathering is greatest in the benches and up against the walls with people sitting and leaning. Their is also a higher density of people in the sunnier spots. People circulate around the pavilion rather than through it as it is acessible from all surounding angles. A commonly traveled path is from the gallery downwards away from the shaded spaces.
6
Threshold The canopy structure and the mirrored tanbark define a zone that can be considered inside the pavilion, its contained but still open to the park. The walls further define these zones by creating a small private zone only acessible from the less circulated shaded side of the pavilion.
Appendix
Process
In the shaded view form in rhino, the pavilion was modeled in the corresponding colour coded layers (above). Once complete I changed this view to rendered (below).
Annotated plan and elevations of the complex pavilion in order to break it down into parts clearly and colour coded so it will be easier to model. All hand written dimentions are in mm
7
Appendix Process
The above pavilion analysis focuses on the circulation of people and the thresholds created by the canopy shadows and the curved walls.
Pavillion Isometric Views: ((left to right) g ) South East, South West, North West, North East
8
Appendix
Process
Diagram process. Circulation (above) and Threshold (below)
The process of creating the shadows for the diagram was done on rhino. Using the sun setting in and imputting the location and time, I then traced the shadow in the isometric view, before making 2D and converting into Adobe Illustrator
9 | 1,193 | 5,635 | {"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} | 3.046875 | 3 | CC-MAIN-2019-13 | latest | en | 0.878369 |
http://davidburchnavigation.blogspot.com/2012/12/great-circle-sailing-by-sight-reduction.html | 1,726,754,459,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00636.warc.gz | 9,630,116 | 16,265 | ## Thursday, December 13, 2012
### Great Circle Sailing by Sight Reduction
It is not often we need to know the route of shortest distance waypoints (great circle route, GC) from one position to another on the globe, because usually wind or other matters dominate the routing. But sometimes we do–more often than not, when we are taking a navigation exam for some license or certification!
The best practical solution is just type the Lat-Lon of departure and destination into an electronic charting system (ECS) or a computer or calculator program and it will give the results immediately. The basic results are the distance between the two points and the initial heading of the route (in GC sailing the heading changes continuously along the route). To apply this route, however, we need waypoints along the route since the heading is changing. This is usually accomplished in these programs by telling them a longitude interval, and then the program tells you the Lat at each of these Lon intervals along the route–in other words, a set of waypoints.
Another parameter often given out is the vertex of the route, which is the Lat-Lon of the highest Lat along the route. Since the GC route only differs notably from the rhumb line route (RL) for high Lat at departure and destination, we often find out that the vertex hits the ice, so we can’t use this route anyway!
If you do not have an ECS device to compute the basic data you can get all the answers by drawing a straight line from departure to destination on what is called a great circle chart or plotting sheet. Then just pull off the waypoints with dividers from the grid on the chart. The distance can also be summed up this way in steps along the route. Places like Captains Nautical Supply sell GC Plotting Sheets.
Without ECS or GC plotting sheets we are left to compute these things on our own. This can be done from the basic spherical trig equations used in cel nav, or we can use the sight reduction tables that celestial navigators use to solve routine position fixing from sextant sights. The best set of tables for this application is Pub. 229, and at this point we will have to assume you are familiar with these tables. This application is standard for the most part, but you will see we need a couple extra interpolations not usually called for, but in principle always a small improvement.
We just do a sight reduction but replace the Assumed Position (AP) with the Departure (dep), and replace the Geographical Position (GP) with the Destination (dest). In other words, the sight reduction process tells us the angular height (Hc) and azimuth (Zn) of a star viewed from the assumed position (a-Lat, a-Lon). And we know that the zenith distance (z = 90º - Hc) is the distance from the AP to the GP, so the GC distance is just z converted from angle to nmi at the rate of 1º = 60 nmi. Zn is then the initial heading (always poleward of the RL heading).
Let us work an example: What is the CG distance and initial heading from 13º 12’ N, 49º 35’ E to 15º 04.6’ N, 54º 49.2’ E? This brings up the point that we can use this procedure to figure the course and distance between two points, even if we do not care if there is any difference between CG and RL. This example is at low Lat and short distance, so there will not be much difference in the two solutions.
Dear Reader: at this point you have to decide if you really care about this subject, because it gets more tedious from here on.
a-Lat = dep. Lat = 13º 12’ N
a-Lon = dep. Lon = 49º 35’ E
dec = dest. Lat = N 15º 04.6’ (We write the N in front when calling it a declination.)
GHA = dest. Lon = 54º 49.2’ E = 305º 10.8’ (see below)
Now we spot a bit of a twist in the process. The GHA is measured 0 to 360 headed W from Greenwich, whereas Lon goes 0 −180 W and 0 − 180 E from Greenwich (Lon = 0). So we have to convert our dest. Lon into the equivalent meridian labeled as if it were a GHA.
(For example, 20º W Lon is just GHA = 20º, but 20º E Lon would be 360 − 20 = 340º GHA. To get to the meridian of 20º E, i have to go west 340º.)
So Lon 54º 49.2’ E = 359º 60’ - 54º 49.2’ is the same as GHA = 305º 10.8’
Then following regular sight reduction procedures, we figure the Local Hour Angle (LHA) = GHA + a-Lon(E), where we choose minutes of a-Lon so that they add to GHA for a whole degree, thus a-Lon = 49º 49.2’ so LHA = 305º 10.8’ + 49º 49.2’ = 354º 60’ = 355º
Next we choose an AP with whole degrees since the tables only have whole degrees of a-Lat and LHA. In this application we can just round them off to get a-Lat = 13º N and LHA = 355º, and we note that since our declination is also N, we have a Same Name solution.
Now we enter Pub 229 with a-Lat = 13 N, dec = N 15, and LHA = 355 same name to get: Hc = 84º 45.2', d = −26.9'* and Z = 067º, and since the Hc is so high, we need to get Z from a-Lat = 14 as well and it is Z = 057.6. In other words, when bodies are high in the sky, any small change in anything changes the bearing, so we will have to interpolate for 13º 12'. (d = altitude difference, and the * means a dsd correction is called for, but we are skipping this for now.)
ie Lat interpolation for consecutive declinations gives:
Zn = Z = 67.0 + [(12/60) x (77.7-67.0)] = 069.14 for dec = 15 at Lat 13º 12'
Zn = Z = 66.9 + [(12/60) x (66.9-57.6)] = 068.76 for dec = 16 at Lat 13º 12'
Now interpolate for the minutes of dec (which at 04.6/60 should be small)
Final Zn = 069.14 - [(4.6/60) x (69.14-68.76)] = 069.1º and that is the initial heading of the GC route we are after.
To find distance we have to finish getting an accurate Hc. With dec min = 4.6' and d = −26.9’ we get the Hc correction in three parts from Pub 229, but we can skip the dsd correction and just add the tens (1.5’) and units (0.5’) corrections to get 2.0’, which is negative, so we have Hc = 84 45.2 − 2.0 = 84 43.2.
Altitude difference (d) of 26.9 means tens = 20, units = 6, decimals = 0.9, and you get total correction as shown below.
Then z = 89 60 − 84 43.2 = 5 16.8, which converted to nmi = 300 + 16.8 = 316.8. But this is GC from the AP not from the departure, so we have to make a correction, which is best done by plotting.
Here we see where we computed from (AP = center of plotting sheet) and where we were at the same time, and thus had to add the projected distance (red line) to what we got. This step could also diminish the earlier result. There are several forms of cel nav plotting sheets that can be used for this. See some in the support page for our cel nav text.
After plotting we see we need to add another 9.5 nmi for a total GC distance of 316+9.5 = 324.5 nmi
——————————
Alternatively, if you know these formulas and have a trig calculator, you can get the result this way:
lat = 13º 12' = 13.2º,
dec = 15º 04.6' = 15.077º
LHA = GHA + Lon = 305º 10.8' + 49º 35.0' = 354º 45.8' = 354.763º (we do not need to use AP when doing a direct computation.)
and sin Hc = sin (13.2) sin(15.077) + cos(13.2) cos(15.077) cos (354.763) = 0.995539
or solve the arcsin to get: Hc = 84.5862 and z = 90-Hc = 5.41378º then x 60 to get GC distance = 324.8 nmi. Our plot and table work was off a hair.)
And:
tan Z = cos(15.077) sin (354.763) / [ cos(13.2) cos(15.077) - sin(13.2) cos(15.077) cos(354.763) ] = -2.6172338039, and solving for the arc tangent:
or Z = Zn = 069.09º, which is what we got from the tables (069.1).
Notation note: cap Z = azimuth angle, lower case z = zenith distance, true bearing or azimuth = Zn.
The above formula is what the programs use. It is obviously much easier to use a calculator that has these formulas already programmed in. We offer a free calculator for this at www.starpath.com/navpubs.
=========
PS. The above note is for doing GC computations over large distances (without a computer!). If the run you need to compute is less than 500 miles or so at low latitude, you can get a good estimate more easily with mid-latitude sailing. That is:
Solve a right triangle with one side = dLat = 15.077 - 13.2 = 1.877 x 60 nmi = 112.62 nmi
and on the other side take the departure as dLon x cos (mid-lat)
convert to decimals
= (54.82 - 49.58) x cos [(15.077+13.2)/2] = 5.062º x 60 = 303.7 nmi.
Then the run is the hypotenuse = sq root (112.62^2 + 303.7^2) = 323.9 nmi compared to 324.8,
and the course is then E xx N, where xx = arc tan (112.6/303.7) so xx = 20.3º, and CMG = 90 -20.3 = 069.7 compared to 069.1
========
Summary: after you pass all of your tests on these subjects, buy a calculator that will do all this for you... and much more. | 2,489 | 8,571 | {"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} | 3.203125 | 3 | CC-MAIN-2024-38 | latest | en | 0.937557 |
http://oeis.org/A231989 | 1,547,722,913,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658928.22/warc/CC-MAIN-20190117102635-20190117124635-00067.warc.gz | 156,957,987 | 3,868 | This site is supported by donations to The OEIS Foundation.
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A231989 Least prime p such that p(0),...,p(n) are all primes, where p(0) = p, then p(i+1) = triangular(p(i))+1. 1
3, 3, 43, 236367611, 31542795419 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Triangular(p) = p*(p+1)/2. LINKS EXAMPLE a(2) = 3 because 3 is the least prime such that the following are two primes: p1 = 3 * 4 / 2 + 1 = 7. p2 = 7 * 8 / 2 + 1 = 29. a(3) = 43 because 43 is the least prime such that the following 3 numbers are prime: p1 = 43 * 44 / 2 + 1 = 947. p2 = 947 * 948 / 2 + 1 = 448879. p3 = 448879 * 448880 / 2 + 1 = 100746402761. CROSSREFS Cf. A000040, A000217, A231847, A231988. Sequence in context: A082394 A086889 A269909 * A059495 A174325 A290366 Adjacent sequences: A231986 A231987 A231988 * A231990 A231991 A231992 KEYWORD nonn,hard,more AUTHOR Alex Ratushnyak, Nov 16 2013 STATUS approved
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Last modified January 17 05:26 EST 2019. Contains 319207 sequences. (Running on oeis4.) | 484 | 1,468 | {"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} | 3.734375 | 4 | CC-MAIN-2019-04 | latest | en | 0.750305 |
http://www.jiskha.com/display.cgi?id=1288646952 | 1,498,234,513,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320070.48/warc/CC-MAIN-20170623151757-20170623171757-00599.warc.gz | 563,688,139 | 3,832 | # Math.
posted by on .
A butcher has 18 2/3 pounds of steak that he wishes to place into packages that average 2 1/3 pounds each. How many packages can he make?
• Math. - ,
(18 2/3) / (2 1/3)
(56/3) / (7/3)
(56/3) * (3/7) = 168/21 = 8 | 95 | 239 | {"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} | 3.53125 | 4 | CC-MAIN-2017-26 | latest | en | 0.963042 |
https://www.physicsforums.com/threads/n-divisible-n-torsionfree.278285/ | 1,521,397,766,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645830.10/warc/CC-MAIN-20180318165408-20180318185408-00463.warc.gz | 852,097,833 | 17,916 | # N-divisible, n-torsionfree
1. Dec 8, 2008
### mathsss2
Let $$n$$ be a nonzero integer. An abelian group $$A$$ is called $$n$$-divisible if for every $$x \in A$$, there exists $$y \in A$$ such that $$x=ny$$. An abelian group $$A$$ is called $$n$$-torsionfree if $$nx=0$$ for some $$x \in A$$ implies $$x=0$$. An abelian group $$A$$ is called uniquely $$n$$-divisible if for any $$x \in A$$, there exists exactly one $$y \in A$$ such that $$x=ny$$.
Let $$\mu_n : A \rightarrow A$$ be the map $$\mu_n(a)=na$$
(a) Prove that $$A$$ is $$n$$-torsionfree iff $$\mu_n$$ is injective and that $$A$$ is $$n$$-divisible iff $$\mu_n$$ is surjective.
(b) Now suppose $$0 \rightarrow A \overset{f}{\rightarrow} B \overset{g}{\rightarrow} C \rightarrow 0$$ is an exact sequence of abelian groups. It is easy to check that the following diagram commutes: [see attachment]
Suppose that $$B$$ is uniquely $$n$$-divisible. Prove that $$C$$ is $$n$$-torsionfree if and only if $$A$$ is $$n$$-divisible.
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2. Dec 8, 2008
### Hurkyl
Staff Emeritus
It's hard to be helpful if I have no idea where you're stuck....
3. Dec 10, 2008
### mathsss2
So I am now thinking to use the snake lemma for part b). However, I am a little confused on how to do this. Also, for part a) these statements seem a bit trivial, I will work on those proofs using the definitions. Do you have any ideas on part a) or b), I never used the snake lemma to prove something yet.
4. Dec 10, 2008
### Hurkyl
Staff Emeritus
I agree, (a) looks easy. I think it's just a matter of writing out the definitions, as you say.
For (b) you have the right idea -- in the two cases (assuming C n-torsion-free and assuming A n-divisible), what did you compute as the sequence you get from the snake lemma?
5. Dec 11, 2008
### mathsss2
So here is what I have for the first part of a).
$$A$$ is $$n$$-torsionfree
$$\Leftrightarrow \forall x\in A\ (nx=0 \Rightarrow x=0)$$
$$\Leftrightarrow \forall x\in A\ (\mu_n(x)=0 \Rightarrow x=0)$$
$$\Leftrightarrow \mu_n$$ is injective
I am confused now on how to do the second part of a).
6. Dec 12, 2008
### mathsss2
How do I do the second part with the snake lemma? I do not know how to do this part.
Here is the picture:
http://i719.photobucket.com/albums/ww191/xianghu21/2.png?t=1229062259" [Broken]
Last edited by a moderator: May 3, 2017
7. Dec 12, 2008
### Hurkyl
Staff Emeritus
Doesn't the same idea work? What went wrong?
Can you be more precise? Is it... that you don't know how to write down the exact sequence given by the snake lemma? Is it... that you don't know what to do with the sequence one you have it? Is it... something else?
Last edited by a moderator: May 3, 2017
8. Dec 13, 2008
### mathsss2
So, I need to take the kernels and cokernels of the vertical mappings and apply the snake lemma. Also, I need to write down the exact sequence to use the snake lemma. I am not sure how to actually do this. I never used the snake lemma to prove something before. I guess I just need to know how to start.
9. Dec 13, 2008
### mathsss2
Here is how to do the second part of part a):
$$\mu_n \text{ is surjective}$$
$$\Leftrightarrow \forall y\in A,\, \exists x\in A \, (\mu (x) = y)$$
$$\Leftrightarrow \forall y\in A,\, \exists x\in A\, (nx = y)$$
$$\Leftrightarrow \text{ A is n-divisible}$$
For part b):
I am still not sure how to do part b).
10. Dec 13, 2008
### Hurkyl
Staff Emeritus
I don't see what the problem is. I presume you know what the 6 groups are -- you even have concrete descriptions of many of them. (As opposed to a purely formal description, such as "$\\ker \left( B \xrightarrow{\mu_n} B \right)$") I presume you even know how those groups are to be organized into an exact sequence. So I don't understand why you are not sure how to write the exact sequence.
11. Dec 14, 2008
### mathsss2
I still do not know how to solve (b) using the snake lemma or any other ways. help!
When I use the snake lemma, how will this help me with what I am trying to show? And for both directions? I am very confused.
just solved it. :-)
Last edited: Dec 14, 2008
12. Dec 14, 2008
### Hurkyl
Staff Emeritus
From the information you've given me, it sounds like you haven't even bothered to write down what the snake lemma tells you in this case! And if you haven't even done that, then of course you don't know how it will help!
It's okay to try something without knowing how it will be useful. In fact, doing so is required for doing mathematics. (In fact, it's a requirement for doing research in any field!)
So... just do it! Write down what the snake lemma tells you, and figure out everything you can from there. If you can't manage to see your way to the end of the problem, then come back here and show us how far you have gotten, and we can help you figure out what you've overlooked.
13. Dec 14, 2008
### mathsss2
I solved it Hurkyl (as I said in my post if you read it thoroughly). And no, I didn't use the snake lemma at all. One step was messing me up, but I got through it.
14. Dec 15, 2008 | 1,516 | 5,070 | {"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} | 3.421875 | 3 | CC-MAIN-2018-13 | latest | en | 0.831437 |
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# Module 2 28 - page 11 / 12
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Module 2 38
d. Graph P(t).
# Polynomial and Rational Inequalities
Solve the following polynomial inequalities by first writing them in general form, factoring (if needed) to identify boundary points, drawing a number line with boundary points labeled, and testing each interval. Express the solution in interval notation.
1. x2 > 7x
2. 12x2 < 37x + 10
3. (x + 2)(x – 5)(x + 7) ≤ 0
Solve the following rational inequalities by first writing them as a single simplified rational expression on one side and zero on the other side of the inequalities. Identify the boundary points by labeling them on a number line and then testing the intervals. Express the solution in interval notation.
4.
5.
6.
Applications
7. The profit from selling x pieces of handmade jewelry can be modeled by the
equation P(x) = –x2 + 130x – 3000. What is the range of profitable orders (P(x) > 0)?
8. The difference between the value of a piece of equipment and what is owed on it is
modeled by the expression d(x) = , where x is the number of years since
the equipment was purchased. When does the amount owed exceed the value of
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Singapore Math in Rotterdam 2 Opleiding Singapore rekenspecialist
Singapore Math in Rotterdam 2 Opleiding Singapore rekenspecialist. This set of slides cover the presentation made on the second day: The Model Method.
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Singapore Math in Rotterdam 2 Opleiding Singapore rekenspecialist
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1. Singapore Math in Rotterdam 2 Opleiding Singapore rekenspecialist • This set of slides cover the presentation made on the second day: • The Model Method De centralelaats van visualisatie in het rekenonderwijskunnenplaatsen. Leren hoe je de Model Methodekunttoepassenbij het oplossen van redactieopgaven. Begrijpen hoe dezeModelbenaderinghelptomzakentevisualiseren en omverbadentezientussen de feiten en de informatie die in de opgavengegevenwordt.
2. Singapore Math in Rotterdam 2 Opleiding Singapore rekenspecialist Review of Day 1 What are some features of Singapore Math and its theoretical underpinnings. On Day 2, we look at the focus on visualization and the model method.
3. Review of Day 1 Yeap Ban Har, Ph.D. Marshall Cavendish Institute Singapore banhar@sg.marshallcavendish.com
4. Variations Tasks are varied in a systematic way to ensure that average and struggling learners can learn well.
5. Math in Focus 2A
6. Math in Focus 2A
7. Math in Focus 2A
8. ZoltanDienes The three lessons include mathematical variations within the same grade. This is referred to as a spiral approach.
9. It is likely that a teacher will start this unit using the sticks. This is followed by the use of base ten blocks. Finally, non-proportionate materials such as coins are used. In each of these lessons, the teacher is likely to introduce the following five notations in turn – place value chart, expanded notation, number in numerals, number in words and the tens and ones notation. The question is what is an appropriate sequence? Should the place value chart be used first? Or the expanded notation? Give your reasons. Place Value Chart Expanded Notation Words Numerals Tens and Ones Notation Primary Mathematics
10. ZoltanDienes This lesson include perceptual variations. This is Dienes’ idea of multiple embodiment. The mathematical concept is constant while the materials used to embody it are varied.
11. Jerome Bruner Bruner advised teachers to use the CPA Approach in teaching mathematics.
12. Richard Skemp Skemp distinguished between instrumental understanding from relational understanding to encourage teachers to teach for conceptual understanding.
13. conceptual skemp’s theory understanding BinaBangsa School, Semarang, Indonesia
14. Example 2 Division in Other Grade Levels
15. My Pals Are Here! Mathematics 3A
16. My Pals Are Here! Mathematics 3A
17. My Pals Are Here! Mathematics 3A
18. My Pals Are Here! Mathematics 3A
19. My Pals Are Here! Mathematics 3A
20. My Pals Are Here! Mathematics 3A
21. My Pals Are Here! Mathematics 3A
22. My Pals Are Here! Mathematics 3A
23. My Pals Are Here! Mathematics 3A
24. Keys Grade School, Manila, The Philippines
25. Keys Grade School, Manila, The Philippines
26. The Bar Model Method de strookmodel Yeap Ban Har, Ph.D. Marshall Cavendish Institute Singapore banhar@sg.marshallcavendish.com
27. Beliefs Interest Appreciation Confidence Perseverance Monitoring of one’s own thinking Self-regulation of learning Attitudes Metacognition Numerical calculation Algebraic manipulation Spatial visualization Data analysis Measurement Use of mathematical tools Estimation Mathematical Problem Solving Reasoning, communication & connections Thinking skills & heuristics Application & modelling Skills Processes Concepts Numerical Algebraic Geometrical Statistical Probabilistic Analytical Mathematics Curriculum Framework
28. visualization Wellington Primary School
29. Primary Mathematics Standards Edition
30. John had 1.5 m of copper wire. He cut some of the wire to bend into the shape shown in the figure below. In the figure, there are 6 equilateral triangles and the length of XY is 19 cm. How much of the copper wire was left? 19 cm x 5 = 95 cm 150 cm – 95 cm = 105 cm
31. There was an interesting discussion on this problem. There was an explanation that a + b + c = 19 cm. Then there was an assumption that a : b : c = 4 : 2 : 1 which was met with rebuttals such as there is no need to know a : b : c as well as the point that a : b : c can be determined by measuring or folding.
32. The Bar Model Method de strookmodel Yeap Ban Har, Ph.D. Marshall Cavendish Institute Singapore banhar@sg.marshallcavendish.com
33. Ali has 3 sweets. Billy has 5 sweets. How many sweets do they have altogether? Ali Billy
34. Ali has 3 sweets. Billy has 5 sweets. How many sweets do they have altogether? Ali Billy
35. Introduction The focus is on the bar model method.
36. Materials developed by Poon Yain Ping
37. Materials developed by Poon Yain Ping
38. Materials developed by Poon Yain Ping
39. Summary The three basic situations are part-whole, comparison and before-after situations.
40. Materials developed by Poon Yain Ping
41. The class decided that this was impossible. The teacher asked the class to change this to another number to make the situation possible. We discussed when it is 3, 4 and 5 times. A student gave an incorrect solution for the second part. The teacher asked students to write a question for which this would be a correct solution. Materials developed by Poon Yain Ping
42. Summary We discussed how to use students’ responses to make the lesson focus on depth. We also saw how a problem can be modified to challenge learners.
43. Materials developed by Poon Yain Ping
44. Materials developed by Poon Yain Ping
45. Materials developed by Poon Yain Ping
46. School Assessment women men
More Related | 1,360 | 5,769 | {"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} | 3.4375 | 3 | CC-MAIN-2024-33 | latest | en | 0.642481 |
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Number of problems found: 783
• Segments
Line segments 62 cm and 2.2 dm long we divide into equal parts which lengths in centimeters is expressed integer. How many ways can we divide?
• Dog II
Journey from a rangers' home to the town is 6 km. Mr. ranger started from his home and went 3 km per hour. His women came from the town and also went 3 km per hour. With ranger started dog and runs in 12 km in 1 hour. The dog went to ranger's women then b
• Tetrapack
How high should be the milk box in the shape of a prism with base dimensions 8 cm and 8.8 cm if its volume is 1 liter?
• Triangle
Determine if it is possible to construct a triangle with sides 28 31 34 by calculation.
• Wire
From wire 53 meters long they cut of 12.1 m and then 13.1 m. How many meters of wire left?
• Aircrafts
Above the town hall tower flew the plane with constant speed 510 km/h and 15 minutes later the second plane at speed of 703 km/h. How long and how far from the town hall will be aircrafts caught up?
• Two cars
Car A1 goes at average speed of 106 km/h and the second car A2 goes at 103 km/h. How many second will it take car A1 to circulate car A2? Assume that both cars are 5 meters long and the safety gap between cars is 1.5 meters.
• Tachometer
Tatra's tachometer shows the initial state 886123 km this morning. Tatra today travel at an average speed of 44 km/h. Determine the function that describes the Tatra's tachometer depending on the time. What is the state of tachometer after 4 hours?
• Knight
Knight passed 13 km long track in 26 minutes, what was his average speed?
• Shaving
Mathematical problem from Sunday evening. If the beard is growing at 0.52 mm per day and that my length is 16.8 mm, how long I don't shave?
• MPG
252 miles a week fills up twice a week. Each fill is 38 gallons. What is gas mileage of the car?
• 30-60-90
The longer leg of a 30°-60°-90° triangle measures 5. What is the length of the shorter leg?
• Scouts
The boys from scout group traveled 5 days distance 115 km. Every day walked 1.5 km less than the previous day. How many kilometers scouts walked in the first day?
• Dog
Man goes with a dog on a long walk 17 km from the house. Man is walking at speed 4.9 km/h and a dog that constantly running between house and man at 12.6 km/h. How many kilometers will run the dog when they reach their destination?
• Carpenter
From wooden block carpenter cut off a small cuboid block with half the edge length. How many percent of wood he cut off?
• Bases
The length of the bases trapezium are in ratio 4:5. Length of midline is 15. How long are the bases of a trapezoid?
• Car
The driver of a car is to get to 608 km distant city. From atlas found that 162 km will have to pass through the cities at average speed 48 km/h. Remainder of the journey pass outside the cities at average speed 116 km/h. Calculate how many hours it will
• Trains
From station 130 km away started passenger train and after 2.2 hours after the express train, which travels 37 km an hour more. Express train finish journey 7 minutes early. Calculate the average speed of this two trains.
• Prism X
The prism with the edges of the lengths x cm, 2x cm and 3x cm has volume 20250 cm3. What is the area of surface of the prism?
• Prism
The base of the prism is a rhombus with a side 30 cm and height 27 cm. The height of the prism is 180% longer than the side length of the rhombus. Calculate the volume of the prism.
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Do you want to convert length units? | 1,007 | 3,906 | {"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} | 3.890625 | 4 | CC-MAIN-2020-45 | latest | en | 0.939258 |
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• Nov 6th 2008, 04:22 AM
oxrigby
trig differential
Im doing this differential and used the chain/product rule up untuil i cannot simplify any further. Its quite messy and ive spent ages looking for a way to do it. The Q was differentiate with respect to x and simplify where possible
$\displaystyle sin^3xcos3x+cos^3xsin3x$
I got down to:$\displaystyle \frac{sin^3xsin3x}{3}+3sin^2xcos3xcosx+\frac{cos^3 xcos3x}{3}-sin3xcos^2xsinx$ a bit long dont you think?
• Nov 6th 2008, 05:16 AM
HallsofIvy
Quote:
Originally Posted by oxrigby
Im doing this differential and used the chain/product rule up untuil i cannot simplify any further. Its quite messy and ive spent ages looking for a way to do it. The Q was differentiate with respect to x and simplify where possible
$\displaystyle sin^3xcos3x+cos^3xsin3x$
I got down to:$\displaystyle \frac{sin^3xsin3x}{3}+3sin^2xcos3xcosx+\frac{cos^3 xcos3x}{3}-sin3xcos^2xsinx$ a bit long dont you think?
It's a bit completely incorrect is what I really think! (and perhaps not long enough.) You seem to be confusing differentiation with anti-differentiation. The derivative of cos 3x, for example is -sin(3x) times 3, not divided by 3.
The derivative of [tex]sin^3(x) cos(3x)[/itex] alone is $\displaystyle (sin^3(x))' cos(3x)+ sin^3(x)(cos(3x))'= (3 sin^2(x)cos(x))(cos(3x))- 3sin^3(x)sin(3x)$ and you do the derivative of $\displaystyle cos^3(x) sin(3x)$ the same way.
• Nov 6th 2008, 05:19 AM
mr fantastic
Quote:
Originally Posted by oxrigby
Im doing this differential and used the chain/product rule up untuil i cannot simplify any further. Its quite messy and ive spent ages looking for a way to do it. The Q was differentiate with respect to x and simplify where possible
$\displaystyle sin^3xcos3x+cos^3xsin3x$
I got down to:$\displaystyle \frac{sin^3xsin3x}{3}+3sin^2xcos3xcosx+\frac{cos^3 xcos3x}{3}-sin3xcos^2xsinx$ a bit long dont you think?
There are a number of careless mistakes (corrected in red below) in your answer.
$\displaystyle {\color{red}-3 }sin^3x sin3x+3sin^2xcos3xcosx+ {\color{red}3} cos^3xcos3x - {\color{red}3} sin3xcos^2xsinx$
It's still long, which is why simplification is asked for.
• Nov 6th 2008, 05:29 AM
Soroban
Hello, oxrigby!
I have no idea how you're getting fractions . . .
Quote:
Differentiate: .$\displaystyle f(x) \:=\:\sin^3\!x\cdot\cos3x + \cos^3\!x\cdot\sin3x$
Product Rule and Chain Rule . . .
$\displaystyle f'(x) \;=\;(\sin^3\!x)\!\cdot\!(\text{-}3\sin3x) + (3\sin^2\!x\cos x)\!\cdot\!(\cos3x) +$ $\displaystyle (\cos^3\!x)\!\cdot\!(3\cos3x) + (\text{-}3\cos^2\!3x\sin x)\!\cdot\!(\sin3x)$
. . . $\displaystyle = \;-3\sin^3\!x\sin3x + 3\sin^2\!x\cos x\cos3x + 3\cos^3\!x\cos3x - 3\sin x\cos^2\!x\sin3x$
$\displaystyle \text{Factor: }\;3\sin^2\!x\cdot\underbrace{(-\sin x\sin3x \;+\; \cos x\cos3x)} \;+\; 3\cos^2\!x\cdot\underbrace{(\cos x\cos3x \;-\; \sin x\sin3x)}$
. . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \nwarrow\;_{\text{ common factor! }}\;\nearrow$
$\displaystyle \text{Factor: }\;(3\sin^2\!x + 3\cos^2\!x)\,(\cos x\cos3x - \sin x\sin3x)$
$\displaystyle \text{Factor: }\;3\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}}\,\underbrace{(\cos x\cos3x - \sin x\sin3x)}_{\text{This is }\cos(x + 3x)}$
Therefore: .$\displaystyle f'(x) \;=\;3\cos4x$ | 1,255 | 3,306 | {"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} | 4.125 | 4 | CC-MAIN-2018-26 | latest | en | 0.880294 |
https://kluedo.ub.uni-kl.de/citationExport/index/download/searchtype/simple/query/%2A%3A%2A/browsing/true/rows/10/facetNumber_subject/all/sortfield/author/sortorder/desc/start/0/yearfq/2009/facetNumber_author_facet/all/subjectfq/binomial+tree/docId/2166/output/ris | 1,571,222,049,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-43/segments/1570986666959.47/warc/CC-MAIN-20191016090425-20191016113925-00476.warc.gz | 588,749,222 | 2,074 | TY - THES A1 - Müller, Stefanie T1 - The Binomial Approach to Option Valuation: Getting Binomial Trees into Shape N2 - This thesis deals with the application of binomial option pricing in a single-asset Black-Scholes market and its extension to multi-dimensional situations. Although the binomial approach is, in principle, an efficient method for lower dimensional valuation problems, there are at least two main problems regarding its application: Firstly, traded options often exhibit discontinuities, so that the Berry- Esséen inequality is in general tight; i.e. conventional tree methods converge no faster than with order 1/sqrt(N). Furthermore, they suffer from an irregular convergence behaviour that impedes the possibility to achieve a higher order of convergence via extrapolation methods. Secondly, in multi-asset markets conventional tree construction methods cannot ensure well-defined transition probabilities for arbitrary correlation structures between the assets. As a major aim of this thesis, we present two approaches to get binomial trees into shape in order to overcome the main problems in applications; the optimal drift model for the valuation of single-asset options and the decoupling approach to multi-dimensional option pricing. The new valuation methods are embedded into a self-contained survey of binomial option pricing, which focuses on the convergence behaviour of binomial trees. The optimal drift model is a new one-dimensional binomial scheme that can lead to convergence of order o(1/N) by exploiting the specific structure of the valuation problem under consideration. As a consequence, it has the potential to outperform benchmark algorithms. The decoupling approach is presented as a universal construction method for multi-dimensional trees. The corresponding trees are well-defined for an arbitrary correlation structure of the underlying assets. In addition, they yield a more regular convergence behaviour. In fact, the sawtooth effect can even vanish completely, so that extrapolation can be applied. KW - Finanzmathematik KW - Option KW - Derivat KW - Bewertung KW - Binomialbaum KW - Approximationsalgorithmus KW - Finanznumerik KW - Multi-Asset Option KW - Konvergenzrate KW - Konvergenzverhalten KW - Sägezahneffekt KW - monotone Konvergenz KW - Extrapolation KW - computational finance KW - multi-asset option KW - option valuation KW - binomial tree KW - rate of convergence KW - convergence behaviour KW - sawtooth effect Y1 - 2009 UR - https://kluedo.ub.uni-kl.de/frontdoor/index/index/docId/2166 UR - https://nbn-resolving.org/urn:nbn:de:hbz:386-kluedo-24627 ER - | 550 | 2,621 | {"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} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.907923 |
https://research.stlouisfed.org/fred2/series/PL2GDPINA621NUPN | 1,448,975,455,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398467979.1/warc/CC-MAIN-20151124205427-00089-ip-10-71-132-137.ec2.internal.warc.gz | 831,345,346 | 19,395 | # Price Level of GDP, average of GEKS-CPDW for India
2010: 36.69611 U.S.=100
Annual, Not Seasonally Adjusted, PL2GDPINA621NUPN, Updated: 2012-09-17 10:23 AM CDT
1yr | 5yr | 10yr | Max
Price Level of GDP is the PPP over GDP divided by the exchange rate times 100. The PPP of GDP or any component is the national currency value divided by the real value in international dollars. The PPP and the exchange rate are both expressed as national currency units per US dollar.The value of price level of GDP for the United States is made equal to 100. Price Levels of the components Consumption, Investment, and Government are derived in the same way as the price level of GDP. While the U.S. = 100 over GDP, this is not true for the component shares. The purchasing power parity in domestic currency per \$US for GDP or any component, may be obtained by dividing the price level by 100 and multiplying by the Exchange Rate. More information is available at http://pwt.econ.upenn.edu/Documentation/append61.pdf.
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: p2
Source: University of Pennsylvania
Release: Penn World Table 7.1
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Graph updated. | 641 | 2,466 | {"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} | 2.75 | 3 | CC-MAIN-2015-48 | latest | en | 0.828233 |
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# Which of the following most logically completes the
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Which of the following most logically completes the [#permalink]
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19 Jun 2006, 08:40
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Which of the following most logically completes the argument?
The irradiation of food kills bacteria and thus retards spoilage. However, it also lowers the nutritional value of many foods. For example, irradiation destroys a significant percentage of whatever vitamin B1 a food may contain. Proponents of irradiation point out that irradiation is no worse in this respect than cooking. However, this fact is either beside the point, since much irradiated food is eaten raw, or else misleading, since _______.
A. many of the proponents of irradiation are food distributors who gain from food’s having a longer shelf life
B. it is clear that killing bacteria that may be present on food is not the only effect that irradiation has
C. cooking is usually the final step in preparing food for consumption, whereas irradiation serves to ensure a longer shelf life for perishable foods
D. certain kinds of cooking are, in fact, even more destructive of vitamin B1 than carefully controlled irradiation is
E. for food that is both irradiated and cooked, the reduction of vitamin B1 associated with either process individually is compounded
E, I guess. but what is wrong with A?
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19 Jun 2006, 08:49
Why not C
_________________
I believe its yogurt!
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### Show Tags
19 Jun 2006, 09:13
alexiswhy wrote:
Which of the following most logically completes the argument?
The irradiation of food kills bacteria and thus retards spoilage. However, it also lowers the nutritional value of many foods. For example, irradiation destroys a significant percentage of whatever vitamin B1 a food may contain. Proponents of irradiation point out that irradiation is no worse in this respect than cooking. However, this fact is either beside the point, since much irradiated food is eaten raw, or else misleading, since _______.
A. many of the proponents of irradiation are food distributors who gain from food’s having a longer shelf life
E. for food that is both irradiated and cooked, the reduction of vitamin B1 associated with either process individually is compounded
E, I guess. but what is wrong with A?
The question is "which....most logically....." , that's the key to eliminate A, i think.
While A provides the reason for " misleading" , it doesn't derive from the logical flow of the argument.
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19 Jun 2006, 09:48
C. cooking is usually the final step in preparing food for consumption, whereas irradiation serves to ensure a longer shelf life for perishable foods
but "usually the final step " is contradicted with the fact that "vegies are usually eaten raw"...
the point is "Proponents of irradiation point out that irradiation is no worse in this respect than cooking" then to logically complete the sentence, you have to refute the proponent's argument.
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19 Jun 2006, 11:22
Will go with C.
We have to show a striking dissimilarility between cooking and irradiation.
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23 Jun 2006, 10:44
I shall go with E
it is misleading that irradiation is no worse because it compounds the effect of loss of Vit B1.
option C says that cooking is the last step but irradiation is done to preserve the foods. and this point is favour of the proponents of Irradiation. The sentence following this uses " However, this fact is either beside the point, since " and "else misleading, since" which implies that these points must weaken the the claim of the proponents of irradiation
E does this effectively because it explains why the claim that irradiation is good is not true.
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23 Jun 2006, 15:48
I am going for E
logically completes the argument needs something that has been stated before and no NEW INFO!
on that basis anything to do with shelves needs to be eliminated, so A and C are out
now, B is the same idea... we were never told about other effects of this irradiation
D: "certain types of cooking..." don't read further... the argument is talking about cooking in general
E holds: you irradite veggies, then cook those -> double trouble...
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23 Jun 2006, 15:48
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# Which of the following most logically completes the
Moderators: GMATNinjaTwo, GMATNinja
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,530 | 6,022 | {"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} | 2.9375 | 3 | CC-MAIN-2017-47 | latest | en | 0.932197 |
http://kitada.com/List/time/199908/0050.html | 1,558,908,356,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259757.86/warc/CC-MAIN-20190526205447-20190526231447-00147.warc.gz | 110,433,197 | 2,491 | # [time 556] Re: [time 553] Re: [time 535]Modeling change with nonstandard numbers
Matti Pitkanen (matpitka@pcu.helsinki.fi)
Mon, 16 Aug 1999 21:46:00 +0300 (EET DST)
Dear Stephen et all,
1. In TGD context quantum jump Einsteinian solution to Zeno paradox
holds is modified somewhat. With respect to geometric time
there is no motion: tortoise becomes 4-dimensional
geometric object. With respect to subjective time
the observed motion of tortoise is discretized with
average time increment of about 10^4 Planck times per quantum
jump: cognitive spacetime sheet jumps by this temporal distance in
each quantum jump and sees new t=constant section of 4-dimensional
tortoise (in good approximation).
2. I only now realized that every infinite prime, whose inverse is
infinitesimal is smaller than 1/0, the largest possible infinity.
3. The concept of lexicon is phantastic but I could not understand
the notion of rational as novelty and subsequent claim that motion
is illusion.
4. I realized a nice manner to represent surreals (or whatever TGD
version about extension of reals is). Consider definition of a finite
real as pinary expansion:
x= SUM(n>n0) x(n)p^(-n)
a) For ordinary reals all *finite* integers n
are present in series
b) For extened reals also infinite integers
n are present. Certainly infinite values of n correspond to infinitesimal
contributions in the expansion of x in negative powers of p.
c) How should one define the part of expansion for which the values of n
are infinite? One can make the expansion unique by following trick: sum
over all n expressible as products of finite and infinite primes!
If one can construct *all* infinite primes (I have constructed quite
many good candidates!) one can make sense of this expansion
at least formally.
5. Riemann zeta function contains product over factors over
all primes. An interesting question is whether one could understand
something about zeta function by allowing
also infinite primes in the product formula
Z(s) = Prod(p prime) [1/(1-p^s)].
Best,
Matti Pitkanen
This archive was generated by hypermail 2.0b3 on Sat Oct 16 1999 - 00:36:29 JST | 526 | 2,134 | {"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} | 2.515625 | 3 | CC-MAIN-2019-22 | latest | en | 0.881886 |
https://math.answers.com/Q/What_is_12_minus_7b_plus_10b_equals_18 | 1,716,554,593,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00146.warc.gz | 341,472,508 | 47,723 | 0
# What is 12 minus 7b plus 10b equals 18?
Updated: 9/19/2023
Wiki User
13y ago
First lets set up the equation.
12 - 7b + 10b = 18.
Now lets combine all like terms on the left hand side.
12 - 7b + 10b = 18
12 + 3b = 18
Now lets combine all like terms using the like terms on the left side and on the right side.
12 + 3b = 18
-12 -12
3b = 6
Now lets divided (since the 3 is being multiplied to the b).
3b/3 = 6/3
b = 2
Wiki User
13y ago | 167 | 454 | {"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} | 4.0625 | 4 | CC-MAIN-2024-22 | latest | en | 0.777246 |
https://www.convertunits.com/from/pygme/to/micron | 1,627,098,483,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150129.50/warc/CC-MAIN-20210724032221-20210724062221-00025.warc.gz | 739,264,346 | 16,571 | ## ››Convert pygme [Greece] to micron
pygme micron
How many pygme in 1 micron? The answer is 2.8901734104046E-6.
We assume you are converting between pygme [Greece] and micron.
You can view more details on each measurement unit:
pygme or micron
The SI base unit for length is the metre.
1 metre is equal to 2.8901734104046 pygme, or 1000000 micron.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between pygme [Greece] and microns.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of pygme to micron
1 pygme to micron = 346000 micron
2 pygme to micron = 692000 micron
3 pygme to micron = 1038000 micron
4 pygme to micron = 1384000 micron
5 pygme to micron = 1730000 micron
6 pygme to micron = 2076000 micron
7 pygme to micron = 2422000 micron
8 pygme to micron = 2768000 micron
9 pygme to micron = 3114000 micron
10 pygme to micron = 3460000 micron
## ››Want other units?
You can do the reverse unit conversion from micron to pygme, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Micron
a metric unit of length equal to one millionth of a meter
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 474 | 1,685 | {"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} | 2.53125 | 3 | CC-MAIN-2021-31 | latest | en | 0.766342 |
https://englishlangkan.com/question/solve-4-3y-5-2-2y-6-by-using-elimination-show-all-work-helpppp-thanks-12937843-93/ | 1,631,971,676,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00323.warc.gz | 288,954,703 | 14,740 | ## SOLVE 4x + 3y = –5 -2x + 2y = 6 BY USING ELIMINATION. SHOW ALL WORK!!! HELPPPP :)))) THANKS! ;)
Question
SOLVE 4x + 3y = –5 -2x + 2y = 6 BY USING ELIMINATION. SHOW ALL WORK!!! HELPPPP :)))) THANKS! 😉
in progress 0
1 week 2021-09-11T01:53:37+00:00 1 Answer 0
x=
y=6x+11
Step-by-step explanation:
4x + 3y = –5 -2x + 2y = 6
– 4x – 3y  –5 -2x + 2y = 6
– 4x – 3y -2x + 2y = 6+5
-4x – 3y -2x +2y =11
-6x-y= 11 | 227 | 429 | {"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} | 4.09375 | 4 | CC-MAIN-2021-39 | latest | en | 0.452858 |
https://software.intel.com/content/www/us/en/develop/documentation/daal-programming-guide/top/algorithms/training-and-prediction/regression/elastic-net.html | 1,603,888,628,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107898499.49/warc/CC-MAIN-20201028103215-20201028133215-00079.warc.gz | 518,660,410 | 25,034 | Contents
# Elastic Net
Elastic Net is a method for modeling relationship between a dependent variable (which may be a vector) and one or more explanatory variables by fitting regularized least squares model. Elastic Net regression model has the special penalty, a sum of
L
1
and
L
2
regularizations, hat takes advantage of both Ridge Regression and LASSO algorithms. This penalty is particularly useful in a situation with many correlated predictor variables [Friedman2010].
## Details
Let (
x
1
, ...,
x
p
) be a vector of input variables and
y
= (
y
1
, ...,
y
k
) be the response. For each
j
= 1, ...,
k
, the Elastic Net model has the form similar to linear and ridge regression models [Hoerl70] with one exception: the coefficients are estimated by minimizing mean squared error (MSE) objective function that is regularized by
L
1
and
L
2
penalties.
Here
x
i
,
i
= 1, ...,
p
are referred to as independent variables,
y
j
,
is referred to as dependent variable or response.
Training Stage
Let (
x
11
, ...,
x
1
p
,
y
11
, ...,
y
1
k
), ..., (
x
n
1
, ...,
x
np
,
y
n
1
, ...,
y
nk
) be a set of training data (for regression task,
n
>>
p
, and for feature selection
p
could be greater than
n
). The matrix
X
of size
n
x
p
contains observations
x
ij
,
i
= 1, ...,
n
,
j
= 1, ...,
p
of independent variables.
For each
y
j
,
j
= 1, ...,
k
, the Elastic Net regression estimates
by minimizing the objective function:
In the equation above, the first term is a mean squared error function, the second and the third are regularization terms that penalize the
L
1
and
L
2
norms of vector
, where
For more details, see [Hastie2009] and [Friedman2010].
By default, Coordinate Descent iterative algorithm is used for minimization of the objective function. SAGA solver is also applicable for minimization. See Analysis > Optimization Solvers > Iterative Solvers.
Prediction Stage
Prediction based on Elastic Net regression is done for input vector (
x
1
, ...,
x
p
) using the equation
where
#### Product and Performance Information
1
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Notice revision #20110804 | 706 | 2,806 | {"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} | 2.875 | 3 | CC-MAIN-2020-45 | latest | en | 0.875035 |
https://help.qlik.com/en-US/sense/November2020/Subsystems/Hub/Content/Sense_Hub/LoadData/join-keep-tables-in-data-manager.htm | 1,653,493,967,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00140.warc.gz | 361,421,902 | 12,228 | Joining tables in Data manager
# Joining tables in Data manager
Join is an operation that can be used to manually combine two tables' data, producing varied results depending on the configuration you select.
This allows more granular control over combined tables than with concatenation. See Concatenating tables in Data manager to learn more about concatenation.
The Join operation takes two tables and combines them into one, which will be a combination of the fields in both original tables, based on the overlap of a common value for one or several common fields. There are multiple operators that can be applied to Join operations: Outer, Inner, Left, and Right.
Warning noteA joined table still occupies the amount of memory as the tables combined in it. Excessive use of joined tables may cause Qlik Sense to slow down. The information that is excluded by a join operation will not be accessible by Qlik Sense until the table is split.
## Join operators
There are four join operators: Outer join, Inner join, Left join, and Right join. The selected operator determines which overlapping fields or values are included or excluded.
Information noteWhen the join operators refer to Left and Right tables, they are referring to the first and second tables respectively, in order of selection.
### Outer join
The Outer join operator contains all possible combinations of values from the two tables, if the overlapping field values are represented in either one or both tables. For a more detailed explanation and examples, see the scripting reference entry: Outer.
Example:
First table
A
B
1
aa
2
cc
3
ee
Second table
A
C
1
xx
4
yy
Joined table
A
B
C
1
aa
xx
2
cc
-
3
ee
-
4
-
yy
### Inner join
The Inner join operator only contains combinations of values from the two tables, if the overlapping field values are represented in both tables. For a more detailed explanation and examples, see the scripting reference entry: Inner.
Example:
First table
A
B
1
aa
2
cc
3
ee
Second table
A
C
1
xx
4
yy
Joined table
A
B
C
1
aa
xx
### Left join
The Left join operator contains combinations of values from the two tables, if the overlapping field values are represented in the first table. For a more detailed explanation and examples, see the scripting reference entry: Left.
Example:
First table
A
B
1
aa
2
cc
3
ee
Second table
A
C
1
xx
4
yy
Joined table
A
B
C
1
aa
xx
2
cc
-
3
ee
-
### Right join
The Right join operator contains combinations of values from the two tables, if the overlapping field values are represented in the second table. For a more detailed explanation and examples, see the scripting reference entry: Right.
Example:
First table
A
B
1
aa
2
cc
3
ee
Second table
A
C
1
xx
4
yy
Joined table
A
B
C
1
aa
xx
4
-
yy
## Joining tables
The Concatenate or join tables pane is accessed by clicking in Data manager, clicking Concatenate or join, and selecting two tables. When tables are selected in Concatenate or join tables, Qlik Sense analyzes the fields and automatically maps any fields together that match. If there are no clear matches, fields are left unmapped. When the join is applied, mapped fields are combined in the joined table. Unmapped fields are either included as individual fields with null values for the rows where there is no corresponding value, or excluded entirely if there are no overlapping instances of the value.
The first table selected in Concatenate or join tables is set as the primary table, the table to which the other table is joined. The joined table uses the table and field names from the primary table unless these are manually renamed. You can change which table is the primary table with the button. Concatenate or join tables arranges fields in two rows, with the primary table fields in the top row and the secondary table fields on the bottom row. You can swap the primary and secondary tables with the button.
You can use Edit mappings to change the default mapping and select which fields to map, leave unmapped, or to exclude from the joined table. Edit mappings contains a drag and drop interface for editing mappings and the Fields pane, which lists all table fields. Fields can be mapped by dragging them beneath a primary table field. Fields can be added as a new unmapped field by beside the field in the Fields pane or dragging them into the top row of fields. Unmapped fields are marked with in the Fields pane. Fields removed from the joined table are not included in the table and are not available for use in Qlik Sense after joins are applied to the table.
Once mappings are applied and the tables are joined, you cannot edit the mapped fields, but they can be removed from the tables by splitting the joined table, which restores the tables to their original state.
To join tables in Data manager, do the following:
1. Select tables for joining.
2. Optionally, rename the joined table and the field names.
3. Optionally, edit mappings.
4. Select the join operator.
5. Join the tables.
### Selecting tables for joining
1. In Data manager, click in the bottom row.
2. Click Concatenate or join.
The Concatenate or join tables pane opens.
3. Select two tables.
The fields of both tables will be mapped or left unmapped in the Concatenate or join tables pane.
4. To preview a sample of unique values in each field, click .
5. To switch the primary and secondary tables, click .
### Renaming the table and field names
1. In the Concatenate or join tables pane, in the table name field, enter a new table name.
2. In a field name field, enter a new field name.
### Editing mappings
1. In the Concatenate or join tables pane, click Edit mappings.
2. To map two fields, click and drag a table field under a primary table field.
3. To add a new unmapped field, click and drag a table field into the upper row of fields.
4. To remove a field from the joined table, in the field click .
5. To return a removed field back to the table, click in the Fields pane, click beside the field.
6. Click Edit mappings to close Edit mappings.
### Selecting the join operator
1. In the Concatenate or join tables pane, click Select action.
2. Select an operator from the list: Outer join, Inner join, Left join, or Right join.
### Joining tables
• In the Concatenate or join tables pane, click Apply.
The tables are now joined.
## Splitting joined tables
In cases where joining is no longer needed, you can split the joined tables into their source tables.
Information note
Splitting a joined table will remove any associations the joined table had as well as any associations the primary and secondary tables had with each other. If you want to preserve your associations while splitting joined tables, click to undo the join instead of splitting the table. You cannot use to undo joins after you load data in Data manager.
### Splitting a joined table
Do the following:
1. Select the joined table.
2. Click .
The table is now split into its source tables. All fields in the source tables and their fields have their pre-join names. Splitting a joined table only splits one level of joining, so that any joined tables that were part of the split joined table have their own join preserved.
Related learning: | 1,609 | 7,316 | {"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} | 2.546875 | 3 | CC-MAIN-2022-21 | latest | en | 0.90863 |
https://www.coursehero.com/file/6633817/q5sol/ | 1,529,553,150,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864019.29/warc/CC-MAIN-20180621020632-20180621040632-00263.warc.gz | 799,639,897 | 47,369 | # q5sol - q x is continuous for x 6 = 0 • g x is continuous...
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MAP 2302 — SOLUTION TO QUIZ 5 — FALL 2002 [6 points ] Use Theorem 2 (The Existence/Uniqueness Theorem for 2nd order linear DE’s) to discuss the existence and uniqueness of a solution to the Initial Value Problem: x 2 d 2 y dx 2 + (1 - x ) dy dx - 3 y = tan x, y (1) = 0 , y 0 (1) = π 2 . Solution : The DE is equivalent to d 2 y dx 2 + (1 - x ) x 2 dy dx - 3 x 2 y = tan x x 2 , provided x 6 = 0. Let p ( x ) = (1 - x ) x 2 , q ( x ) = - 3 x 2 , g ( x ) = tan x x 2 . Then p ( x ) is continuous for x 6 = 0.
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Unformatted text preview: q ( x ) is continuous for x 6 = 0. • g ( x ) is continuous for x 6 = 0, ± π 2 , ± 3 π 2 ... . The three functions p ( x ), q ( x ) and g ( x ) are all continous on the open interval (0 , π 2 ), which contains x = 1. The Theorem implies that the IVP has a unique solution on the open interval (0 , π 2 ). 1...
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Jill Tulane University ‘16, Course Hero Intern | 578 | 2,025 | {"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} | 3.515625 | 4 | CC-MAIN-2018-26 | latest | en | 0.890427 |
http://www.numbersaplenty.com/432202 | 1,596,605,155,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735909.19/warc/CC-MAIN-20200805035535-20200805065535-00500.warc.gz | 154,426,503 | 3,260 | Search a number
432202 = 2316971
BaseRepresentation
bin1101001100001001010
3210221212111
41221201022
5102312302
613132534
73450031
oct1514112
9727774
10432202
112757a1
1218a14a
13121954
14b3718
15880d7
hex6984a
432202 has 8 divisors (see below), whose sum is σ = 669312. Its totient is φ = 209100.
The previous prime is 432199. The next prime is 432203. The reversal of 432202 is 202234.
Adding to 432202 its reverse (202234), we get a palindrome (634436).
It is a sphenic number, since it is the product of 3 distinct primes.
It is not an unprimeable number, because it can be changed into a prime (432203) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 3424 + ... + 3547.
It is an arithmetic number, because the mean of its divisors is an integer number (83664).
2432202 is an apocalyptic number.
432202 is a deficient number, since it is larger than the sum of its proper divisors (237110).
432202 is a wasteful number, since it uses less digits than its factorization.
432202 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 7004.
The product of its (nonzero) digits is 96, while the sum is 13.
The square root of 432202 is about 657.4207176535. The cubic root of 432202 is about 75.6070437527.
The spelling of 432202 in words is "four hundred thirty-two thousand, two hundred two".
Divisors: 1 2 31 62 6971 13942 216101 432202 | 445 | 1,473 | {"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} | 3.5 | 4 | CC-MAIN-2020-34 | latest | en | 0.867619 |
https://www.numbersaplenty.com/1211213 | 1,643,407,498,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00319.warc.gz | 964,440,670 | 3,466 | Cookie Consent by FreePrivacyPolicy.com
Search a number
1211213 = 9471279
BaseRepresentation
bin100100111101101001101
32021112110202
410213231031
5302224323
641543245
713203143
oct4475515
92245422
101211213
11758003
124a4b25
133353c3
14237593
1518dd28
hex127b4d
1211213 has 4 divisors (see below), whose sum is σ = 1213440. Its totient is φ = 1208988.
The previous prime is 1211207. The next prime is 1211227. The reversal of 1211213 is 3121121.
1211213 is nontrivially palindromic in base 9.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 1211213 - 212 = 1207117 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 1211194 and 1211203.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (1211813) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 308 + ... + 1586.
It is an arithmetic number, because the mean of its divisors is an integer number (303360).
21211213 is an apocalyptic number.
It is an amenable number.
1211213 is a deficient number, since it is larger than the sum of its proper divisors (2227).
1211213 is an equidigital number, since it uses as much as digits as its factorization.
1211213 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 2226.
The product of its digits is 12, while the sum is 11.
The square root of 1211213 is about 1100.5512255229. The cubic root of 1211213 is about 106.5958199253.
Adding to 1211213 its reverse (3121121), we get a palindrome (4332334).
It can be divided in two parts, 121121 and 3, that multiplied together give a palindrome (363363).
The spelling of 1211213 in words is "one million, two hundred eleven thousand, two hundred thirteen".
Divisors: 1 947 1279 1211213 | 607 | 2,013 | {"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} | 3.1875 | 3 | CC-MAIN-2022-05 | latest | en | 0.875489 |
http://ocham.blogspot.com/2007/07/is-every-man-is-animal-true-when-there.html | 1,467,197,443,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397696.49/warc/CC-MAIN-20160624154957-00008-ip-10-164-35-72.ec2.internal.warc.gz | 224,823,443 | 16,197 | ## Saturday, July 14, 2007
### Is 'every man is an animal true' when there are no men?
A splendid new page in the Logic Museum here.
Ken said...
The question was how could 'every man is an animal' or 'three and four are seven' possibly be false?
“Every man is an animal” would be false if “No man is an animal” is true. This could be that animal and man are separate within a single category or that no man exists.
“Three and four are seven” would be false under a number of conditions. If the three and four are binary numbers that use the and operation such as 3 and 4, this would actually equal 0. The assumption is that “and” is acting like an “add” which may not be valid.
Augustine indicates that the number 6 is perfect if those “things” did or did not exist. This appears to say that the number six is perfect without any other statements necessary.
Because the number 6 is perfect, then other results will be true or false because 6 perfect but the same propositions may not result in the same true or false result if replaced with 5 or 7 because these are not perfect.
Is this what he is saying?
Ocham said...
Augustine actually says if the 'things numbered' do not exist. I.e. if there are not six men, not six planets, and so on. Clearly other statements are necessary if 6 is to be a perfect number. | 313 | 1,323 | {"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} | 3.09375 | 3 | CC-MAIN-2016-26 | longest | en | 0.950958 |
https://questions.llc/questions/1642883 | 1,702,055,801,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100762.64/warc/CC-MAIN-20231208144732-20231208174732-00221.warc.gz | 530,400,862 | 6,671 | # Karen is applying for the nursing school at the University of Davis. Before applying, she received a breakdown of the tuition and fees from the university.
Tuition: \$611 per semester plus \$250 per class
Books: \$82 per class
Nursing School Fee: \$19 per class
Parking: \$58 per semester
Facilities Fee: \$50 per semester
The total cost of a semester for c classes is represented by the following expression.
Which statement best describes the coefficient of c?
A.
It is the sum of the cost of books and the nursing school fee.
B.
It is the total semester cost of c classes at the university.
C.
It is the sum of the tuition, parking, and facilities fee.
D.
It is the sum of the additional cost of tuition, books, and nursing school fee.
I think B
## \$351c + \$719 <<< not an equation
B is not the correct answer. | 194 | 827 | {"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} | 3.125 | 3 | CC-MAIN-2023-50 | latest | en | 0.948076 |
http://mathhelpforum.com/calculus/124881-what-does-mean-find-dy-dt-question-parametrics.html | 1,481,432,676,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544097.11/warc/CC-MAIN-20161202170904-00322-ip-10-31-129-80.ec2.internal.warc.gz | 174,421,960 | 12,057 | # Thread: what does it mean to find dy/dt in this question?(parametrics)
1. ## what does it mean to find dy/dt in this question?(parametrics)
Original question:
A particle moves in the xy-plane so that at any time t ³ 0 its position (x,y) is given by
x = e^t + e^-t and y = e^t - e^-t .
(a) Find the velocity vector for any t ³ 0.
I know how to do part a it is simply (derivative of x, derivative of y)
(b) Find lim (dy/dt)/(dx/dt)
(t approaching infinity)
What does this (dy/dt)/(dx/dt) tell us?
I know dy/dt is going to be y component of the velocity and dx/dt is the x component of the velocity. Yet, simply dividing y component of the velocity at certain t by x component of the velocity at certain t ... what does this mean?
So, i just attempted to find the limit of (dy/dt)/(dx/dt) as t approaches infinity without actual understanding.
(e^t - e^-t) /(e^t + e^-t) ;the first one, the derivative of y, the second, the derivative of x
In this i run into the problem of the function being undefined. I fiddled with it a little but could not find a way to avoid the problem
....
Help me deal with my stupidity,
(c) The particle moves on a hyperbola. Find an equation for this hyperbola in
terms of x and y.
i do not know how to isolate t in either function x or function y (in order to subsitute that for t in the other function )
(d) On the axes provided, sketch the path of the particle showing the velocity vector
I need your help.
Thanks.
2. Originally Posted by hangainlover
Original question:
A particle moves in the xy-plane so that at any time t ³ 0 its position (x,y) is given by
x = e^t + e^-t and y = e^t - e^-t .
(a) Find the velocity vector for any t ³ 0.
I know how to do part a it is simply (derivative of x, derivative of y)
ok
(b) Find lim (dy/dt)/(dx/dt)
(t approaching infinity)
What does this (dy/dt)/(dx/dt) tell us?
I know dy/dt is going to be y component of the velocity and dx/dt is the x component of the velocity. Yet, simply dividing y component of the velocity at certain t by x component of the velocity at certain t ... what does this mean?
So, i just attempted to find the limit of (dy/dt)/(dx/dt) as t approaches infinity without actual understanding.
(e^t - e^-t) /(e^t + e^-t) ;the first one, the derivative of y, the second, the derivative of x
In this i run into the problem of the function being undefined. I fiddled with it a little but could not find a way to avoid the problem
....
actually, $\frac {dy/dt}{dx/dt} = \frac {dy}{dx} = \frac {e^t~{\color{red} +}~ e^{-t}}{e^t ~{\color {red}-}~ e^{-t}}$
i will tell you the answer, the limit is 1. how would you get there?
Help me deal with my stupidity,
(c) The particle moves on a hyperbola. Find an equation for this hyperbola in
terms of x and y.
i do not know how to isolate t in either function x or function y (in order to subsitute that for t in the other function )
take the hint, it's a hyperbola. consider what $x^2 - y^2$ would be (it would actually be nicer to think of what $\left( \frac x2 \right)^2 - \left( \frac y2 \right)^2$ is. do those expressions look familiar? think hyperbolic functions)
(d) On the axes provided, sketch the path of the particle showing the velocity vector
after doing part (c), this should be ok
3. ## how part b is applicable in reality
Originally Posted by Jhevon
ok
actually, $\frac {dy/dt}{dx/dt} = \frac {dy}{dx} = \frac {e^t~{\color{red} +}~ e^{-t}}{e^t ~{\color {red}-}~ e^{-t}}$
i will tell you the answer, the limit is 1. how would you get there?
take the hint, it's a hyperbola. consider what $x^2 - y^2$ would be (it would actually be nicer to think of what $\left( \frac x2 \right)^2 - \left( \frac y2 \right)^2$ is. do those expressions look familiar? think hyperbolic functions)
after doing part (c), this should be ok
so for part (b), I can find the limit.
Yet, i still have problem understanding the concept.
dy /dx is going to be relation between dy and dx
Its not like it is going to give you the magnitude of the velocity
Just for the sake of the argument, let's say the velocity, at t =3, is (4,5) in some other function.
Then, the magnitude of the velocity is square root of 4^2 + 5^2
but if you do dy/dx, it is simply the slope, 5/4.
So, in practicality, is there any point doing dy/dx?
4. Originally Posted by hangainlover
so for part (b), I can find the limit.
Yet, i still have problem understanding the concept.
dy /dx is going to be relation between dy and dx
Its not like it is going to give you the magnitude of the velocity
Just for the sake of the argument, let's say the velocity, at t =3, is (4,5) in some other function.
Then, the magnitude of the velocity is square root of 4^2 + 5^2
but if you do dy/dx, it is simply the slope, 5/4.
So, in practicality, is there any point doing dy/dx?
the magnitude of velocity vector is speed. dy/dx gives you the slope of the velocity vector. of course there are reasons for finding dy/dx. it is another way of thinking about direction
5. ## acceleration at that t
Originally Posted by Jhevon
the magnitude of velocity vector is speed. dy/dx gives you the slope of the velocity vector. of course there are reasons for finding dy/dx. it is another way of thinking about direction
Oh dy/dx is the acceleration at that specific t.
Is this correct?
I believe it is..
6. Originally Posted by hangainlover
Oh dy/dx is the acceleration at that specific t.
Is this correct?
I believe it is..
no. acceleration is the second derivative: $\frac {d^2y}{dx^2}$
7. ## (dy/dt)/(dx/dt) is...
Originally Posted by Jhevon
no. acceleration is the second derivative: $\frac {d^2y}{dx^2}$
(dy/dt)/(dx/dt) is the slope of (x component of velocity, y component of the velocity) at that specific t
in velocity function, if you find a tangent line at one particular t, the slope of the tanget line is the acceleration. Isn't it?
I don't see how (dy/dt)/(dx/dt) is different from that slopeof the tangent line in this case
(i know the second derivative is usually acceleration in s(t). Maybe its parametrics that confuses me)
8. ## (dy/dt)/(dx/dt) being unitless seems problematic....
Originally Posted by hangainlover
(dy/dt)/(dx/dt) is the slope of (x component of velocity, y component of the velocity) at that specific t
in velocity function, if you find a tangent line at one particular t, the slope of the tanget line is the acceleration. Isn't it?
I don't see how (dy/dt)/(dx/dt) is different from that slopeof the tangent line in this case
(i know the second derivative is usually acceleration in s(t). Maybe its parametrics that confuses me)
Unlike slope of a tangent line in v(t) function, where the unit of slope is m/s^2, (dy/dt)/(dx/dt) is unitless. what am i doing wrong. is (dy/dt)/(dx/dt) denoting something other? | 1,871 | 6,755 | {"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": 8, "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} | 3.8125 | 4 | CC-MAIN-2016-50 | longest | en | 0.895005 |
https://learnerstests.in/80155/paste-cut-worksheets-quark.php | 1,660,783,613,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00638.warc.gz | 342,142,805 | 32,228 | ## After sick people who are a whole
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What place value cut paste pockets is randomly created using. Friend Facebook. | 1,626 | 8,784 | {"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} | 2.78125 | 3 | CC-MAIN-2022-33 | latest | en | 0.926412 |
https://archivemore.com/archives/565 | 1,659,898,806,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00043.warc.gz | 141,371,221 | 10,181 | ## What happens when I try to divide by zero?
Less fruit, more math But what about division by zero? If we say that 10/div 0 is some number, then that number times 0 would have to be 10. That’s impossible, because anything times zero has to be zero — so again, division by zero is undefined, and the situation looks pretty bleak.
## What happens when you attempt a division by zero in C?
The default one does nothing, so if division by zero is detected, the division function returns zero.
## Can u rename Siri?
You can also call her any other personal assistant, like Cortana or Android! You can’t really change her name, but you can tell her your name by making a contact of yourself or telling her your name!!
## What happens when you say hyphen 5 times to Siri?
Many people have found that after saying the fifth “hyphen,” their phone’s iOs immediately “crashes” and either restarts their phone completely or closes their current screen and brings them back to their home one. So if you do, don’t worry – your phone will be exactly as it was before you said “hyphen” five times.
## What does Siri do when you say 000?
If you genuinely need emergency services you can say 000 to Siri or simply say “dial emergency services”. Siri will then give you a five-second countdown and a chance to cancel or call before then.
## Should I charge my phone to 100 %?
The key is to not store or keep your phone’s battery at a 100% charge for extended periods. Instead, Schulte said that “it would be very good to charge the phone in the morning or whenever, but don’t store the phone overnight at 100%.”
## Should I only charge my phone to 80?
Don’t charge to 100 percent A good rule of thumb seems to be to never charge your phone up to more than 80 percent of capacity. Some research shows that after 80 percent, your charger must hold your battery at a constant high voltage to get to 100 percent, and this constant voltage does the most damage.
## What is the best percentage to charge your phone?
What’s the best way to charge your smartphone? Avoid full cycle (zero-100 percent) and overnight charging. Instead, top-up your phone more regularly with partial charges. Ending a charge at 80 percent is better for the battery than topping all the way up to 100 percent.
## Is it OK to leave phone charging overnight?
Android phone manufacturers, including Samsung, say the same. “Do not leave your phone connected to the charger for long periods of time or overnight.” Your battery will automatically stop charging when it’s full, but in some cases, once it drops to 99%, it will need more energy to get back to 100.
Charging My iPhone Overnight Will Overload the Battery: FALSE. Once the internal lithium-ion battery hits 100% of its capacity, charging stops. If you leave the smartphone plugged in overnight, it is going to use a bit of energy constantly trickling new juice to the battery every time it falls to 99%.
## Do fast chargers shorten battery life?
Unless there’s some technical flaw with your battery or charger electronics, however, using a fast charger won’t do your phone’s battery any long-term damage. That’s because during the first phase of charging, batteries can absorb a charge quickly without major negative effects on their long-term health.
## Is there a downside to fast charging?
The Disadvantages of Fast Charging With fast charging, people are less likely to care for their batteries extensively, which causes them to excessively charge and drain the batteries, shortening their lifespan. The second critical disadvantage is the way phones can overheat due to this phone charger technology.
## Should I turn off fast charging?
Fast charging will shorten the lifespan of your battery. Anyone who charges overnight would benefit from disabling fast charge, as it doesn’t matter how long it takes since you’re asleep.
## Is it bad to charge your phone multiple times a day?
It’s fine to plug and unplug it multiple times a day. Not only does this keep your smartphone’s battery performing optimally for longer, but it also keeps it topped up throughout the day.
## Why is phone not fast charging?
If you have checked everything and it turns out that the fast charging feature is still not working. So the only best way is to do a Factory Reset. Restore all settings of your phone to factory settings. The reason is, the damage to fast charging is a sign that something is wrong with the Android software system. | 948 | 4,460 | {"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} | 3.015625 | 3 | CC-MAIN-2022-33 | latest | en | 0.949271 |
https://www.hackmath.net/en/math-problem/15301?tag_id=13 | 1,585,748,251,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505731.37/warc/CC-MAIN-20200401130837-20200401160837-00529.warc.gz | 843,528,792 | 10,723 | # Wall height
Calculate the height of a regular hexagonal pyramid with a base edge of 5 cm and a wall height w = 20 cm.
Result
h = 19.526 cm
#### Solution:
$a=5 \ \text{cm} \ \\ n=6 \ \\ w=20 \ \text{cm} \ \\ \ \\ w_{1}=\sqrt{ a^2 - (a/2)^2 }=\sqrt{ 5^2 - (5/2)^2 } \doteq 4.3301 \ \text{cm} \ \\ \ \\ h=\sqrt{ w^2 - w_{1}^2 }=\sqrt{ 20^2 - 4.3301^2 } \doteq 19.5256 \doteq 19.526 \ \text{cm}$
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A cell tower is located at coordinates (-5, -7) and has a circular range of 12 units. If Mr. XYZ is located at coordinates (4,5), will he be able to get a signal? | 1,173 | 4,398 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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} | 4.34375 | 4 | CC-MAIN-2020-16 | latest | en | 0.866778 |
https://nutriverse.io/nipnTK/articles/as.html | 1,675,745,545,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500384.17/warc/CC-MAIN-20230207035749-20230207065749-00200.warc.gz | 420,900,366 | 16,354 | ## Age and sex distributions (children’s data)
Age heaping is the tendency to report children’s ages to the nearest year or adult ages to the nearest multiple of 5 or 10 years. Age heaping is very common. It is a major reason why data from nutritional anthropometry surveys are often analysed and reported using broad age-groups. The commonest age-groups used with children’s data are 6 to 17 months, 18 to 29 months, 30 to 41 months, 42 to 53 months, and 54 to 59 months (see figure below). These are known as year-centred age-groups. Note that the last age-group covers only six months but is nominally centred at five years. Other age-groups may be used for specific analyses. The techniques presented here can be adapted to work with other age- groups.
We will retrieve a survey dataset:
svy <- read.table("dp.ex02.csv", header = TRUE, sep = ",")
head(svy)
#> psu age sex weight height muac oedema
#> 1 1 6 1 7.3 65.0 146 2
#> 2 1 42 2 12.5 89.5 156 2
#> 3 1 23 1 10.6 78.1 149 2
#> 4 1 18 1 12.8 81.5 160 2
#> 5 1 52 1 12.1 87.3 152 2
#> 6 1 36 2 16.9 93.0 190 2
The dataset dp.ex02 is a comma-separated-value (CSV) file containing anthropometric data from a SMART survey in Kabul, Afghanistan.
### Tabulation and visualisation
The NiPN data quality toolkit provides an R language function called recode() that makes it easy to recode and group data. We will use the recode() function to group the data in the age variable (age in months) into year-centred age-groups.
svy$ycag <- recode(svy$age, "6:17=1; 18:29=2; 30:41=3; 42:53=4; 54:59=5")
#> psu age sex weight height muac oedema ycag
#> 1 1 6 1 7.3 65.0 146 2 1
#> 2 1 42 2 12.5 89.5 156 2 4
#> 3 1 23 1 10.6 78.1 149 2 2
#> 4 1 18 1 12.8 81.5 160 2 2
#> 5 1 52 1 12.1 87.3 152 2 4
#> 6 1 36 2 16.9 93.0 190 2 3
A tabular analysis can be performed:
table(svy$ycag, svy$sex)
#>
#> 1 2
#> 1 101 106
#> 2 102 96
#> 3 126 115
#> 4 78 82
#> 5 31 36
prop.table(table(svy$ycag, svy$sex)) * 100
#>
#> 1 2
#> 1 11.569301 12.142039
#> 2 11.683849 10.996564
#> 3 14.432990 13.172967
#> 4 8.934708 9.392898
#> 5 3.550974 4.123711
The table() function performs a cross-tabulation. The first variable specified (svy$ycag in this example) is the row variable. The second variable specified (svy$sex in this example) is the column variable.
It is useful to examine row percentages and column percentages in tables of age-group by sex. We should look at row percentages:
prop.table(table(svy$ycag, svy$sex), margin = 1) * 100
This returns:
#>
#> 1 2
#> 1 48.79227 51.20773
#> 2 51.51515 48.48485
#> 3 52.28216 47.71784
#> 4 48.75000 51.25000
#> 5 46.26866 53.73134
Which shows approximately equal proportions of males and females in each year-centred age-group. We specified margin = 1 with the prop.table() function because we wanted row percentages.
We should also look at column percentages:
prop.table(table(svy$ycag, svy$sex), margin = 2) * 100
This returns:
#>
#> 1 2
#> 1 23.059361 24.367816
#> 2 23.287671 22.068966
#> 3 28.767123 26.436782
#> 4 17.808219 18.850575
#> 5 7.077626 8.275862
We expect there to be approximately equal proportions of children in the age-groups centred at 1, 2, 3, and 4 years and a smaller proportion (i.e. about half that in the other age-groups) in the age-group centred at 5 years. We specified margin = 2 with the prop.table() function because we wanted column percentages.
A graphical analysis using a population pyramid can be useful. The NiPN data quality toolkit provides an R language function called pyramid.plot() for plotting population pyramids:
pyramid.plot(svy$ycag, svy$sex)
We can make a more informative plot by specifying a title and axis labels:
pyramid.plot(svy$ycag, svy$sex,
main = "Distribution of age by sex",
xlab = "Frequency (Males | Females)",
ylab = "Year-centred age-group")
pyramid.plot(svy$ycag, svy$sex,
main = "Distribution of age by sex",
xlab = "Frequency (Males | Females)",
ylab = "Year-centred age-group",
col = c("grey80", "white"))
or colours:
pyramid.plot(svy$ycag, svy$sex,
main = "Distribution of age by sex",
xlab = "Frequency (Males | Females)",
ylab = "Year-centred age-group",
col = c("lightblue", "pink"))
We expect there to be approximately equal numbers of children in the age-groups centred at 1, 2, 3, and 4 years and a smaller number (i.e. about half the number in the other age-groups) in the age-group centred at 5 years. There should also be approximately equal numbers of males and females. This is what we see in the population pyramid below.
pyramid.plot(svy$ycag, svy$sex,
main = "Distribution of age by sex",
xlab = "Frequency (Males | Females)",
ylab = "Year-centred age-group")
The pyramid.plot() function uses the values of the grouped age variable as y-axis value labels.
We can assign descriptive text values using the recode() function. For example:
svy$ageLabel <- recode(svy$age, "6:29='< 30 months'; 30:hi='30 month or older'")
#> Warning in recode(svy$age, "6:29='< 30 months'; 30:hi='30 month or older'"): NAs #> introduced by coercion pyramid.plot(svy$ageLabel,
svy$sex, main = "Distribution of age by sex", xlab = "Frequency (Males | Females)", ylab = "Age-group") We can also use a factor type variable. This type of variable allows labels to be specified: svy$ageLabel <- factor(svy$ycag, labels = c("6:17", "18:29", "30:41", "42:53", "54:59")) pyramid.plot(svy$ageLabel,
svy$sex, main = "Distribution of age by sex", xlab = "Frequency (Males | Females)", ylab = "Year-centred age-group") The cut() function may also be used: svy$ageCuts <- cut(svy$age, breaks = c(0, 17, 29, 41, 53, 59)) pyramid.plot(svy$ageCuts,
svy$sex, main = "Age-group (months) ", xlab = "Frequency (Males | Females)", ylab = "Year-centred age-group", cex.names = 0.9) The cut() function is a versatile grouping function. It is explained in more detail later in this section. The cex.names parameter of the pyramid.plot() function allows us to change the size of the value labels on the y-axis. The value for cex.names is a magnification factor. Values above one make the labels larger than the default. Values below one make the labels smaller than the default. ## Simple testing It is possible to perform a formal test on the distribution of age-groups by sex. A simple test is: chisq.test(table(svy$ycag, svy$sex)) This yields: #> #> Pearson's Chi-squared test #> #> data: table(svy$ycag, svy$sex) #> X-squared = 1.2675, df = 4, p-value = 0.8669 In this example the p-value is not below 0.05 so we accept the null hypothesis that there is no significant association between age and sex. This is an important test as it tests whether the distribution of ages is similar for males and females. It does not, however, test whether the age structure in the sample meets expectations. This requires a test that compares observed numbers with expected numbers derived from an external source (e.g. census data) or from a demographic model. ### A model of the expected age structure A simple model-based method for calculating expected numbers is exponential decay in a population in which births and deaths balance each other and with a 1:1 male to female sex ratio. Under this model the proportion surviving in each group at each year can be calculated as: $p ~ = ~ e ^ {-zt}$ in which e is the base of the natural logarithm (approximately 2.7183), z is the mortality rate associated with each time period, and t is time. Time (t) starts at zero for the purposes of computation. Age can be used as a measure of time since birth. We should use 0 for the first year-centred age-group, 1 for the second year-centred age-group, and so-on. This is the rationale for us using t <- 0:4 below. With five year-centred age-groups and a mortality rate of 1 / 10,000 / day, the expected proportions surviving at each year can be calculated as: z <- (1 / 10000) * 365.25 t <- 0:4 p <- exp(-z * t) p This yields the following survival probabilities: z <- (1 / 10000) * 365.25 t <- 0:4 p <- exp(-z * t) p #> [1] 1.0000000 0.9641340 0.9295544 0.8962149 0.8640713 We need to specify the duration (i.e. the number of years) represented by each age-group: d <- c(1, 1, 1, 1, 0.5) We can then calculate expected proportions of children in each age-group: ep <- d * p / sum(d * p) ep This gives: #> [1] 0.2368580 0.2283628 0.2201724 0.2122757 0.1023311 We can now calculate expected numbers: expected <- ep * sum(table(svy$ycag)) names(expected) <- 1:5
expected
giving:
#> 1 2 3 4 5
#> 206.77703 199.36076 192.21049 185.31667 89.33505
A formal test would compare the observed numbers with the expected numbers. The observed numbers can be found using:
observed <- table(svy$ycag) observed This gives: #> #> 1 2 3 4 5 #> 207 198 241 160 67 It can be useful to examine observed and expected numbers graphically: par(mfcol = c(1, 2)) barplot(observed, main = "Observed", xlab = "Age group", ylab = "Frequency", ylim = c(0, 250)) barplot(expected, main = "Expected", xlab = "Age group", ylab = "Frequency", ylim = c(0, 250)) We will calculate a Chi-squared test statistic: $\chi ^ 2 ~ = ~ \sum \frac{(\text{observed} - \text{expected}) ^ 2}{\text{expected}}$ using: X2 <- sum((observed - expected) ^ 2 / expected) which yields a Chi-Squared test statistic of: We can find the p-value using: pchisq(X2, df = 4, lower.tail = FALSE) This gives: #> [1] 0.000259395 In this example the age distribution is significantly different from expected numbers calculated using a simple demographic model. Note that we specify the degrees of freedom (df) for the Chi-Squared test as the number of age-groups minus one. As we have five age-groups we specify df = 4. The degrees of freedom (df) that we need to specify will depend on the number of age-groups that we use. It is always the number of age-groups minus one. If, for example, there are ten age-groups we would need to specify df = 9. The NiPN data quality toolkit provides an R function called ageChildren() that performs the model- based Chi-Squared test: ageChildren(svy$age, u5mr = 1)
which returns:
#>
#> Age Test (Children)
#>
#> X-squared = 21.4366, df = 4, p = 0.0003
Note that we specified the under five years mortality rate as 1 / 10,000 / day using u5mr = 1. Another, more appropriate, rate may be specified.
The ageChildren() function calculates year-centred age-groups for children aged between six and fifty-nine months by default. This is a standard survey population and is used in SMART and many other surveys. The use of year-centred age-groups is also standard practice. The commands that are given above can, however, be adapted for use with different age-groups.
The output of the ageChildren() function can be saved for later use:
ac <- ageChildren(svy$age, u5mr = 1) The saved output contains the Chi-squared test results and tables of observed and expected values. These can be accessed using: ac #> #> Age Test (Children) #> #> X-squared = 21.4366, df = 4, p = 0.0003 ac$X2
#> [1] 21.43662
ac$df #> [1] 4 ac$p
#> [1] 0.000259395
ac$observed #> 1 2 3 4 5 #> 207 198 241 160 67 ac$expected
#> 1 2 3 4 5
#> 206.77703 199.36076 192.21049 185.31667 89.33505
The saved results may also be plotted:
plot(ac)
The ageChildren() function can be applied to each sex separately. To males:
acM <- ageChildren(svy$age[svy$sex == 1], u5mr = 1)
acM
#>
#> Age Test (Children)
#>
#> X-squared = 15.8496, df = 4, p = 0.0032
plot(acM)
and to females:
acF <- ageChildren(svy$age[svy$sex == 2], u5mr = 1)
acF
#>
#> Age Test (Children)
#>
#> X-squared = 6.8429, df = 4, p = 0.1444
plot(acF)
An easier way of doing this is:
by(svy$age, svy$sex, ageChildren, u5mr = 1)
#> svy$sex: 1 #> #> Age Test (Children) #> #> X-squared = 15.8496, df = 4, p = 0.0032 #> #> ------------------------------------------------------------ #> svy$sex: 2
#>
#> Age Test (Children)
#>
#> X-squared = 6.8429, df = 4, p = 0.1444
The test statistics should be interpreted with caution. A significant test result may, for example, be due to the use of an inappropriate model to generate expected numbers.
A significant result in this particular test may be due to:
• Specifying an inappropriate under five years mortality rate: This is a particular problem because the specified under five years mortality rate is assumed to have applied for five years prior to data being collected.
• The assumption of a 1:1 male to female sex ratio: This is a particular problem in setting in which there is sex-selective abortion and sex-selective infanticide.
The model is crude. Mortality is related to age. Younger children have a greater mortality risk than older children and only an average under five years mortality rate is used. A more sophisticated model could be used but, in many settings, we will not have the data required to use such a model.
It should also be noted that the sample sizes used in most survey can cause tests to yield statistically significant results for small differences between observed and expected numbers.
## Use of census data
The use of simple demographic models is far from ideal. It is usually better to calculate the expected proportions from census data. A useful source of census data is the United States Census Bureau’s International Data Base:
https://www.census.gov/data-tools/demo/idb/informationGateway.php
The population in single year age-groups for 0, 1, 2, 3, and 4 years for Afghanistan in 2015 was:
Age Both Sexes Males Females
0 1148379 584276 564103
1 1062635 539589 523046
2 1015688 515793 499895
3 981288 498365 482923
4 950875 482926 467949
We can calculate expected values from these data:
pop <- c(1148379, 1062635, 1015688, 981288, 950875)
ep <- pop / sum(pop)
With a sample size of $$n = 900$$ the expected number in each age-group would be:
expected <- ep * 900
expected
#> [1] 200.3427 185.3841 177.1939 171.1925 165.8868
These expected values can be used in a Chi-squared test as is illustrated above.
Census data may also be used to estimate the under five years’ mortality rate (U5MR) which can then be used with the ageChildren() function.
The model of exponential decay in a population in which births and deaths balance each other with a 1:1 male to female sex ratio:
$p ~ = ~ e ^ {-zt}$
means that we can, given an age-distribution, estimate mortality by fitting the model:
$\log_e(n) ~ = ~ \alpha ~ + ~ \beta t$
where $$n$$ is the count of children in each age-group.
The absolute value of the β coefficient is the point estimate of the mortality rate (z). Using the 2015 population data for Afghanistan:
t <- 0:4
lm(log(pop) ~ t)
This gives:
#>
#> Call:
#> lm(formula = log(pop) ~ t)
#>
#> Coefficients:
#> (Intercept) t
#> 13.93601 -0.04571
The value reported under t is the $$\beta$$ coefficient (-0.04571). The absolute value of the $$\beta$$ coefficient (i.e. the value without the sign) is 0.04571. This is the point estimate of the mortality rate. Expressed as the number of deaths / 10,000 persons / day:
(0.04571 / 365.25) * 10000
this is:
#> [1] 1.251472
We can use this estimate with the ageChildren() function:
ageChildren(svy$age, u5mr = 1.251472) #> #> Age Test (Children) #> #> X-squared = 20.4744, df = 4, p = 0.0004 ### The age ratio A much simpler and less problematic age-related test of survey and data quality is the age ratio test. The age ratio is defined as: $\text{Age ratio} ~ = ~ \frac{\text{number of children aged between 6 and 29 months}}{\text{number of children aged between 30 and 59 months}}$ We will use the recode() function from NiPN data quality toolkit to create the relevant age-groups: svy$ageGroup <- recode(svy$age, "6:29=1; 30:59=2") head(svy) #> psu age sex weight height muac oedema ycag ageLabel ageCuts ageGroup #> 1 1 6 1 7.3 65.0 146 2 1 6:17 (0,17] 1 #> 2 1 42 2 12.5 89.5 156 2 4 42:53 (41,53] 2 #> 3 1 23 1 10.6 78.1 149 2 2 18:29 (17,29] 1 #> 4 1 18 1 12.8 81.5 160 2 2 18:29 (17,29] 1 #> 5 1 52 1 12.1 87.3 152 2 4 42:53 (41,53] 2 #> 6 1 36 2 16.9 93.0 190 2 3 30:41 (29,41] 2 The observed age ratio is: sum(svy$ageGroup == 1) / sum(svy$ageGroup == 2) which gives: #> [1] 0.8653846 It is often easier to work with proportions than with ratios so we only need to calculate the proportion in the younger age-group: sum(svy$ageGroup == 1) / sum(table(svy$ageGroup)) which gives: #> [1] 0.4639175 We can calculate an expected value using census data or a simple demographic model. The simplest approach is to use a standard value. SMART surveys often use the ratio 0.85:1. We only need to calculate the expected proportion in the younger group. For the ratio 0.85:1 this is: p <- 0.85 / (0.85 + 1) This gives: #> [1] 0.4594595 The observed proportion (0.4639175) and expected proportion (0.4594595) are so similar that a formal test of statistical significance is not required in this case. Formal testing can be done using a Chi-squared test: prop.test(sum(svy$ageGroup == 1), sum(table(svy$ageGroup)), p = 0.4594595) This returns: #> #> 1-sample proportions test with continuity correction #> #> data: sum(svy$ageGroup == 1) out of sum(table(svy$ageGroup)), null probability 0.4594595 #> X-squared = 0.053062, df = 1, p-value = 0.8178 #> alternative hypothesis: true p is not equal to 0.4594595 #> 95 percent confidence interval: #> 0.4304994 0.4976573 #> sample estimates: #> p #> 0.4639175 The age ratio in the example data is not significantly different from the expected age ratio. The NiPN data quality toolkit provide an R function called ageRatioTest() that performs the age ratio test: ageRatioTest(svy$age, ratio = 0.85)
This returns:
#>
#> Age Ratio Test (children's data)
#>
#> Expected age ratio = 0.8500
#> Expected proportion aged 6 - 29 months = 0.4595
#>
#> Observed age ratio = 0.8654
#> Observed proportion aged 6 - 29 months = 0.4639
#>
#> X-squared = 0.0531, p = 0.8178
The ratio parameter of the ageRatioTest() function allows you to specify an expected age ratio other than 0.85:1.
Note that the ageRatioTest() function applies the test to data from children aged between 6 and 59 months only (all other ages are ignored).
The age ratio test might be applied to data from both sexes (as above) and to each sex separately:
by(svy$age, svy$sex, ageRatioTest, ratio = 0.85)
#> svy$sex: 1 #> #> Age Ratio Test (children's data) #> #> Expected age ratio = 0.8500 #> Expected proportion aged 6 - 29 months = 0.4595 #> #> Observed age ratio = 0.8638 #> Observed proportion aged 6 - 29 months = 0.4635 #> #> X-squared = 0.0145, p = 0.9041 #> #> ------------------------------------------------------------ #> svy$sex: 2
#>
#> Age Ratio Test (children's data)
#>
#> Expected age ratio = 0.8500
#> Expected proportion aged 6 - 29 months = 0.4595
#>
#> Observed age ratio = 0.8670
#> Observed proportion aged 6 - 29 months = 0.4644
#>
#> X-squared = 0.0247, p = 0.8750
The example data meets expectations regarding the age ratio for all children and for male and female children separately.
## Age and sex distributions : Adults and general population surveys
A key test of survey quality is whether the survey data represents the population in terms of the age and sex distribution. We can test this by comparison with census data.
We will retrieve some example data:
svy <- read.table("as.ex01.csv", header = TRUE, sep = ",")
head(svy)
#> age sex
#> 1 44 2
#> 2 1 2
#> 3 15 2
#> 4 7 1
#> 5 14 1
#> 6 14 1
These data are taken from household rosters collected as part of a household survey in Tanzania. We will use census data taken from the Wolfram|Alpha knowledge engine:
http://www.wolframalpha.com/input/?i=Tanzania+age+distribution
Another useful source of census data is the United States Census Bureau’s International Data Base:
https://www.census.gov/data-tools/demo/idb/informationGateway.php
The pyramid plot produced by Wolfram|Alpha is shown in the figure below.
The table produced by Wolfram|Alpha was downloaded and stored in a CSV file:
ref <- read.table("as.ex02.csv", header = TRUE, sep = ",")
ref
#> age Males Females All
#> 1 [0,5) 4043000 3969000 8012000
#> 2 [5,10) 3336000 3284000 6620000
#> 3 [10,15) 2775000 2742000 5517000
#> 4 [15,20) 2386000 2372000 4758000
#> 5 [20,25) 2076000 2073000 4149000
#> 6 [25,30) 1753000 1750000 3503000
#> 7 [30,35) 1453000 1432000 2885000
#> 8 [35,40) 1142000 1099000 2241000
#> 9 [40,45) 873000 846000 1719000
#> 10 [45,50) 673000 699000 1372000
#> 11 [50,55) 538000 601000 1139000
#> 12 [55,60) 433000 503000 936000
#> 13 [60,65) 357000 426000 783000
#> 14 [65,70) 266000 319000 585000
#> 15 [70,75) 182000 222000 404000
#> 16 [75,80) 108000 137000 245000
#> 17 [80,85) 51000 68000 119000
#> 18 [85,90) 17000 25000 42000
#> 19 [90,95) 3000 6000 9000
#> 20 [95,100) 0 1000 1000
The age-groups are expressed using the form specified in ISO 31-11, an international standard that applies to mathematical symbols. The form [a,b) expresses the interval $$a ≤ x < b$$. For example, [30,35) is used to indicate the set {30, 31, 32, 33, 34} of ages in years. The form [a,b) is said to be closed on the left and open on the right.
The reference data (ref) uses five-year age-groups. We will create the same age-groups in the example dataset.
We should first check the range of ages in the example data:
range(svy$age) which returns: #> [1] 0 93 The R language provides a function that makes it easy to create ISO 31-11 groupings from raw data: svy$ageGroup <-cut(svy$age, breaks = seq(from = 0, to = 95, by = 5), include.lowest = TRUE, right = FALSE) Using include.lowest = TRUE tells the cut() function to include the lowest breaks value (zero in this case). Using right = FALSE tells the cut() function to use groupings that are closed on the left. This combination of parameters creates the same “closed on the left” and “open on the right” age-groups as are used in the reference (ref) data: table(svy$ageGroup)
#>
#> [0,5) [5,10) [10,15) [15,20) [20,25) [25,30) [30,35) [35,40) [40,45) [45,50)
#> 1598 1268 1072 808 870 575 580 385 424 258
#> [50,55) [55,60) [60,65) [65,70) [70,75) [75,80) [80,85) [85,90) [90,95]
#> 284 128 165 82 98 51 60 18 12
A tabular analysis of age-group by sex can be produced using:
table(svy$ageGroup, svy$sex)
#>
#> 1 2
#> [0,5) 821 777
#> [5,10) 637 631
#> [10,15) 547 525
#> [15,20) 389 419
#> [20,25) 342 528
#> [25,30) 343 232
#> [30,35) 250 330
#> [35,40) 177 208
#> [40,45) 206 218
#> [45,50) 125 133
#> [50,55) 162 122
#> [55,60) 70 58
#> [60,65) 87 78
#> [65,70) 33 49
#> [70,75) 47 51
#> [75,80) 22 29
#> [80,85) 24 36
#> [85,90) 10 8
#> [90,95] 1 11
A visual inspection is useful:
pyramid.plot(svy$ageGroup, svy$sex)
We can make this easier to read:
pyramid.plot(svy$ageGroup, svy$sex,
main = "Age-group by sex",
xlab = "Number (Males | Females)",
ylab = "",
las = 1,
cex.names = 0.9)
Note that we specified ylab = "" because it is clear that the category labels represent age-groups and to prevent the y-axis label from obscuring the category labels, which happens with:
pyramid.plot(svy$ageGroup, svy$sex,
main = "Age-group by sex",
xlab = "Number (Males | Females)",
ylab = "Age-group",
las = 1,
cex.names = 0.9)
It is possible to alter the number of lines of text in margins of the plot, reduce the size of the age-group labels, and place the y-axis label on a specific line in the left margin of the plot in order to make a clearer plot:
par(mar = c(5, 5, 4, 2))
pyramid.plot(svy$ageGroup, svy$sex,
main = "Age-group by sex",
xlab = "Number (Males | Females)",
ylab = "",
las = 1,
cex.names = 0.8)
title(ylab = "Age-group", line = 4)
The easiest way of checking whether the survey data represents the general population in terms of the age and sex distribution is to compare the observed (figure on right) and expected (figure on left) distributions.
The general shapes of the two distributions are similar. Some of the lumpiness in figure on the right is due to age heaping in the adult ages at decades and half-decades:
ah <- ageHeaping(svy$age, divisor = 10) plot(ah, main = "Remainder of age / 10") A more formal test of the age structure can be made by comparing observed and expected numbers. We can do this graphically: ref <- ref[1:19, ] expectedProportions <- ref$All / sum(ref$All) expectedNumbers <- expectedProportions * sum(table(svy$ageGroup))
mp <- barplot(table(svy$ageGroup), main = "Observed and expected numbers", ylim = c(0, max(expectedNumbers)), las = 2) lines(mp, expectedNumbers, lty = 2, lwd = 2) The observed and expected numbers are similar to each other. The lumpiness in the observed numbers is due to age heaping. See Figure ASA04. Formal testing can be performed: chisq.test(table(svy$ageGroup),
p = expectedProportions)
This gives:
#> Warning in chisq.test(table(svy$ageGroup), p = expectedProportions): Chi-squared #> approximation may be incorrect #> #> Chi-squared test for given probabilities #> #> data: table(svy$ageGroup)
#> X-squared = 248.41, df = 18, p-value < 2.2e-16
The warning is due to small expected numbers (i.e. n < 5) in the older age-groups. R provides a more robust “Monte Carlo” test:
chisq.test(table(svy$ageGroup), p = expectedProportions, simulate.p.value = TRUE) This may take a few seconds to compute and yields: #> #> Chi-squared test for given probabilities with simulated p-value (based #> on 2000 replicates) #> #> data: table(svy$ageGroup)
#> X-squared = 248.41, df = NA, p-value = 0.0004998
The test results need to be interpreted with caution. The sample size ($$n = 8736$$) is large in this example. This means that small differences, which may be due to age heaping, become statistically significant. This test cannot be considered to be good evidence that the age-structure of the sample differs from the expected age-structure of the population. We also need to examine the sex ratio of the sample. A sex ratio test can be performed using the sexRatioTest() function from the NiPN data quality toolkit and the sex ratio observed in the census data:
censusM <- sum(ref$Males) censusF <- sum(ref$Females)
sexRatioTest(svy\$sex,
codes = c(1, 2),
pop = c(censusM, censusF))
This yields:
#>
#> Sex Ratio Test
#>
#> Expected proportion male = 0.4988
#> Observed proportion male = 0.4914
#> X-squared = 1.8770, p = 0.1707
There is no evidence that the sex ratio in the sample differs much from the expected sex ratio in the population.
The techniques outlined in this section are illustrative. This is because many surveys, other than nutritional anthropometry surveys in young children, are not standardised. A survey may sample only women of child-bearing age. The sample may be restricted to women aged between 15 and 45 years.
In this case the age-structure can be examined using the techniques outlined above but it would make no sense to examine the sex ratio. Care should be taken when examining data from surveys that may have deliberately oversampled specific age-groups. | 8,719 | 27,528 | {"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": 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} | 2.53125 | 3 | CC-MAIN-2023-06 | latest | en | 0.794966 |
https://subscanner.com/biK6_fWLDRc | 1,632,701,836,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00505.warc.gz | 573,424,151 | 3,472 | # Curvature of a cycloid
- [Voiceover] So let's do another curvature example.
This time, I'll just take a two-dimensional curve,
so it'll have two different components, x of t and y of t
and the specific components here will be
t minus the sine of t,
t minus sine of t,
and then one minus cosine of t.
One minus cosine of t.
This is actually the curve,
if you watched the very first video that I did
about curvature introducing it, this is that curve.
This is the curve that I said,
"Imagine that it's a road and you're driving
"along it, and if your steering wheel gets stuck,
"you're thinking of the circle that you trace out
"as a result and it various different points
"you're gonna be turning it various different amounts,"
so the circle that your car ends up tracing out
would be of varying sizes.
So if the curvature's high, if you're steering a lot,
radius of curvature is low and things like that.
So here, let's actually compute it.
And in the last example I walked through
thinking in terms of the derivative
of the unit-tangent vector with respect to arc length
but in this case, instead of doing that,
I just want to show what it looks like
when we take the explicit formula that looks like
x prime times y double-prime,
minus y prime times x double-prime,
and then all of that divided by
x prime squared plus y prime squared.
And then I'm writing x prime and y prime and such
and all of these you should think of as
taking in the variable t.
I'm must being a little too lazy to write it.
And you take that to the three-halves power.
So, this was the formula
and I'm not a huge fan of memorizing formulas
and then hoping to apply them later.
I really do think the one thing you should take away
from curvature is the idea that
it's the derivative of the unit-tangent vector
with respect to arc length
and if you need to, you can just look up
a formula like this but it's worth pointing out
that it makes some things easier
to compute because finding the tangent vector
and everything can be kind of like reinventing
the wheel when you already have the results here.
So, first thing to do is just find
x prime, y double-prime,
y prime and x double prime.
So let's go ahead and write those out.
So the first derivative of x of t
if we go up here that's t minus sine of t.
So its derivative is one minus cosine of t.
And the derivative of the y component
of one minus cosine t,
y prime of t,
is gonna be, derivative of cosine is negative sine
so negative derivative of that is sine,
and that one goes to a constant,
and then when we take the second derivatives of those guys,
so maybe change the color for the second derivative here,
x double prime of the,
so now we're taking the derivative of this,
which actually we just did because
by coincidence the first derivative of x
is also the y component
so that also equals sine of t.
And then y double prime is just the derivative of sine here
so that's just gonna be cosine, cosine of t.
So now, when we just plug those four values
in for kappa, for our curvature, what we get
is x prime was one minus cosine of t,
multiplied by y double prime is cosine of t.
Cosine of t.
We subtract off from that y prime,
which is sine of t,
sine of t,
multiplied by x double prime.
So x double prime is also sine of t,
so I could just say
sine of t squared,
and the whole thing is divided by x prime squared
so x prime was one minus cosine of t,
minus cosine of t,
squared, plus y prime squared
so y prime was just sine,
so that's just gonna be sine squared of t.
And that whole thing to the power three halves. | 857 | 3,558 | {"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} | 4.3125 | 4 | CC-MAIN-2021-39 | latest | en | 0.960426 |
https://www.k8mathsense.com/january-2023-blog | 1,723,736,930,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00159.warc.gz | 642,133,315 | 11,675 | Create Expressions with 2, 0, 2, and 3
Start the new year with a versatile activity that can be modified easily! Have elementary or middle school students write expressions using 2, 0, 2, and 3. Click below to download a FREE worksheet with instructions and a table for values up to 23. This activity can help students review 5 key concepts, explained below.
1. Adding or Multiplying with Zero
2. Order of Operations
3. Using Exponents
4. Expressions with Parentheses
5. Making Organized Lists
1. Adding or Multiplying with Zero
What properties can you apply?
• Multiplying any number by 0 gives a result of 0. This is called the Zero Property of Multiplication.
• Adding 0 to any number does not change the number. This is called the Zero Property of Addition.
Try these:
Have students write other examples that include multiplication or addition with 0. If you are using the worksheet, have students use the digits 2, 0, 2, and 3 exactly once. See how many different expressions they can write with these restrictions and without using parentheses.
More examples: 23 × 2 × 0 [0], 22 × 3 × 0 [0], 3 × 0 + 22 [22], 2 × 0 + 23 [23]
2. Order of Operations
In expressions without parentheses or exponents, what is the correct order operations?
• Do multiplication and division from left to right.
• Then do addition and subtraction from left to right.
Try these:
You may want to encourage students to underline the multiplication and division expressions to do first.
3. Using Exponents
How do you evaluate expressions with exponents?
• An exponent tells how many times to use the base number as a factor.
• Any number with a zero exponent has a value of 1.
• When exponents appear along with other operations, evaluate the exponents first.
Try these:
When an expression has several parts, encourage students to copy each simplified expression below the previous expression, and not to leave out any parts. Here is a way to show the work:
4. Expressions with Parentheses
How can parentheses impact the order of operations?
• Do the parts within parentheses first.
• If there are nested parentheses, do the innermost ones first.
• Inserting parentheses can cause the expression to have a different value.
Try these pairs of expressions:
Discuss why the parentheses in the first expression of each pair are not necessary. [The expressions inside parentheses would be computed first due to the rules for order of operations.]
5. Making Organized Lists
How do you organize the ways to find expressions?
• Organize the possible numbers made of the digits 2, 0, 2, and 3.
• Organize operations to put between the numbers in expressions.
Examples for organizing numbers:
Examples for organizing operations:
By trying the possibilities in an organized way, students should be able to find dozens of expressions using the digits 2, 0, 2, and 3. If you are using the worksheet, you may want to have them limit their answers to expressions that equal whole numbers up to 23.
Summary and Variations
As you can see, students can review many concepts while evaluating expressions. The free download includes an answer key with at least one expression for each value from 1 to 23.
Here are some ideas for creating variations.
• For lower grades, have students use only addition and subtraction with 1-digit and 2-digit numbers. Perhaps allow students to include expressions that use two of three of the digits rather than all four digits.
• For more challenge, have students find expressions with values up to 100. Some may not be possible. Have teams compete for finding the most expressions.
Happy Holidays!
Angie | 797 | 3,613 | {"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} | 4.65625 | 5 | CC-MAIN-2024-33 | latest | en | 0.872439 |
https://askinglot.com/what-is-the-difference-between-part-to-part-and-part-to-whole-ratios | 1,620,319,135,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988758.74/warc/CC-MAIN-20210506144716-20210506174716-00211.warc.gz | 134,540,108 | 6,772 | asked in category: General Last Updated: 15th June, 2020
# What is the difference between part to part and part to whole ratios?
Ratios can compare a part to the whole. An example of a part to a whole ratio is the number of females in a class to the number of students in the class. Ratios can compare parts to parts. An example of a part to a part ratio is where the number of females in a class is compared to the number of males.
Likewise, people ask, what is a part to part and a part to whole ratio?
Ratios can be divided into part-to-part ratios and part-to-whole ratios. Part-to-part ratios provide the relationship between two distinct groups. For example, the ratio of men to women is 3 to 5, or the solution contains 3 parts water for every 2 parts alcohol.
what is a part ratio? part-part ratio. • a ratio that compares a selected number of parts. to a number of other parts in a whole.
Keeping this in consideration, what does part to whole mean in math?
Part-Whole is a ratio or a fraction that represents a relationship between a part and its whole.
What is ratio formula?
Ratio Formula. When we compare the relationship between two numbers dealing with a kind, then we use the ratio formula. It is denoted as a separation between the number with a colon (:). Sometimes a division sign is also used to express ratios.
10
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Diagram 1.6: Diagram showing tension force in the string (from 0 cm increasing in 1 cm until a displacement of 15 cm has been attained) This diagram shows the manner in which the tension force applied to the string was calculated. Starting at 0 centimeters, forces were provided in the string such that the string would extend one additional centimeter from the previous measurement. Using the Capstone® software, the force applied to the string was calculated at the given horizontal…
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It begins with a couple of pedal strokes. Round and Round the wheels turn and your off like a shot. Sounds straightforward? Traveling across rural scenery. I love it. Edinburgh should be called the Happy Trails city once you get away out of the city there are a million of places to run, walk, hike or bike. I go mountain biking every Monday night with the club. For the first 5 minutes my muscles burn, it feels like I am dragging myself through mud and being stewed alive all at the same time. Then…
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Rene Descartes as a famous philosopher, mathematician and scientist in seventieth century has introduced a new method for recognising and distinguishing truth from untruth. Descartes believed that this method could be useful for every body since reason is same for all people. The aim of this essay is to demonstrate Descartes’ method in three parts. In the first part it will explain that what Descartes’ method is. In the second part, it will explain the philosophical meaning of this method. And…
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Introduction Sporting code of choice is Netball Netball is a sport played with a ball and two teams of seven players. Netball games are played on a rectangular court surface with two raised goal posts with rings on the top. The rings are used to shoot the ball in. Each player has a specific position to play on the court and certain players are restricted to a certain place. Each team has a Goal Shooter, Goal Attack, Wing Attack, Centre, Wing Defence, Goal Defence and Goal Keeper. A…
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Figure (4.2): Failure pattern of cylinder specimen (CY1). Under pure axial compression loading for cube specimen (C1), the failure cracks generated are approximately parallel to the direction of load as shown in figure (4.3) with some cracks formed at an angle to the applied load. As a result, lateral shearing stress is produced in the cube specimen. The effect of this shear decrease towards the center of cubes; so these sides of the cube have near vertical cracks at the cube center.…
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In gymnastics, the physics involved in energy, Newton’s Laws of Motion, and angular momentum each contribute immensely to being successful in the Olympics. Learning how to master these physical science concepts can aid in becoming a strong olympic gymnast. To begin, there is the running start; this is where energy plays a major role. During a gymnast’s running start for a vault, she gains kinetic energy, which is, simply put, the energy of motion. Looking at the equation for kinetic energy (KE =…
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• Page 1 2 3 4 5 6 7 8 9 50 | 1,281 | 5,688 | {"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} | 3.546875 | 4 | CC-MAIN-2023-50 | latest | en | 0.912281 |
https://www.airmilescalculator.com/distance/vbs-to-fmm/ | 1,606,768,331,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00026.warc.gz | 567,628,553 | 27,447 | # Distance between Brescia (VBS) and Memmingen (FMM)
Flight distance from Brescia to Memmingen (Brescia Airport – Memmingen Airport) is 177 miles / 285 kilometers / 154 nautical miles. Estimated flight time is 50 minutes.
Driving distance from Brescia (VBS) to Memmingen (FMM) is 296 miles / 477 kilometers and travel time by car is about 5 hours 14 minutes.
## Map of flight path and driving directions from Brescia to Memmingen.
Shortest flight path between Brescia Airport (VBS) and Memmingen Airport (FMM).
## How far is Memmingen from Brescia?
There are several ways to calculate distances between Brescia and Memmingen. Here are two common methods:
Vincenty's formula (applied above)
• 176.877 miles
• 284.657 kilometers
• 153.702 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 176.925 miles
• 284.733 kilometers
• 153.743 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Brescia Airport
City: Brescia
Country: Italy
IATA Code: VBS
ICAO Code: LIPO
Coordinates: 45°25′44″N, 10°19′50″E
B Memmingen Airport
City: Memmingen
Country: Germany
IATA Code: FMM
ICAO Code: EDJA
Coordinates: 47°59′19″N, 10°14′22″E
## Time difference and current local times
There is no time difference between Brescia and Memmingen.
CET
CET
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 51 kg (113 pounds).
## Frequent Flyer Miles Calculator
Brescia (VBS) → Memmingen (FMM).
Distance:
177
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
177
Round trip? | 483 | 1,812 | {"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} | 2.6875 | 3 | CC-MAIN-2020-50 | latest | en | 0.788121 |
https://www.wiredfaculty.com/question/UTBKVFJVVk9UVUV4TVRBeE16RTFOQT09 | 1,718,353,788,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00833.warc.gz | 974,937,639 | 12,718 | Permutations And Combinations
Question
# Find four numbers in GP. whose sum is 85 and product is 4096.
Solution
Let the four numbers in G.P. be
Since their product is 4096.
∴
Since their sum is 85
∴
Put
∴ or
Putting 2t = x, we get
By inspection x = 5 satisfies the equation. So x - 5 is a factor of . Other roots are given by
x is imaginary
∴ x = 5 is the only real value
So, 2t = 5,
When a = 8, r = 2, numbers are
When a = 8, numbers are or 64, 16, 4, 1 | 177 | 533 | {"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} | 3.890625 | 4 | CC-MAIN-2024-26 | latest | en | 0.824474 |
https://payilagam.com/blogs/xerago-fresher-interview-questions-part-i/ | 1,556,228,505,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578742415.81/warc/CC-MAIN-20190425213812-20190425235812-00486.warc.gz | 502,425,836 | 19,342 | # Xerago – Fresher Interview Questions – Part – I
## Here in this blog we are discussing about Fresher Interview Questions in Xerago, Chennai .
Some of them are :
1) On what dates of April, 2001 did Wednesday fall?
A)1st, 8th, 15th, 22nd, 29th B.2nd, 9th, 16th, 23rd, 30th
C.3rd, 10th, 17th, 24th D.4th, 11th, 18th, 25th
SOLUTION :
Since 1 st April 2001 = ( 2000 years + Period from 1.1.2001 to 1.4.2001 )
January + February + March + April = 31 + 28 + 31 + 1
= 91 days = 13 weeks + 0 days
As there was no odd days , 1 st April falls on Sunday.
Then 4 th April will be Wednesday
Hence In April 2001 , Wednesday falls on 4 th , 11 th , 18 th and 25 th .
______________________________________________
2) The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
A.40% B.42% C.44% D.46%
SOLUTION :
Let the sides of a rectangle be x cm and y cm
Then the area of the rectangle = xy cm2
If the sides are increased by 20 % then the new sides will be
( x + 20 % of x ) and ( y + 20% of y ) = ( 6/5 ) x and ( 6/5 ) y
New area = (6 / 5)x (6 / 5) y = ( 36 / 25 ) xy
Difference in Area = ( 36 / 25 ) xy – xy = 11 / 25
Percentage in increase in Area = ( 11 / 25 ) * 100
= 11 * 4 = 44 %
_______________________________________________
3) Choose the alternative which is closely resembles the mirror image of the given combination.
SUPERVISOR
______________________________________________
4)Choose the alternative which is closely resembles the water-image of the given combination.
NUCLEAR
________________________________________________
5) Statements: No women teacher can play. Some women teachers are athletes.
Conclusions:
1. Male athletes can play. 2. Some athletes can play.
A.Only conclusion I follows
B.Only conclusion II follows
C.Either I or II follows
D.Neither I nor II follows
E.Both I and II follow | 609 | 2,025 | {"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} | 4.3125 | 4 | CC-MAIN-2019-18 | latest | en | 0.768521 |
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Chapter 3 58
# Chapter 3 58 - 208 CHAPTER 3 DIFFERENTIATION RULES 4...
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Unformatted text preview: 208 CHAPTER 3 DIFFERENTIATION RULES 4. Substituting the values of h and 3 in Problem 3 into P(:c) = —:—I:w3 + \$3332 gives us P(:c) 2 aa:3 + b352, where a m 4.937 x 10-5 and b m 4.78 X 10-3. 7 0 64.52 APPLIED PROJECT Building a Better Roller Coaster 1. (a) f(\$)=a\$2+bm+c 2 f'(\$):2ax+b. The origin is at P : f(0) : 0 => c = 0 The slope of the ascent is 0.8: f'(0) = 0.8 => b = 0.8 The slope of the drop is ~16: f'(100) = ‘1.6 => 200a + b 2 —1.6 (b) b = 0.8, so 200a —l— b : —1.6 => 200a -I— 0.8 = —1.6 :> 200a = —2.4 => a 2 —% = #0012. Thus, f(m) = A0.012x2 + 0.8m. (c) Since L1 passes through the origin with slope 0.8, it has equation y = 0.8m. The horizontal distance between P and Q is 100, so the y-coordinate at Q is f(100) = —0.012(1OO)2 + 0.8(100) = —40. Since L2 passes through the point (100, -40) and has slope —1.6. it has equation y + 40 = ~1.6(\$ # 100) or y : —1.6x + 120. (d) The difference in elevation between P (0, 0) and Q(100, —40) is 0 — (—40) : 40 feet. 2. (a) Interval Function First Derivative Second Derivative —l (—007 0) L1 (as) : 0.8m L'1(:c) : 0.8 L'1’(:c) : O [0.10) g(a:) : [€333 + lm2 + mm + n g’(w) : 1&ka + 2lw + m g"(a:) = 6km -— 2l [10, 90] q(a:) = (1:32 + bx + c q’(9:) = 2am + b q"(m) : 2a (90,100] h(:1:) : p363 + qm2 + rm + s h’(\$) = 3pm2 + 2qan + r h"(:c) : 6px —— 2q (100, 00) L2(a:) : —1t6m + 120 L’2(m) : -1.6 L’2’ (51:) = 0 There are 4 values of :L' (0, 10, 90, and 100) for which we must make sure the function values are equal, the first ...
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https://www.teachoo.com/7042/939/Polynomials/category/Introduction/ | 1,685,851,911,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00718.warc.gz | 1,112,706,676 | 33,372 | Introduction
Chapter 2 Class 10 Polynomials
Concept wise
Equations of the type
x 3 − 3x 2 + 4x + 10
are called polynomials.
Here,
x 3 , −3x 2 , 4x , 10 are called terms of polynomials.
A Polynomial has
• Variables like x, y with powers as whole numbers (0, 1, 2, 3, ....). If Power is not a whole number , the equation is not a polynomial.
• constants like 10
## Polynomial in one variable
In this chapter,
We will be studying polynomials which are in one variable, i.e. only has x or y, but not both.
Example - x 2 + 2x - 3 is a polynomial in one variable as there is only x
But x 2 + 2y - 3x = 0 is not a polynomial in one variable as there is both x and y.
Is x 2 + 24y + 1 a polynomial in one variable?
-a-
Since there are two variables – x and y
It is not a polynomial in 1 variable
-ea-
Is y 4 – 10 a polynomial in one variable?
-a-
y 4 – 10 = y 4 – 10y 0
Here power are 4 and 0.
Since all powers are whole numbers, it is a polynomial
And since there is only one variable y,
It is a polynomial in one variable
-ea-
Is 5x -2 + 1 a polynomial in one variable?
-a-
Since power is -2
And -2 is not a whole number,
It is not a polynomial in one variable
-ea-
Is 3x 1/2 + 2 a polynomial in one variable?
-a-
Since power is 1/2
And 1/2 is not a whole number,
It is not a polynomial in one variable
-ea-
Is 3 a polynomial in one variable?
-a-
3 = 3x 0
Since power is 0
And 0 is a whole number,
It is a polynomial.
And since there is only 1 variable x,
It is a polynomial in one variable
-ea-
Note:
Constant numbers are also polynomials.
Example: 2, 3, 100, -988999 are all polynomials
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#### Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo. | 586 | 1,970 | {"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} | 4.5625 | 5 | CC-MAIN-2023-23 | longest | en | 0.920907 |
http://forums.wolfram.com/mathgroup/archive/2004/Jun/msg00158.html | 1,521,705,850,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647782.95/warc/CC-MAIN-20180322073140-20180322093140-00045.warc.gz | 108,124,054 | 7,979 | Re: Re: Random Matrix of Integers
• To: mathgroup at smc.vnet.net
• Subject: [mg48615] Re: [mg48612] Re: Random Matrix of Integers
• From: Oleksandr Pavlyk <pavlyk at phys.psu.edu>
• Date: Tue, 8 Jun 2004 00:48:06 -0400 (EDT)
• Organization: Penn State University; Department of Physics
• References: <200406040851.EAA23928@smc.vnet.net> <c9scag\$bt6\$1@smc.vnet.net> <200406070933.FAA11076@smc.vnet.net>
• Reply-to: pavlyk at phys.psu.edu
• Sender: owner-wri-mathgroup at wolfram.com
```Peter Pein wrote:
>
> Really?
>
> In[2]:=
> Timing[Do[RandomArray[DiscreteUniformDistribution[8], {4, 4}] + 1,
> {k, 1, 10000}]]
> Timing[Do[Table[Random[Integer, {2, 9}], {4}, {4}], {k, 1, 10000}]]
> Out[2]=
> {0.22100000000000009*Second, Null}
> Out[3]=
> {3.4039999999999995*Second, Null}
Dear Peter,
I can not confirm your timings ratio.
On my PC and with my copy of Mathematica 5.0.1
Table[Random[Integer,{2,9}],{4},{4}]
consistently outperforms RandomArray.
In[37]:=
Quit
In[8]:=
\$Version
Out[8]=
"5.0 for Microsoft Windows (November 18, 2003)"
In[1]:=
<< "Statistics`"
In[2]:=
Timing[Do[RandomArray[
DiscreteUniformDistribution[8], {4, 4}] + 1,
{k, 1, 10000}]]
Out[2]=
{2.924*Second, Null}
In[3]:=
Timing[Do[Table[Random[
Integer, {2, 9}], {4},
{4}], {k, 1, 10000}]]
Out[3]=
{0.601*Second, Null}
It occurred to me that we might be using different Mathematica
versions, so I ran this code on Solaris Mathematica 4.1
Mathematica 4.1 for Sun Solaris
Copyright 1988-2000 Wolfram Research, Inc.
-- Terminal graphics initialized --
In[1]:= <<Statistics`
In[2]:= Timing[Do[RandomArray[
DiscreteUniformDistribution[8], {4, 4}] + 1,
{k, 1, 10000}]]
Out[2]= {11.23 Second, Null}
In[3]:= Timing[Do[Table[Random[
Integer, {2, 9}], {4},
{4}], {k, 1, 10000}]]
Out[3]= {4.88 Second, Null}
with essentially the same outcome: Random[Integer,{2,9}]
is faster. It makes perfect sense to me that StandardPackage
function is slower than the built in one, otherwise it would
be wise to make the former built-in.
Are your results consistent ? I mean can you reproduce them
in the freshly started kernel ? what is your cpu and mathematica
version ?
Regards,
Sasha
> "Oleksandr Pavlyk" <pavlyk at phys.psu.edu> schrieb im Newsbeitrag
> news:c9scag\$bt6\$1 at smc.vnet.net...
>
>>Hi Bruce,
>>
>>Random is using uniform distribution, so
>>the it should be used with RandomArray.
>>
>>In[1]:=
>><< "Statistics`"
>>In[11]:=
>>RandomArray[
>> DiscreteUniformDistribution[8], {4, 4}] + 1
>>Out[11]=
>>{{4, 3, 4, 4}, {7, 2, 2, 6},
>> {5, 2, 7, 4}, {6, 8, 2, 9}}
>>
>>However note that RandomArray is much slower
>>
>>In[2]:=
>>Timing[Do[RandomArray[
>> DiscreteUniformDistribution[8], {4, 4}] + 1,
>> {k, 1, 10000}]]
>>Out[2]=
>>{3.636*Second, Null}
>>In[3]:=
>>Timing[Do[Table[Random[
>> Integer, {2, 9}], {4},
>> {4}], {k, 1, 10000}]]
>>Out[3]= {0.6609999999999996*Second, Null}
>>
>>Sasha
>>
>>
```
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• Next by thread: Re: Random Matrix of Integers | 1,144 | 3,133 | {"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} | 3.0625 | 3 | CC-MAIN-2018-13 | longest | en | 0.706066 |
https://stats.stackexchange.com/questions/616977/explainability-of-groups-mix-affect-on-overall-mean | 1,721,613,623,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00522.warc.gz | 467,520,462 | 37,750 | # Explainability of groups mix affect on overall mean
I am stuck on understanding the best way to approach this problem from a purely statistical/mathematical way.
Say that I have this data in the table below. It shows the comparison of prices for one hotel between two years with a breakdown of the board type and room type.
Hotel Board Room 2023 Price 2024 Price
1 HB A 1000 1100
1 HB B 1600 1850
1 FB A 1500 1750
1 FB B 1900
1 FB C 2250 2250
1 ALL A 2000
1 ALL B 2600 2750
Within this data, as a human I can easily tell that most of the prices go up year on year and from the averages below I can also understand this too
2023 average - 1808
2024 average - 1950
However when eyeballing the data I can also see there are some mix differences that will be affecting the overall mean, for example
Board type of "ALL" averages:
2023 average - 2600
2024 average - 2375
Here we can see that because of a lack of data in 2023 for the room type "A" for the board type "ALL" there looks to be a step back in price. This then may actually be artificially increasing the overall mean for this hotel and thus showing a smaller increase in price year on year.
What I want to try and understand - is there a technique out there that will be able to show and explain the drivers affecting the yearly overall means across all groups? | 339 | 1,332 | {"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} | 2.96875 | 3 | CC-MAIN-2024-30 | latest | en | 0.929997 |
https://www.wyzant.com/resources/answers/809799/i-did-everything-else-i-just-need-the-rest-of-part-3 | 1,632,653,098,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00423.warc.gz | 1,085,649,774 | 14,571 | James G.
# I did everything else, I just need the rest of part 3
### Part Two: Representation of Data with Plots
1. Using graphing software of your choice, create a scatter plot of your data. Predict the line of best fit, and sketch it on your graph.
2. Copy and paste your scatter plot into a word processing document.
### Part Three: The Line of Best Fit
Include your scatter plot and the answers to the following questions in your word processing document.
1. Which variable did you plot on the x-axis, and which variable did you plot on the y-axis? Explain why you assigned the variables in that way.
2. Write the equation of the line of best fit using the slope-intercept formula $y = mx + b$. Show all your work, including the points used to determine the slope and how the equation was determined.
3. What does the slope of the line represent within the context of your graph? What does the y-intercept represent?
4. Test the residuals of two other points to determine how well the line of best fit models the data.
5. Use the line of best fit to help you to describe the data correlation.
6. Using the line of best fit that you found in Part Three, Question 2, approximate how tall is a person whose arm span is 66 inches?
7. According to your line of best fit, what is the arm span of a 74-inch-tall person? 101 inches
Part One: Measurements
Data Record
Height vs. Arm Span
Height (x) in. Arm span ( y) in.
Part Two: Representing Data with Plots
Please refer to the attached document for the graphs.
Part Three: The Line of Best Fit
Using excel, the two points that were used to draw the lines of best fits were (62, 63 ) and (70, 72 ).
## 1 Expert Answer
By:
Laura M. answered • 01/14/21
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Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 480 | 2,018 | {"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} | 3.5 | 4 | CC-MAIN-2021-39 | latest | en | 0.928379 |
https://graduatesassignmenthelp.com/25956-2/ | 1,624,096,285,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487647232.60/warc/CC-MAIN-20210619081502-20210619111502-00329.warc.gz | 274,496,188 | 13,680 | formulas
Part 1: Basics
You are living on your own space. Your own car. Your own BILLS!! Use Microsoft Excel to create your own budget as in Figure 1 below.
Figure 1
• Open MS Excel
• Create a blank workbook
• Save your Excel workbook as “Homework 3”.
• Create a worksheet exactly like in figure 1. Change the title of the sheet to “your name’s” Expenses.
• Use Auto fit, Auto fill, and Auto complete features in Excel to create the sheet faster.
• Rename sheet 1 to “Expenses”.
• Add all borders to the range A2:M8. (Border options are located in the “Home” tab-font section)
• Save, DO NOT close the workbook, and move on to part 2
Part 2: Formulas
• Rename the new Worksheet to “Math operations”
• Merge and Center cells A1:E1 and type “Working with Excel Math” in there
• Apply Title style to cell A1(Home tabàStylesàcell stylesàTitle)
• Type “Addition” in cell A2
• Type “Subtraction” in cell A3
• Type “Multiplication” in cell A4
• Type “Division” in cell A5
• Auto-fit the contents in column A
• Insert a row above row #2
• Type “Basic Math” in cell A2
• Merge and Center the text in cell A2 to the range A2:E2
• Apply the Calculation Style to the range A3:A6(Home tabàStylesàcell stylesàCalculation)
• Type = 4 + 6 / 2 in cell B3 then type = (4 + 6) / 2 in cell D3
• Type = 6 – 6 * 2 in cell B4 then type = (6 – 6) * 2 in cell D4
• Type = 2 * 2 + 6 in cell B5 then type = (2 * 2) + 6 in cell D5
• Type = 9 / 3 + 4 in cell B6 then type = ( 9 / 3 ) + 4 in cell D6
• Add all borders to the range A1:E6
• Here is how your sheet would look like.
• press Ctrl + ~ and this what would see
• Save, DO NOT close the workbook, and move on to part 3
Part 3: Import and analyze data
• Download the Expenses.txt file; it’s attached to Homework 3 on blackboard.
• Back to Excel workbook “Homework 3”
• Add a new worksheet to “Homework 3” workbook.
• Name the added worksheet “My Budget”.
• Import the data from Expenses.txt into “My Budget”
• Insert a table in the Range A2:M11
• Merge and center the cells A1:P1
• Format cell A1: font color=red, font size=18, Fill color=yellow and make the text bold.
• Rename sheet 1 to “Expenses”.
• Insert the word “Total” in cells N2 and A12. In column N, insert formulas to find the total of each row. In row 12, insert formulas to find the total of each column, including column N. Once you have completed this sheet, cell N12 should show the grand total of all expenses for the year.
• Insert the text “Monthly min” in cell A14. In row 14, insert formulas in cells B14:M14 to find the minimum expense of each month.
• Insert the text “Monthly Max” in cell A15. In row 15, insert formulas in cells B15:M15 to find the Maximum expense of each month.
• Insert the text “Occurrences” in cell A16. In cell B16, insert formulas to count the occurrences of the number 0 in range B3:M11(Use the countif function)
• Insert the text “sum if value>800” in cell A17. In cell B17, insert formulas to find the total of all the cells in the range B3:M11 that contain values greater than or equal \$800 (Use the sumif function)
• In O2 add the text “Expense Ratio”
• In cells O3:O11 insert formulas to insert “High Expense” if the total in column N is more than 400 and “Okay Expense” if the total is not greater than 400.(use if function)
• In P2 add the text “Expense average”
• In cells P3:P11 insert formulas to find the average of each expense.
• Add all borders to the range A1: P15. (Border options are located in the “Home” tab-font section)
• Create a 3-D Column chart that shows monthly expenses for rent, utilities, food, travel, tuition, and car insurance for the month of June.
• Type “Monthly Expenses for June” in the chart title.
• Move the chart to a new worksheet titled “June”.
• Save and submit your workbook under Homework3 on the blackboard. | 1,089 | 3,797 | {"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} | 3.796875 | 4 | CC-MAIN-2021-25 | longest | en | 0.640614 |
https://stackoverflow.com/questions/52923540/matplotlib-3d-workaround-for-plot-order | 1,696,189,631,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510924.74/warc/CC-MAIN-20231001173415-20231001203415-00027.warc.gz | 601,725,994 | 41,142 | # Matplotlib 3D workaround for plot order
I know that matplotlib 3D is not reliable for plotting multiple 3D objects (planes, lines, points) in the right order: please see Matplotlib 3D plot zorder issue and How to draw intersecting planes?. However these questions seem quite old, so the proposed solutions. Thus, I would like to know if there are some new developments, tools, workarounds or hard-coded solutions for the following specific simple scenario:
``````import mpl_toolkits.mplot3d as a3
import matplotlib.pylab as plt
import numpy as np
fig = plt.figure()
ax = a3.Axes3D(fig)
# create an orizontal plane
corners = [[0,0,0],[0,5,0],[5,5,0],[5,0,0]]
tri = a3.art3d.Poly3DCollection([corners], alpha=1)
tri.set_color('w')
tri.set_edgecolor('k')
# plot a vector
ax.plot([2,2],[2,2],[0,4], c = 'r')
# plot some points
ax.scatter([1,3],[1,3],[1,3], c = 'r')
ax.set_xlim([0, 5.0])
ax.set_ylim([0, 5.0])
ax.set_zlim([0, 2.5]);
plt.show()
``````
In this image you can see the visualization issues: the vector it is not starting from the plane, as it should since his intiali point is (2,2,0)
• Your code produces the correct result for me. The red line starts at [2, 2, 0] as expected, and is not clipped. Matplotlib 2.1.1 and QT5Agg backend. Maybe your backend is at fault? Oct 22, 2018 at 7:42
• Works for me here too, red line starts at [2, 2, 0] as expected. `matplotlib`2.2.3, `Qt5Agg` backend Oct 22, 2018 at 9:39
• Thank you for the answer, but unfortunately it does not solve the issue. I tried also with both 'Qt5Agg' and 'WxAgg' Oct 22, 2018 at 10:07
• I'm sorry to read that - and the matplotlib version is also recent enough? Oct 22, 2018 at 10:26
• Thank you so much. Here it is github.com/matplotlib/matplotlib/issues/12620 Oct 24, 2018 at 18:21 | 567 | 1,772 | {"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} | 2.65625 | 3 | CC-MAIN-2023-40 | latest | en | 0.872076 |
https://blog.ezyang.com/2013/05/the-ast-typing-problem/ | 1,624,279,953,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488273983.63/warc/CC-MAIN-20210621120456-20210621150456-00346.warc.gz | 142,797,834 | 10,176 | the arc of software bends towards understanding
## The AST Typing Problem
This Lambda the Ultimate post (dated 2010) describes a rather universal problem faced by compiler writers: how does one go about adding “extra information” (e.g. types) to an AST? (The post itself divides the problem into three components: adding the information to the data types, using the information to inform the construction of the node, and using the information to inform the destruction of a node—but I’m really only interested in the question of how you define your data type, not do things to it.) In this post, I want to sum up ways of solving the problem which were described in this post, and also take a look at what some real world compilers do. The running example lambda calculus looks like the following:
```data Exp = Num Int
| Bool Bool
| Var Var
| If Exp Exp Exp
| Lambda Var Exp
| App Exp Exp
data Type = TyInt | TyBool | TyArrow Type Type
```
### Separate IR where nodes are decorated with types
The low-tech solution: if you need a new version of the IR with more information, just define a new IR type where each node can also carry the information. A trick to make these definitions more concise is to make a mutually recursive data structure. [1]
```type TExp = (TExp', Type)
data TExp' = TNum Int
| TBool Bool
| TVar Var
| TIf TExp TExp TExp
| TLambda Var TExp
| TApp TExp TExp
```
Despite (or perhaps because of) it’s simplicity, this approach is extremely popular among many compilers, especially in the ML community. A few examples include OCaml (parsetree/typedtree), MLton (AST/CoreML) and Ikarus Scheme. Part of the reason for this is that the transition from frontend language to typed language also comes with some other changes, and when a new AST is defined those changes can be combined in too.
### Nullable field
The unprincipled solution: use one AST, but have an optional field in which you can slot in the information. [2]
```type TExp = (TExp', Maybe Type)
data TExp' = TNum Int
| TBool Bool
| TVar Var
| TIf TExp TExp TExp
| TLambda Var TExp
| TApp TExp TExp
```
Presented without further comment.
### Explicit typing
While closely related to the separate IR solution, an explicitly typed IR takes the approach of not decorating each node with a type, but arranging that the type of any given node can be quickly computed using only local information. [3]
```data TExp = TNum Int
| TBool Bool
| TVar Var
| TIf TExp TExp TExp
| TLambda Var Type TExp
| TApp TExp TExp
```
Here, the difference between TExp and Exp is very slight; the TLambda is annotated with an explicit type for the binder. As far as type-checking is concerned, this makes a world of difference: we no longer need to look outside a lambda to figure out what the binder could be.
Forcing your IR to be explicitly typed is often a good idea for metatheoretic reasons, as complicated type systems often don’t have decidable inference algorithms. Both GHC’s core IR, Ur/Web's core and Coq are explicitly typed in this way.
### Two-level types
By deferring when you tie the knot of a recursive data-structure, you can arrange for the base functor to do double-duty for the untyped and typed representations. [4]
```data ExpF a = Num Int
| Bool Bool
| Var Var
| If a a a
| Lambda Var a
| App a a
newtype Exp = Exp (ExpF Exp)
newtype TExp = TExp (ExpF TExp, Type)
```
The Coq kernel uses this to define its expression type, although it doesn’t use it to define an untyped variant.
### (Lazy) Attribute grammars
I don’t claim to understand this approach too well, but essentially it is a programming model distinct from usual algebraic data types which associates attributes over nodes of a tree. In some sense, it can be thought as a memoized function from AST nodes to the attributes. Many compilers do utlize maps, but only for top-level declarations. [5]
### Closing remarks
There were a few things that I did not mention here which came up in the discussion. One participant suggested using polymorphic variants to define the data type; this doesn’t help much with adding extra information but allows for different ways of writing traversal functions. Indeed, traversal is one of the big concerns, and the mention of generic programming also is targeted at this problem.
As for my preference? It’s hard to say. I’ve worked with compilers mostly written in the “define a new IR style”, and while the initial outlay of defining two data structures is quite annoying, it is mostly a fixed cost. What’s yours?
Also, a question. Is there a presentation of the conventional set of annotations needed to get explicitly typed System F?
### 18 Responses to “The AST Typing Problem”
FWIW, Ur/Web actually uses “explicit typing,” and I think Coq does too, though I’m less sure on the latter.
2. Good catch, I think you’re right on both.
3. Gabriel says:
You seem to have forgotten to delete (or rewrite) the paragraph starting with “GHC – lots of IRs”. It looks like a bunch of scribbled notes.
4. Haha, thanks. “That’s what I get for writing blog posts at 4am in the morning.”
5. Pseudonym says:
The two-level type approach is under-used, I think. I’ve used it for making mutable structures (by recursing through an MVar or IOVar) and to implement hash consing. You could also do value numbering using that approach.
6. Pseudonym says:
Oh, having said that, a common scenario is that rich expressive IRs (where a close match between the source program and the IR must be maintained, say for reporting errors to the user) are typically compiled down to desugared lower-level IRs (where fewer operations is better). If you’re using multiple IRs anyway, there’s no point in making things too generic.
7. José Pedro Magalhães says:
Generic programming can make the “two-level types” approach easier to use, in particular allowing you to define only the original Exp and Type datatypes, and getting ExpF for free. For example:
8. exDM69 says:
I’ve encountered this problem too. And I’ve used yet another kind of solution for this.
I used a type declaration expression `(x :: Int)`, so I added another type of expression to the AST, `data Exp = TypeDecl Type Exp | Num Int | If Exp Exp Exp | ....` After the type checking, I rebuild the AST and add TypeDecl nodes to every branch.
I also tried not putting the type information in the AST in the first place, and passing the type and type environment along with the expression to the code generator tree walk and doing partial type checking as I go. This is crazy inefficient but it worked.
9. Pseudonym: Hmm, I wonder if this approach would work for types, so that types can be temporarily be replaced with IORefs for unification.
10. heatsink says:
The two-level type approach makes it awkward when your AST is several mutually recursive types, because each type in the recursive cycle becomes a type parameter. Suppose that you wanted a separate AST type for functions. Now everything is parameterized over two types.
data ExpF e f = … | Lambda f | App e e
data FunF e f = Fun Var e
newtype Exp = Exp (ExpF Exp Fun)
newtype Fun = Fun (FunF Exp Fun)
newtype ExpT = Exp (ExpF ExpT FunT, Type)
newtype FunT = Fun (FunF ExpT FunT, Type)
The two type parameters cannot vary independently, which makes me think there may be a single-parameter solution using type families.
11. Pseudonym says:
Edward, you’ve inspired me to try it out. I’ve been looking for something small but nontrivial to try with automatically extracting Haskell from Coq. This looks perfect.
12. Pseudonym says:
heatsink: That’s the sort of situation where you really want O’Caml modules. :-)
13. Jimmy Koppel says:
heatsink:
There is indeed a single-parameter solution. You’ve shown how to construct a two-sorted AST using a fixpoint for pairs. In this encoding, separate constructions would be needed for more sorts. However, pairs (*,*) are isomorphic to functions (2 -> *), and an n-tuple is isomorphic to (n -> *). If we approximate n with *, we can get away with the single, mono-kinded higher order fixpoint of kind ((* -> *) -> (* -> *)) -> (* -> *). We can thus rewrite this:
newtype Fix f i = f (Fix f l) i
data (:+:) f g e l = L (f e l) | R (g e l)
data ExpL
data FunL
data ExpF e i where
Lambda :: e FunL -> ExpF e ExpL
App :: e ExpL -> e ExpL -> ExpF e ExpL
data FunF e i where
Fun :: Var -> e ExpL -> FunF e FunL
type Sig = FunF :+: ExpF
type Term = Fix Sig
type Exp = Term ExpL
type Fun = Term FunL
type (:&:) f a e i = (f e i) :&: a
type TypeTerm = Fix (Sig :&: Type)
Here, Exp has the rather pleasing interpretation of labeled trees whose topmost label is “ExpL.”
This idea is laid out in more detail in “Generic programming with fixed points for mutually recursive datatypes” (ICFP ’09), and is used in the multirec and compdata libraries.
14. mpickering says:
Note that now with Pattern Synonyms (in haskell since 7.8) have made it much easier to maintain an interface whilst performing these changes than before.
15. Rich says:
I’ll give you another solution I tried, with the caveat that it’s *not* a good solution:
Keep a weak hashtable around which maps your original (untyped) tree nodes, by address, to your node types.
As I said, I tried this approach for an HTML/CSS parser I was writing, but I couldn’t get the implementation of the weak hashtable to work well. In particular for whatever reason my hashtable never freed any entries when the tree was freed.
16. […] Can you define your AST in a way that all semantic errors become type errors? This blog post The AST Typing Problem by Edward Z. Yang is a good introduction. I might write another post sometime later explaining a […]
17. Edward Kmett says:
Tying this together with some of your more recent work:
This encodes rather nicely if you adapt the “Trees that Grow” approach to Backpack!
18. Benedict Gaster says:
Edward Kmett said:
> This encodes rather nicely if you adapt the “Trees that Grow” approach to Backpack!
I’d been thinking about a solution to this problem for the last couple of days, found this blog via GHC comments and then found this paper via your comment, which was just the solution I was looking for. A couple of hours this morning refactoring… Thanks! | 2,449 | 10,223 | {"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} | 2.671875 | 3 | CC-MAIN-2021-25 | latest | en | 0.877759 |
http://www.perlmonks.org/index.pl?node_id=963260 | 1,503,036,599,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104565.76/warc/CC-MAIN-20170818043915-20170818063915-00374.warc.gz | 638,875,870 | 11,297 | Perl: the Markov chain saw PerlMonks
### Re^5: [OT] The statistics of hashing.
by BrowserUk (Pope)
on Apr 03, 2012 at 15:01 UTC ( #963260=note: print w/replies, xml ) Need Help??
in reply to Re^4: [OT] The statistics of hashing.
in thread [OT] The statistics of hashing.
Thanks. As may be (becoming) obvious, much of that is over my head for now :)
However, Now I know how to derive the constants, I have this which I can substitute for your ex_4() & ex_10() by supplying the power as the first argument. Its output matches those two exactly for all the argument combinations I've tried:
```sub PTn {
my @row;
for( 1 .. shift ) {
push @row, 1;
\$row [\$_] += \$row [\$_ - 1] for reverse 1 .. @row - 2;
}
return @row;
}
sub expN {
my( \$P, \$N, \$X ) = @_;
my \$xDIVn = \$X / \$N;
my @coefs = PTn( \$P+1 );
my \$rv = 0;
my \$toggle = -1;
for my \$p ( reverse 1 .. \$P ) {
\$rv += \$toggle * \$coefs[ \$p ] * \$N / \$p * exp( -\$p * \$xDIVn );
\$toggle *= -1;
}
return \$rv + \$X;
}
That allowed me to investigate the affect of using fewer or more hashes without having to hand code a new function for each.
And the results are quite interesting. This shows that each new (pair?) of hashes added does increase the discrimination substantially, though the gains obviously fall off fairly rapidly. But the really interesting part are the numbers for odd numbers of hashes:
```C:\test>rbtcs-form-verify -H=10 -B=32
Using N x 2**32 vectors:
12 : 24 0 24 0 24
+ 0 24 0 24 0
16 : 32 0 31 0 31
+ 0 31 0 31 0
24 : 48 0 48 0 48
+ 0 48 0 48 0
32 : 64 0 64 0 64
+ 0 64 0 64 0
48 : 95 0 95 0 95
+ 0 95 0 96 0
64 : 127 0 127 0 127
+ 0 128 0 128 0
96 : 191 0 191 0 192
+ 0 192 0 192 0
128 : 255 0 256 0 256
+ 0 256 0 256 0
192 : 383 0 383 0 383
+ 0 384 0 384 0
256 : 511 0 512 0 512
+ 0 512 0 512 0
384 : 767 0 768 0 768
+ 0 768 0 768 0
512 : 1023 0 1024 0 1024
+ 0 1024 0 1024 0
768 : 1535 0 1536 0 1536
+ 0 1536 0 1536 0
1024 : 2047 0 2048 0 2048
+ 0 2048 0 2048 0
1536 : 3071 0 3072 0 3072
+ 0 3072 0 3072 0
2048 : 4095 0 4096 0 4096
+ 0 4096 0 4096 0
3072 : 6143 0 6144 0 6144
+ 0 6144 0 6144 0
4096 : 8191 0 8192 0 8192
+ 0 8192 0 8192 0
6144 : 12287 0 12288 0 12288
+ 0 12288 0 12288 0
8192 : 16383 0 16384 0 16383
+ 0 16383 0 16384 0
12288 : 24575 0 24576 0 24576
+ 0 24576 0 24576 0
16384 : 32767 0 32767 0 32767
+ 0 32767 0 32768 0
24576 : 49151 0 49151 0 49152
+ 0 49152 0 49152 0
32768 : 65535 0 65536 0 65536
+ 0 65536 0 65536 0
49152 : 98303 0 98304 0 98304
+ 0 98303 0 98303 0
65536 : 131071 0 131072 0 131071
+ 0 131072 0 131072 0
98304 : 196606 0 196608 0 196607
+ 0 196608 0 196608 0
131072 : 262142 0 262144 0 262144
+ 0 262143 0 262144 0
196608 : 393211 0 393216 0 393216
+ 0 393215 0 393216 0
262144 : 524280 0 524288 0 524287
+ 0 524287 0 524288 0
393216 : 786414 0 786431 0 786431
+ 0 786432 0 786432 0
524288 : 1048544 0 1048575 0 1048576
+ 0 1048576 0 1048576 0
786432 : 1572792 0 1572863 0 1572864
+ 0 1572864 0 1572864 0
1048576 : 2097024 0 2097151 0 2097151
+ 0 2097152 0 2097152 0
1572864 : 3145440 0 3145727 0 3145728
+ 0 3145727 0 3145728 0
2097152 : 4193792 0 4194303 0 4194304
+ 0 4194304 0 4194304 0
3145728 : 6290304 0 6291455 0 6291455
+ 0 6291455 0 6291455 0
4194304 : 8386560 1 8388607 0 8388608
+ 0 8388608 0 8388608 0
6291456 : 12578306 4 12582911 0 12582912
+ 0 12582912 0 12582912 0
8388608 : 16769029 10 16777215 0 16777216
+ 0 16777215 0 16777216 0
12582912 : 25147409 35 25165823 0 25165823
+ 0 25165824 0 25165824 0
16777216 : 33521706 85 33554431 0 33554431
+ 0 33554431 0 33554432 0
25165824 : 50258063 286 50331646 0 50331647
+ 0 50331648 0 50331648 0
33554432 : 66978132 678 67108860 0 67108863
+ 0 67108864 0 67108864 0
50331648 : 100369532 2283 100663276 0 100663295
+ 0 100663296 0 100663296 0
67108864 : 133696160 5397 134217665 0 134217727
+ 0 134217728 0 134217728 0
100663296 : 200156106 18111 201326276 5 201326591
+ 0 201326591 0 201326592 0
134217728 : 266359979 42681 268434469 24 268435455
+ 0 268435455 0 268435455 0
201326592 : 398007464 142383 402648282 179 402653177
+ 0 402653183 0 402653184 0
268435456 : 528654369 333608 536855705 738 536870874
+ 1 536870911 0 536870911 0
402653184 : 787008255 1100214 805232175 5328 805305969
+ 30 805306365 0 805306367 0
536870912 : 1041542872 2548692 1073515796 21337 1073739728
+ 211 1073741802 2 1073741823 0
805306368 : 1539620713 8218836 1609548824 146448 1610591774
+ 3082 1610612273 70 1610612725 1
1073741824 : 2023785226 18623993 2144354575 558473 2147380065
+ 19731 2147479815 755 2147483497 30
1610612736 : 2953695056 57532294 3207472286 3485537 3220308576
+ 247526 3221157361 19018 3221220099 1532
2147483648 : 3837421596 125076314 4257101987 12129165 4290939237
+ 1371778 4294491373 167491 4294907677 21417
3221225472 : 5487393872 357203949 6295545197 63685472 6413893311
+13112112 6436324179 2901774 6441061714 670967
4294967296 : 7009904423 721946381 8229596479 188903097 8487725572
+56541428 8558136669 18112153 8579512265 6047518
I suspect it is a coding error on my behalf, but I guess it could be a quirk of the numbers?
Here's a set using 1 .. 16 2**16 bit hashes:
```C:\test>rbtcs-form-verify -H=16 -B=16
Using N x 2**16 vectors:
12 : 23 0 23 0 24 0 24 0
+ 24 0 23 0 23 0 23 0
16 : 31 0 31 0 31 0 32 0
+ 32 0 32 0 32 0 32 0
24 : 47 0 47 0 48 0 47 0
+ 48 0 47 0 47 0 47 0
32 : 63 0 63 0 64 0 64 0
+ 64 0 64 0 64 0 64 0
48 : 95 0 95 0 95 0 95 0
+ 95 0 96 0 96 0 96 0
64 : 127 0 127 0 128 0 128 0
+ 128 0 127 0 128 0 127 0
96 : 191 0 191 0 192 0 192 0
+ 192 0 192 0 191 0 191 0
128 : 255 0 255 0 256 0 255 0
+ 256 0 256 0 256 0 256 0
192 : 383 0 383 0 383 0 384 0
+ 384 0 384 0 384 0 384 0
256 : 511 0 511 0 511 0 511 0
+ 512 0 511 0 512 0 512 0
384 : 766 0 767 0 767 0 768 0
+ 768 0 767 0 768 0 767 0
512 : 1022 0 1023 0 1023 0 1024 0
+ 1024 0 1024 0 1024 0 1024 0
768 : 1531 0 1535 0 1535 0 1536 0
+ 1536 0 1536 0 1536 0 1536 0
1024 : 2040 0 2047 0 2047 0 2048 0
+ 2048 0 2048 0 2048 0 2048 0
1536 : 3054 0 3071 0 3071 0 3071 0
+ 3072 0 3072 0 3072 0 3072 0
2048 : 4064 0 4095 0 4095 0 4095 0
+ 4095 0 4096 0 4096 0 4096 0
3072 : 6073 2 6143 0 6143 0 6143 0
+ 6144 0 6144 0 6144 0 6144 0
4096 : 8066 5 8191 0 8191 0 8191 0
+ 8191 0 8191 0 8191 0 8192 0
6144 : 12008 16 12286 0 12287 0 12287 0
+12287 0 12287 0 12288 0 12287 0
8192 : 15892 38 16380 0 16383 0 16383 0
+16383 0 16383 0 16383 0 16383 0
12288 : 23492 125 24559 2 24575 0 24575 0
+24575 0 24575 0 24575 0 24576 0
16384 : 30880 284 32720 8 32766 0 32767 0
+32767 0 32767 0 32767 0 32767 0
24576 : 45069 877 48942 53 49138 3 49150 0
+49151 0 49151 0 49151 0 49151 0
32768 : 58554 1908 64958 185 65474 20 65528 2
+65535 0 65535 0 65535 0 65535 0
49152 : 83730 5450 96062 971 97868 200 98210 44
+98282 10 98299 2 98302 0 98303 0
65536 : 106962 11016 125573 2882 129512 862 130586 276 1
+30912 92 131018 31 131053 11 131065 3
Sorry for the wrapping.
With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.
The start of some sanity?
Replies are listed 'Best First'.
Re^6: [OT] The statistics of hashing. (odd)
by tye (Sage) on Apr 03, 2012 at 15:43 UTC
I suspect you need to start \$toggle reversed for odd values (or for even values, though you note that you verified against the two even cases). Though, I would have expected that particular mistake to lead to negative values (but you don't show most of the code so I didn't go much further in looking). The numbers are supposed to be f(...)**\$P where f() is positive (and monotonic), so, yes, odd values of \$P should give you results between the results for \$P-1 and \$P+1, not like the results you posted.
Sorry for the wrapping.
You might want to put the big tables into READMORE tags so that they don't impact the rendering of the whole thread (though that should only happen for people who have code wrapping badly configured).
- tye
I suspect you need to start \$toggle reversed for odd values
Indeed. Using my \$toggle = \$P & 1 ? 1 : -1;, the output makes much more sense:
you don't show most of the code so I didn't go much further in looking
The entire code looks like this:
```#! perl -slw
use strict;
sub PTn {
my @row;
for( 1 .. shift ) {
push @row, 1;
\$row [\$_] += \$row [\$_ - 1] for reverse 1 .. @row - 2;
}
return @row;
}
sub expN {
my( \$P, \$N, \$X ) = @_;
my \$xDIVn = \$X / \$N;
my @coefs = PTn( \$P+1 );
my \$rv = 0;
my \$toggle = \$P & 1 ? 1 : -1;
for my \$p ( reverse 1 .. \$P ) {
\$rv += \$toggle * \$coefs[ \$p ] * \$N / \$p * exp( -\$p * \$xDIVn );
\$toggle *= -1;
}
return \$rv + \$X;
}
our \$H //= 10;
our \$B //= 32;
print "Using N x 2**\$B vectors:";
for my \$inserts ( map{ 2**\$_*3/4, 2**\$_ } 4 .. \$B ) {
printf "%10d : ", \$inserts;
for my \$h ( 1 .. \$H ) {
printf "%6d ", expN( \$h, 2**\$B, \$inserts ) - expN( \$h, 2**\$B,
+0 );
}
print '';
}
With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.
The start of some sanity?
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Measure the two sections of the diagonal and compare their lengths to the diagonal's total length.
Is this what you expected?
Create or draw some more $10$cm squares and repeat the process.
Do you always end up with the same answer?
What fractions does the second fold appear to divide the diagonal into?
Does this appear to be the case for squares of different sizes?
Can you produce a convincing mathematical argument or proof that justifies what you have found?
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The complex tax dispute between the Covered Bridge Mall and Harris Township is not likely to be adjudicated for several years, and, in the meantime, [u]both sides are intent on creating difficulties for the other.[/u](A) both sides are intent on creating difficulties for the other
(B) both sides are intent on creating difficulties for each other
(C) each side is intent on creating difficulties for the other
(D) each side is intent on creating difficulties for one another
(E) the sides are both intent on creating difficulties for each other
The complex tax dispute between the Covered Bridge Mall and [#permalink] 11 Apr 2007, 05:35
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Display posts from previous: Sort by | 553 | 2,233 | {"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} | 2.6875 | 3 | CC-MAIN-2016-36 | longest | en | 0.928674 |
https://economictimes.indiatimes.com/blogs/policypundit/survey-s-growth-projections-are-contestable/ | 1,680,064,310,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00245.warc.gz | 269,791,927 | 27,665 | Economic Survey rarely has anything to do what reforms a government introduces. What Survey proposes, government disposes. Nevertheless, we retain fond hopes and continue to read Survey. Survey 2011-12 is no different. Though I have used the word Survey, most people read and discuss the first two chapters and I am also going to focus on these. Take Chapter 1 first. And there is a confession there.
Last year, Survey had projected GDP growth of 9% in 2011-12, centred around a band of 0.25% on either side. The confession is that we will end the year with 6.9%. So, apart from blaming the rest of the world, one would have expected some kind of post-mortem to explain what went wrong with that projection and why we should believe Survey on forecasts for 2012-13.
Here is Chapter 1 on these forecasts. “Calculations based on tracking several statistical indicators and projections of incremental capital-output ratios lead to a forecast of the growth rate of real GDP for 2012- 13 to be 7.6 (+/-0.25) per cent.”
Rare is the person who expects growth to be that high. RBI’s collation of independent forecasts has a figure of 7.3%. Rare is the person who expects more than 7%. The difference between 7% and 7.6% may seem to be minor, but there is a crucial difference.
There are those who believe we are stuck in a band between 6.5 and 7%, probably closer to 6.5%. And there are those who believe we are getting out of that band. Survey belongs to the latter category and hence, we not only have 7.6% in 2012-13, but 8.6% in 2013-14. That’s the reason the difference is crucial. In that event, wouldn’t we like to know what these calculations based on several statistical indicators are? Wouldn’t we like to know what the incremental capital-output ratios are? However, on this, Survey is very reticent, suggesting there is no sanctity to the modeling at all.
Chapter 1 also adds, “These projections are based on assumptions regarding factors like normal monsoons, reasonably stable international prices, particularly oil prices, and global growth somewhere between where it now stands and 0.5 per cent higher.” Fair enough. But we still don’t get an answer as to why we are likely to inch up to almost 9%.
It gets more curious with Chapter 2. We have the standard propositions about monetary policy tightening and fiscal deficits hurting growth. Is Survey now confident there will be fiscal consolidation and expenditure-cutting?
More interesting is this quote from paragraph 2.19. “A third factor, even though it is hard to quantify and for that reason is contestable, has to do with some slackening in the pace of reforms.
One consequence of increased awareness of high-profile corruption scandals in different parts of India and welcome civil-society activism has been a sense of caution among civil servants in taking the final decision. Since one way to avoid the charge of an ill-considered or, worse, ill-intentioned decision is to take no decision, it is arguable that some civil servants in this season of charges and countercharges have resorted to precisely this strategy. This would cause a slowdown in decision making.
In addition, coalition politics and federal considerations played their role in holding up economic reforms on several fronts, ranging from diesel and LPG pricing and taxation reform like the goods and services tax (GST) and direct taxes code (DTC), to FDI in retail and reform of the APMC Act.” Precisely, and add to that list forest and environmental clearances and problems with land acquisition and mining. Just because something is difficult to quantify doesn’t make it contestable. It is these, rather than monetary policy tightening, that have hurt investments.There is no reason why this should change.
At least, Survey doesn’t tell us if these have been “modeled” in. And that’s also the reason why Survey’s growth projections are not believable and contestable. As earlier, Survey will again be proved wrong.
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Views expressed above are the author's own. | 966 | 4,204 | {"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} | 2.53125 | 3 | CC-MAIN-2023-14 | latest | en | 0.954524 |
https://edgeinducedcohesion.blog/2020/08/08/book-review-experiencing-geometry/ | 1,624,345,673,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488512243.88/warc/CC-MAIN-20210622063335-20210622093335-00319.warc.gz | 215,832,299 | 28,844 | ## Book Review: Experiencing Geometry
Experiencing Geometry In Euclidian, Spherical, And Hyperbolic Spaces, by David W. Henderson
The increasing popularity of flat earth thinking in certain circles has led me to ponder about the deficiency of my own education in geometry in spherical spaces, and seeking a better way to understand this sort of issue and how it is that a failure to know spherical geometry can lead to negative consequences in how one sees the world, I thought it would be worthwhile to bone up on some geometry, since the last time I did coursework on it was in high school and that was focused on plane geometry. This book is written for students in Cornell, and the author clearly is dealing with an intended audience that is both intensely practical when it comes to modeling more complicated geometries but also familiar with the basics of geometry. This is the sort of book that I can well recommend to someone who wants to be more familiar with advanced geometry, at least advanced undergraduate-level geometry, so as to be able to develop a better informed intuition on geometry based on reading and modeling. Not everyone is going to find a book like this interesting, of course, but at the same time those who do will get a lot out of it.
This book is more than 300 pages long and has twenty-two chapters. The book begins with a preface and some suggestions on how to read the book depending on the interests of the reader as well as a discussion on the nature of proofs. After that the author discusses the meaning of straight (1), before looking at what it means in spherical geometry (2), as well as cylinders and cones (4), and hyperbolic planes (5). The author also examines the question of what an angle is (3), discussing after that triangles and congruencies (6), and problems on SSS, ASS, SAA, and AAA (9). The author also discusses area and holonomy (7), parallel transport (8), and some postulates that relate to parallel lines (10). He shows himself interested in isometries and patterns (11), dissection theory (12), and the geometry of square roots, Pythagoras, and similar triangles (13). There are chapters on circles in the plane (14), projections of a sphere onto a plane, useful for maps (15), projections of hyperbolic planes (16), geometric 2-manifolds and coverings (17), geometric solutions of quadratic and cubic equations (18), trigonometry and duality (19), 3 spheres and hyperbolic 3-spaces (20), polyhedra (21), 3-manifolds (22), as well as appendices on Euclid’s definitions, postulates, and common notions (i) and square roots in the Sulbasutram (ii), an annotated bibliography, and an index.
There are many insights that this book provides, but space and time constraints prevent me from mentioning more than a few of them, so here goes. The author’s focus is on informing readers (and his students) by developing an intuitive grasp of geometry and its importance in algebra and trigonometry, for example, rather than focusing on the formalist approach that is most common in schools. Among the useful tips that the author encourages is recognizing that when it comes to spherical geometry we are surface dwellers whose intrinsic experience of being on the surface of an oblong spheroid (which we call earth) is different than the extrinsic knowledge we have about spheres from a formal perspective. The author also suggests better understanding hyperbolic space through crocheting a surface and then working with it to model the space and see its quirks, which is an inventive solution for mathematical matters. In reading this book I was struck by the melancholy thought that an approach to algebra and trigonometry that utilized geometry and that defended the essential value of standard measurements (as opposed to metric ones) for their geometric value would have led my father to have been far better at math because of his inability to deal with the abstractions of higher math. Alas, math is full of melancholy thinking in such ways. | 855 | 4,017 | {"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} | 2.703125 | 3 | CC-MAIN-2021-25 | latest | en | 0.96022 |
http://slidegur.com/doc/27405/slides | 1,477,677,556,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988725451.13/warc/CC-MAIN-20161020183845-00384-ip-10-171-6-4.ec2.internal.warc.gz | 228,725,057 | 15,239 | ### Slides
```The Zone Theorem
The Cutting Lemma Revisited
Tom Jurgenson
1
The Zone Theorem
2
Definitions reminders
Is a sub-space of d-1 dimensions.
Is a partition of
into relatively open convex sets.
Are 0/1/(d-1)-dimension faces (respectively) in
Are the connected components of
3
.
also called d-faces
Example d=2 (Plane)
Vertices are points, edges are segments, and facets are also
segments since 2-1=1.
Cells are 2-dimesional connected components in
The blue lines are the Hyperplanes in H
Their arrangement creates:
Vertices – in red
Edges\Facets – blue segments(excluding red)
Cells – light blue areas.
4
The Zone Theorem - Motivation
Motivation – we would like to bound the number of faces
that can see any part of a given hyperplane. For that we’ll
need to define what ‘see’ means, and define the set of objects
(faces) that ‘see’ g (the zone).
5
Definition for ‘sees’
A face F can see hyperplane g of H (set of hyperplanes), if
there are points
such that the open segment
xy does not intersect any hyperplane of H.
Important Note: It does not matter which point x we
choose, either all of them can see g or none can.
In this example, the red line represents a segment between
the thick line – g, and the face – the cell. The blue line
exemplifies the fact that no segment connects that cell with
the hyperplane g.
6
Zone Definition
Zone – The zone of hyperplane g, is the set of the faces of the
arrangement of H that can see g.
Here is the zone (dark grey area) for hyperplane g (the thick line)
and hyperplane set H (all the thin lines). Each face (vertex, edge
and cell) in the zone of g can see g.
7
The Zone Theorem
The number of faces in the zone of any hyperplane in an
arrangement of n hyperplanes in
is
with the
constant of proportionality depending on d.
we will prove this theorem in an inductive manner
8
Base case, d=2 (The plane)
Let H be a set of n lines in the plane in general position. We
consider the zone of line g.
In this case the faces would be: vertices (0-faces), edges (1faces), and the cells are convex polygons that may or may not
be bounded(1-faces).
The bound for the number of faces in the zone in this case
should be
9
Bound for Edges is enough
The number of vertices is proportional to the number of
edges since in a convex polygon the number of vertices is
equal to the number of edges. And in open cells, the number
of edges is at most twice the number of vertices.
It is clear that the number of polygons is bounded by the
number of vertices by using Euler’s formula: F=v-e+2.
Therefore proportional to the number of edges.
10
Bound for Edges
Let g be a horizontal line.
We’ll count the number of edges that see g that are also
above g.
The number of edges that intersect g that are also above g is
bounded by n. Since each line (hyperplane in H) out of the n
lines can only intersect g in one point.
The other edges are disjoint from g, and in the next slides we
are going to find a bound for these edges.
11
Bound for the disjoint edges
Let uv be an edge that sees g but is disjoint from it.
Let
be the line containing uv. And let a be the
intersection of g and h.
Since u is a vertex - it is the intersection of two lines – so let
be the other line that passes through u. Let b be the
intersection of l and g.
12
Right Edge of Line l
Definition: uv is right edge of l if b is on the right of a. uv is
left edge of l if b is on the left of a.
We’ll show that for each such
there is at most one
right edge.
Let’s assume uv is a right edge for l.
13
Only one right edge
Assume by contradiction that there are two right edges for l.
Let uv and xy be the two right edges, and assume that uv lies
below xy.
By definition xy should see some point of g.
However, every point to the left of a on g is obscured by the
line l. And every point to the right of a on g is obscured by
the line h.
This contradiction shows that only one right edge can exist
for each line.
14
Only one right edge (2)
15
Conclusion for d=2
Each line in H has at most one right edge, and by symmetry
also at most one left edge. There is also an edge that
intersects g. The same can be argued for edges below g, for
the lower right and lower left edges, and an edge that
intersects. Altogether there are at most 6 edges for each
line in H in the zone.
There are n lines in H so the number of edges is at most 6n.
Linear in n.
We already argued that the total number of faces is linear in
the number of edges therefore the number of faces in the
zone when d=2 is:
as needed.
16
Inductive Step from d-1 to d
Assume that the total number of faces of a zone in
is
Mark
as the max. number of (d-1) faces in the zone in
an arrangement of n hyperplanes in
.
Let H be an arrangement of hyperplanes and g a base
hyperplane s.t
is attained for them.
17
Motivation: number of blue facets
We color one hyperplane in H red, let it be
18
. The rest
n-1 hyperplanes in H we color blue.
A Blue Facet is a facet that lies in a blue hyperplane.
Since every facet has a chance of
becoming blue
(choose any blue hyperplane out of n blue hyperplanes), the
expected number of blue facets is
Now we bound the expected number of blue facets in a
different way and use it to estimate
Example in the next slide…
Blue and Red Facets
19
Bound for Expected Blue Facets
Consider the arrangement of blue hyperplanes only. Since
there are n-1 such hyperplanes the number of blue facets is
bounded by
In the next stage we add the red hyperplane h and see how it
affects the number of blue facets in the zone of g.
20
The increase in number of blue facets
caused by h
The red hyperplane h may split an existing blue facet F into
two new facets – F1 and F2. If both new facets can see g then
we get an increase in the number of facets.
Claim:
is visible from
(we’ll prove it later)
If we intersect all blue hyperplanes with g and with h we get
a d-1 dimensional arrangement in which
is a facet in
a zone of the d-2 dimensional hyperplane
. Using the
induction assumption we get
such “splits”. The
total number of facets becomes:
21
Split Example
22
Proof for the above claim
Reminder: We wanted to show the following claim:
is visible from
Proof: Let C be a cell of the zone in the arrangement of blue
hyperplanes having F on the boundary.
Because F1 and F2 see g we get that
sees
and
sees
The interior of
is contained in C and we get that
the intersection of
with the hyperplane h
contains a segment witnessing the visibility of
from
23
Split Example (II)
24
Conclusion for facets
On the one hand we calculated the expected number of blue
facets:
On the other hand we found an upper bound for the number
of blue facets:
We combine the two and get:
25
Conclusion for facets (II)
And as we can see:
as needed.
Next we are going to discuss d-k faces and not just d-1 faces (or
facets)
26
Expected number of d-k faces
Denote
27
the max. possible number of j-faces in the
zone for an arrangement of n hyperplanes of dimension d.
Let H be an arrangement of n hyperplanes where
is attained.
Again a random hyperplane h in H is colored red and the rest
blue.
Reminder: a d-k face is the intersection of k hyperplanes.
A d-k face is blue if its relative interior is disjoint from the
red hyperplane h.
Expected number of d-k faces (2)
The probability for a d-k face to be blue is
:
[For the selection of first blue hyperplane, then the second on
so on until we select the k-th hyperplane]
As in the facet case we get that the expected number of blue
d-k faces in the required zone is:
28
Upper bound for d-k faces
By adding the red hyperplane h the number of blue d-k faces
can increase by at most
by the inductive
hypothesis.
Using the same arguments in lower dimensions as the facets
case we get to the conclusion that:
And in the next slide we’ll show:
29
Algebraic proof
30
Bound for edges and vertices
For the case k=d-1 (edges) by using this method we only get
the bound:
So the number of edges and vertices must be bound
separately:
Vertices: Each vertex is contained in some3-face of the zone.
Within such 3-face, the number of vertices is at most 3 times
the number of 2-faces, because the 3-face is a 3D convex
polyhedron. Since H is simple each 2-face is contained in a
bounded number of 3-face. It follows that the total number of
vertices is at most proportional to
31
Final details
Edges are proportional to vertices.
In conclusion, for every k=1…d-2 we get
For vertices and edges we get
In conclusion we get that the zone has
32
Zones in other arrangements
The max. complexity of a zone can be investigated for
objects other than hyperplanes, and that leads to many
related problems that we won’t discuss here.
33
Cutting Lemma Revisited
34
General Plan
First, we’ll prove the cutting lemma for the planar case with
a tight bound. We already saw a proof for a bound that was
not tight that uses random sampling, and a different proof
that is tight but does not use random sampling.
The following proof is both tight and uses random
sampling in a way that makes it easy to be generalized
to higher dimensions.
But first we’ll do a review for some cutting-lemma related
definitions.
35
Review of Definitions (I)
A generalized triangle is the intersection of 3 half planes.
A cutting of the plane into generalized triangles, is the
subdivision of the plane to a disjoint set generalized triangles
that their union covers the entire plane.
36
Review of Definitions (II)
A
cutting of the plane with a set of n of lines H – is a
cutting of the plane into generalized triangles
s.t the interior of each such
is intersected
by at most
lines from H.
Note – the triangles’ edges may or may not include lines
from H.
37
½ cutting example
In the next example we have a ½ cutting. The red lines are
lines in H, the black lines are the lines that create the
triangulation.
38
The Cutting Lemma
For every set H of n lines in the plane and every r>1 there
exists a 1/r cutting for H the size
In other words, there is a subdivision of the plane into
generalized triangles s.t the interior of each triangle is
intersected by at most n/r lines of H.
39
The sampler
Goal – sample line. Once the lines are sampled their
intersection (possibly with some modifications) would result
in a triangulation.
In the previous lecture when we wanted to select a subgroup
S of H, we picked |S| lines out of H with repetitions. This
procedure may result with a subgroup of size smaller than
|S|.
In order to simplify the calculation, we’ll choose S by
independent Bernoulli trials. We fix the probability of p=s/n,
and include a line from H in S using that probability.
For this point on, let s=|S|
40
Why sampling and triangulation does
not work?
In a previous lecture we already saw a sampling algorithm
that outputs a triangulation that with probability close to 1
non of the triangles in the outputted triangulation is
intersected by more than
where is some
constant (Weak Cutting Lemma)
We will demonstrate that a similar statement with a
is not generally true. So our method would have to include
more than just sampling in order to achieve the desired
triangulation.
To see that we’ll examine a simpler case than the plane – 1dimensional situation.
41
The 1-D case
H is a set of n points.
A generalized triangle in this case is the segment between two
points.
We would like to count the longest consecutive count of
unselected points of H. Let k be our target number for such a
count.
In example below, n=30 and s=15 so p=0.5. The black dots
are selected points (In this example there are exactly 15
selected dots, but it doesn’t have to be this way)
42
k consecutive unselected points
We’ll show that for a fixed k it is very likely that k
consecutive unselected points show up in a sequence of n
points, where n is sufficiently large
For simplicity assume n is divisible by k and divide our n
points in to blocks the length of k. Notice that this is an even
more restrictive case than a general consecutive count.
In each such a block there is a probability of
for the entire block to be unselected. Therefore the
probability of not obtaining any block as entirely unselected
is
and this probability is
exponentially small for k proportional to
43
Why is it exponentially small?
p is a constant, let p=0.5. The probability becomes:
Now set k=0.5logn:
And that exponentially small as n goes to infinity.
44
Conclusion of 1D
A sequence of k that is proportional to log of n is very likely
to appear using this sampling method. This is why this
method is not good enough to get the 1/r cutting we require.
Of course, in the 1D case we can define a sampler that selects
every n/s point. However it is not clear how to transform
this selection to the plane (Where the lines are not ordered).
45
2nd attempt – Two level decomposition
General plan: Instead of trying to improve the sampling
algorithm to generate a 1/r cutting from the start:
1. we’ll take a random sample (obtained as before with
probability p=s/n independent Bernoulli trials)
2. We just saw that this is not good enough, so we’ll take
every triangle that intersects more lines than required and
divide it so each new triangle will be good for our 1/r
cutting.
Finally, we’ll show that the total number of triangles is
46
But first, more definitions
Let T be the collection of triangles after triangulating the
random sample.
Let
denote the set of lines of H that intersect the
triangle
and
number of lines that
intersect.
Let the excess of
be
Note: if
so the triangle
can be included in the 1/r cutting. If
we subdivide it
to a collection of finer triangles.
47
Subdivision of triangles with excess
greater than 1
We consider the arrangement of
we construct a
cutting for it. (How? We’ll see later...)
We intersect this triangulation with . This process may
produce triangles, but also quadrilaterals, pentagons, and
hexagons. Each of these convex polygons is subdivided into
triangles.
Each triangle in the
cutting is intersected by at most
triangles, and therefore are valid for the 1/r cutting.
48
Example:
Up – the process of subdividing a triangle with excess
greater than 1.
Left – An example for a hexagon created by combining
the two triangulations.
49
But How? The suboptimal cutting
lemma.
As of now, we can’t subdivide
because the cutting lemma is
not valid yet. So let’s use a weaker claim:
The Suboptimal Cutting Lemma
for every finite collection of lines and any u>1, there exists a
1/u cutting consisting of at most:
triangles, where K is a suitable constant.
We’ll prove the suboptimal cutting lemma later.
50
Using the suboptimal cutting lemma
If we use the suboptimal cutting lemma for producing the
cuttings, we can estimate that the number of triangles
in the 1/r cutting is bounded by:
Note that we get one if the excess is less or equal to 1. If the
excess is greater than 1 we get the bound of the yet unproven
suboptimal cutting lemma times 4.
The times 4 is for the triangulation of a
possible hexagon as shown here:
51
Using the suboptimal cutting lemma (II)
We have two goals now:
1. Prove the suboptimal cutting lemma.
2. Show that although we typically have triangles
as large as about log(r) the expected number of triangles in
T with excess t or larger decreases exponentially as a
function of t.
52
We’ll prove the first claim by demonstrating a triangulation
method that achieves that bound – the vertical
decomposition (next slide). The second claim would be
included in the proof of the Cutting Lemma.
Vertical decomposition - reminder
We erect vertical segments upwards and downwards from
each vertex in the arrangement of S and extends them until
they meet another line (or all the way to infinity)
We get trapezoids, but those can be split into 2 triangles.
Let Trpz(S) denote the set of generalized trapezoids in the
vertical decomposition of S.
53
Proposition:
Trapezoids with large excess are rare
Proposition: Let H be a fixed set of n lines in general
position, let p=r/n, where
, let S be a
random sample drawn from H by independent Bernoulli
trials with success probability of p, and let
be a real
parameter. Let
denote the set of trapezoids in
with excess at least t
Then the expected number of trapezoids in
is bounded as follows:
For a suitable constant C.
54
Notes for this proposition
If r is not smaller than n/2 than we can easily find a cutting,
as we saw in the cutting lemma lecture it’s only gets
interesting when r is proportional to logn.
Reminder: How to construct such a cutting? If r is
proportional to n one can select all lines in the H, as shown
before this creates the desired subdivision.
We use this limitation to find a finer cutting only because this
is needed in the proof of this proposition (but we’ll get there
much later)
55
Proof of the suboptimal cutting lemma
Set
for a sufficiently large constant
A.
Assume that the trapezoid proposition holds – and choose a
sample S accordingly.
Activate the proposition twice – once for
and once
for
56
First case
For
by using the proposition:
for some constant B.
Also note that
Finally:
57
Second case
For
by using the proposition:
for some constant C. We’ll group the constants together:
Since:
there exists A large enough s.t
58
Proof of the Suboptimal Cutting
Lemma – Conclusion
From these two cases and the linearity of expectation we get:
There exist a sample were both parts of the sum are smaller
than 2/3. Notice that the second part is at least 1 and
therefore 0.
So there exists a sample S with both
and
59
Proof of the Suboptimal Cutting
Lemma – Conclusion (II)
That means that there exists a 1/u cutting into
triangles.
Suboptimal Cutting Lemma proof is complete.
60
Proof of the Cutting Lemma (I)
Reminder: To produce a 1/r cutting we pick a sample S with
probability p=r/n as described before. We let T be the
collection of triangles of the vertical decomposition of S
(actually – trapezoids, but that has no effect). For every
triangle t in T with excess greater than 1 we refine the
triangulation with a
cutting. We’ll estimate the
expected number of triangles:
61
Proof of the Cutting Lemma (II)
As
and
Since each part of the sum is at least 1. Also group by
order
62
‘s
Proof of the Cutting Lemma (III)
Notice that: if
then
so
Also, the summing over triangles with
is less then summing over
63
Proof of the Cutting Lemma (IV)
Using the suboptimal cutting lemma and the trapezoid proposition
for some M large enough
The ratio test shows that the series converges to some
number. So we get that the total number of triangles is
as needed.
64
Just one more
Now we just need to prove the trapezoid proposition, for
that we’ll need some more definitions:
is the set of all trapezoids that can ever appear in the vertical
decomposition for some subset S of H (including the empty
set).
is the set of lines of H incident to at least one vertex
of . Notice that
(all the possible cases, up to
symmetry are shown in the next slide)
65
D cases
The following shows that
66
Some Properties of Vertical
Decomposition
C0: As we saw:
. Moreover, any subset S of H is
the defining set for at most a constant number of trapezoids
in Reg.
C1: For any
we have
(the
defining set must be present) and
(no
intersecting lines may be present).
C2: For any
and any
s.t
and
we have
C3: For every
we have
67
C3: Think of the iterative process of adding one vertical line
at a time. Each splits an existing region in two. Since the lines
are in general position the number of intersections is about
C2 is the most interesting - it means that the vertical
decomposition is defined “locally”:
is present in
whenever it is not
excluded for simple local reasons.
68
Proof of the Trapezoid Proposition
First case
:
From C3 we get
is the sum of independent random variables.
and so we get:
69
where |S|
First case continued:
For now we have:
for
some constant M independent of t.
On the other hand
by
combining these two and setting C=2M we get:
as required.
This concludes the case of
70
Second case
For now we assume
. Let
be a random sample
as defined before with probability p. From C1 and C2 we get:
(a trapezoid appears iff all
his defining lines are selected into S, and no line that
intersects it are selected into S).
Let the set of all possible trapezoids with excess at least t:
The expected number of trapezoids can be written as:
71
Second case continued:
We now define p’=p/t and since t>1 we get: p’<p
Let S’ be an a sample drawn of H with probability p’
From what we saw earlier:
Also:
And finally:
The last transition holds from limiting the sum.
72
Second case continued (II):
Now be expanding with p:
Where R is defined:
As we’ll see in the next slide R is bounded by:
Finally we derive:
for a sufficiently large constant C (and than choosing the max
between it and the first case) the proposition is proved!
73
Lower bound for R
And since
For every real p:
Since p’<p<0.5:
74
and
we get:
Lower bound for R (II)
By the last slide we see that:
Since R is defined:
it will always be
75
Cutting Lemma conclusion
Since the trapezoid decomposition lemma is proved the
entire proof is done.
As mentioned before this Lemma can be generalized to
higher dimensions, and the proof we have just presented can
also be generalized to prove it (but it won’t be covered now).
Cutting Lemma for Higher Dimensions: Let d>0 be a fixed
integer, let H be a set of n hyperplanes in
and let r be a
parameter
.Then there exists a 1/r cutting for H the
size of
generalized simplexes s.t the interior of each simplex
is intersected by at most n/r hyperplanes of H.
76
``` | 5,775 | 21,747 | {"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} | 4.6875 | 5 | CC-MAIN-2016-44 | latest | en | 0.884465 |
https://www.fembestpractices.com/2020/10/design-analysis-of-tower-bolts-by-kotur.html?showComment=1641359010787 | 1,659,913,487,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570730.59/warc/CC-MAIN-20220807211157-20220808001157-00457.warc.gz | 676,233,560 | 34,942 | ### Design Analysis of Tower Bolts by Kotur Raghavan
In this article we will look into design requirements of tower bolts which we find as integral part of the doors of residential buildings. Design of exterior doors will be assessed from security point of view. Typical tower bolt arrangements are shown in the figure below.
Fig. 1
The main part of the bolt is fastened to the door through set screws. When the door is to be secured the bolt will engage into the socket which is fastened to the frame through set screws. When all the bolts are in engaged position the door and the frame will be nearly integral. The security concern is that a burglar may use force from outside and break open the door.
When any pushing force is applied to the door from outside, the load gets transferred to the frame partly at hinge locations and partly at socket locations. The latter force will have tendency to pull the sockets from the frame. In the entire assembly the set screw connections between the sockets and the frame happen to be the weakest points. For a given value of the applied force, the force coming on to the sockets can be considered to be ‘figure of merit’. Lesser the better. Rest of this article will focus on development of analysis model for computation of forces coming on to the sockets.
Structural Aspects and Simulation Model
There are two main structural parts. The assembly of bolts and the door is one. The other is the assembly of frame and the sockets. When the load is applied from outside, it is transferred from the door assembly to the frame assembly at hinges and bolt locations. At the hinge locations, the load transfer is through rigid connections. At the bolt locations, the load transfer is through contacts. At this point, we take a look at the purpose of analysis. The main results of importance are the forces transferred from the socket to the frame. As the sockets and the frame are in series, these forces will be equal to the reactions at set screw locations if the sockets are constrained. It is even more so as the frame, being firmly built into the wall, is relatively, very stiff and can be considered to be rigid. Likewise, the door can be considered to be constrained at hinge location. Elimination of the frame from the analysis model is the first approximation.
We use the same logic to simplify simulation of socket. The bolt pulls the socket through contact and the same pulling force is experienced by the screws. Thus we arrive at the second approximation resulting in elimination of the sockets and associated contacts from the model. Consequently, the tower bolts happen to be structural elements bonded to the door and constrained at the (socket) screw locations.
As the door thickness (about 30 mm) is very small relative to the door dimensions (1200 * 2000 mm), it is adequate to use shell elements in the simulation model.
F. E. Model and Results
The finite element model is shown in the figure below.
Fig. 2
An arbitrary load of 500 newtons is applied. This is representative of the force generated when a strong man kicks the door from outside. The load is assumed to be uniformly distributed over a region of 100*100. For the purpose of comparison four different analyses are carried out corresponding to four different load application locations. The locations are in indicated as L1, L2 etc. in the figure below. In the figure are also shown reaction points R1 to R6 for reference.
Fig. 3
Typical results are shown in the figure below. These are for load application at L1. It contains deformation contours and reaction values. R2 and R4 are positive reactions whereas R1 and R3 are negative reactions. This indicates that there is local bending effect. Total reaction force from six constraints is about 340 newtons. Difference between this and the applied load will be equal to total reaction force at hinge locations.
Fig. 4
The computed reaction forces for other loading conditions are given in the figure below. Qualitatively they appear to be correct.
Fig. 5
The foregoing is for qualitative understanding only. As we do not know the the strength of set screws and also because the load applied is arbitrary, no design inferences are intended.
The foregoing results are obtained for bolt length of 100 mm. Tower bolts are available in different sizes. Typical options are shown in the figure below.
Fig. 6
We note that for all the bolt sizes, the socket dimensions are unchanged. In all the cases the socket has provision for two set screws only. Intuitively it can be felt the longer bolts may not be advantageous from a security point of view.
Fig. 7
Reaction forces for bolt length of 200 mm are presented in the figure above. These are for the centrally applied loading condition (L2). The values corresponding to 100 mm length are in the inner box. We see practically no difference between the two sets of results.
Summary
1. This study was taken up just as an exercise in simulation. Simulation means mathematical representation of a physical process. The model developed will be useful for assessing the adequacy of tower bolts. Based on the approach one can carry out parametric studies to arrive at optimum configurations.
2. An important lesson has been on how to develop simulation models keeping in mind the purpose of analysis and the results of design importance.
3. One needs to find justification for paying more money for longer tower bolts. Purely from a point of view of security the additional money spent may not be worth.
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35. I'm Delighted to Share This Useful and Trustworthy Information About Tower Bolts. Your Post Contains Excellent Information That Demonstrates Your Knowledge of the Subject. for More Information, Visit Uns S30400 Bolt & Ss 316 Bolts, Manufacturers of Nace Mr0175 Fastener. I Appreciate You Making This Article Available to Me. I Am Thankful. | 2,939 | 13,899 | {"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} | 3.140625 | 3 | CC-MAIN-2022-33 | latest | en | 0.942282 |
https://en.wikipedia.org/wiki/Strange%E2%80%93Rahman%E2%80%93Smith_equation | 1,532,259,001,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593208.44/warc/CC-MAIN-20180722100513-20180722120513-00379.warc.gz | 654,276,894 | 11,527 | # Strange–Rahman–Smith equation
The Strange–Rahman–Smith equation is used in the cryoporometry method of measuring porosity. NMR cryoporometry [1][2][3] is a recent technique for measuring total porosity and pore size distributions. NMRC is based on two equations, the Gibbs–Thomson equation, that maps the melting point depression to pore size, and the Strange–Rahman–Smith equation [1] that maps the melted signal amplitude at a particular temperature to pore volume.
## Equation
If the pores of the porous material are filled with a liquid, then the incremental volume of the pores ${\displaystyle \Delta v}$ with pore diameter between ${\displaystyle x}$ and ${\displaystyle x+\Delta \,x}$ may be obtained from the increase in melted liquid volume for an increase of temperature between ${\displaystyle T}$ and ${\displaystyle T+\Delta T}$ by:[1]
${\displaystyle {\frac {dv}{dx}}={\frac {dv}{d\,T}}{\frac {k_{GT}}{x^{2}}}}$
Where: ${\displaystyle k_{GT}}$ is the Gibbs–Thomson coefficient for the liquid in the pores.
## References
1. ^ a b c Strange, J.H.; Rahman, M.; Smith, E.G. (Nov 1993), "Characterization of Porous Solids by NMR", Phys. Rev. Lett., 71 (21): 3589–3591, Bibcode:1993PhRvL..71.3589S, doi:10.1103/PhysRevLett.71.3589, PMID 10055015
2. ^ Mitchell, J.; Webber, J. Beau W.; Strange, J.H. (2008), "Nuclear Magnetic Resonance Cryoporometry", Phys. Rep. (Review), 461: 1–36, Bibcode:2008PhR...461....1M, doi:10.1016/j.physrep.2008.02.001
3. ^ SE-10044 Stockholm Sweden. [Furo Istvan Royal Inst TechnolDiv Phys Chem Dept Chem SE-10044 Stockholm Sweden. Royal Inst Chem SE-10044 Stockholm Sweden. [Furo Istvan] Royal Inst Technol Technol Ind NMR Ctr Dept Chem SE-10044 Stockholm Sweden. Div Phys Chem Dept Chem SE-10044 Stockholm Sweden. Royal Inst], "NMR cryoporometry: Principles applications and potential" | 537 | 1,832 | {"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} | 2.703125 | 3 | CC-MAIN-2018-30 | latest | en | 0.721234 |
https://leanprover-community.github.io/archive/stream/113489-new-members/topic/congr.20under.20an.20existential.html | 1,620,774,959,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990419.12/warc/CC-MAIN-20210511214444-20210512004444-00501.warc.gz | 401,776,401 | 3,989 | ## Stream: new members
### Topic: congr under an existential
#### Yakov Pechersky (Sep 21 2020 at 22:03):
Is there an existing lemma that says the following? Is it a pi lemma?
example {α β γ δ : Type} {x y : α} {f : α → β → γ} {g : γ → δ} (h : ∃ z, f x z = f y z) :
∃ z, g (f x z) = g (f y z) :=
begin
obtain ⟨z, hz⟩ := h,
use z,
rw hz,
end
#### Yakov Pechersky (Sep 21 2020 at 22:17):
I guess the reverse, with a hypothesis about injectivity of g is what I really want.
#### Yakov Pechersky (Sep 21 2020 at 22:20):
Ah, a variant of exists.imp
#### Yakov Pechersky (Sep 21 2020 at 22:27):
So, why does the first example work but not the second?
lemma exists_inj {α β γ δ : Type} {x : α} {f : α → β} {g : β → γ} (hg : function.injective g)
(h : ∃ z, g (f x) = g z) : ∃ z, f x = z :=
Exists.imp (λ z H, @hg (f x) z H) h
lemma exists_inj {α β γ δ : Type} {x : α} {f : α → β} {g : β → γ} (hg : function.injective g)
(h : ∃ z, g (f x) = g z) : ∃ z, f x = z :=
Exists.imp (λ z H, hg H) h
#### Mario Carneiro (Sep 21 2020 at 22:37):
function.injective is a definition that wraps something like \forall {{x y}}, f x = f y -> x = y
#### Yakov Pechersky (Sep 21 2020 at 22:38):
So it can't infer the (f x) or z. Got it. In any case, this isn't leading me down to where I want because of the "exists can only elim into prop" issue.
#### Mario Carneiro (Sep 21 2020 at 22:38):
Oh, it's the elab order
#### Mario Carneiro (Sep 21 2020 at 22:38):
lemma exists_inj {α β γ δ : Type} {x : α} {f : α → β} {g : β → γ} (hg : function.injective g)
(h : ∃ z, g (f x) = g z) : ∃ z, f x = z :=
h.imp (λ z H, hg H)
this works fine
#### Mario Carneiro (Sep 21 2020 at 22:39):
What are you trying to do?
#### Yakov Pechersky (Sep 21 2020 at 22:40):
I'm basically trying to do the equivalent of an apply_fun on both sides of an equality that is behind an existential.
but you did that
#### Mario Carneiro (Sep 21 2020 at 22:42):
so that must not be the problem
#### Yakov Pechersky (Sep 21 2020 at 22:42):
I'm going to try something, will be back later with attempts.
#### Mario Carneiro (Sep 21 2020 at 22:43):
Generally, you should open an existential if you want to work on the contents
Last updated: May 11 2021 at 22:14 UTC | 809 | 2,236 | {"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} | 2.921875 | 3 | CC-MAIN-2021-21 | latest | en | 0.774262 |
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