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https://thermalfluidscentral.org/encyclopedia/index.php/Numerical_Solution_of_1-D_Steady_Conduction | 1,696,209,925,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510942.97/warc/CC-MAIN-20231002001302-20231002031302-00695.warc.gz | 597,674,959 | 10,413 | # Numerical Solution of 1-D Steady Conduction
## Discretization of Governing Equations
To keep our discussion as general as possible, let us consider a one-dimensional steady-state heat conduction problem with temperature-dependent thermal conductivity and internal heat generation (Tao, 2001):
$\frac{1}{{A(x)}}\frac{d}{{dx}}\left( {kA(x)\frac{{dT}}{{dx}}} \right) + S = 0 \qquad \qquad(1)$
where A(x) is the area of heat conduction. Equation (1) is valid for heat conduction in Cartesian (A(x) = 1), cylindrical (x = r,A(r) = r), and spherical (x = r, A(r) = r2) coordinate systems, as well as heat transfer from extended surfaces (A(x) is the cross-sectional area of the fin). The source term, S, can be caused by internal heat source/sink, convection from the extended surface, or metabolic heat generation and blood perfusion in bioheat transfer problems. The source term in eq. (1) is often a function of temperature as in the cases of heat transfer from extended surfaces and bioheat transfer problems. It is desirable to reflect this temperature-dependence in our discretized governing equation. While the dependence of the source term on the temperature can be of a very complicated form (e.g., radiation heat loss from the extended surface), it is ideal to express the source term as a linear function of temperature because the discretized equation will be a linear algebraic equation. We can thus express the source term as
$S = {S_C} + {S_P}T \qquad \qquad(2)$
where SC is a constant and SP is a coefficient.
Substituting eq. (2) into eq. (1), and multiplying the resulting equation by A(x) yields
$\frac{d}{{dx}}\left( {kA(x)\frac{{dT}}{{dx}}} \right) + ({S_C} + {S_P}T)A(x) = 0$
Figure 1: Control volume for one-dimensional heat conduction
Integrating the above equation over the control volume P (shaded area in Fig. 1), one obtains
${\left( {kA\frac{{dT}}{{dx}}} \right)_e} - {\left( {kA\frac{{dT}}{{dx}}} \right)_w} + \int_w^e {({S_C} + {S_P}T)Adx} = 0 \qquad \qquad(3)$
Assuming the temperature T and conduction area A for each control volume are represented by their value at a grid point, the integral in eq. (3) becomes
$\int_w^e {({S_C} + {S_P}T)Adx} = {S_C}{A_P}{(\Delta x)_P} + {S_P}{T_P}{A_P}{(\Delta x)_P}$
Assuming the temperature distribution between any two neighboring grid points is piecewise linear, the first two terms on the left-hand side of eq. (3) become
${\left( {kA\frac{{dT}}{{dx}}} \right)_e} = {k_e}{A_e}\frac{{{T_E} - {T_P}}}{{{{(\delta x)}_e}}}$
${\left( {kA\frac{{dT}}{{dx}}} \right)_w} = {k_w}{A_w}\frac{{{T_P} - {T_W}}}{{{{(\delta x)}_w}}}$
where ke and kw are thermal conductivities at the faces of the control volume. To ensure that the heat flux across the faces of the control volume is continuous, it is important that the following harmonic mean conductivity at the faces are used (see Problem 3.43), i.e.,
$\frac{{{{(\delta x)}_e}}}{{{k_e}}} = \frac{{{{(\delta x)}_{e - }}}}{{{k_P}}} + \frac{{{{(\delta x)}_{e + }}}}{{{k_E}}},{\rm{ }}\frac{{{{(\delta x)}_w}}}{{{k_w}}} = \frac{{{{(\delta x)}_{w - }}}}{{{k_W}}} + \frac{{{{(\delta x)}_{w + }}}}{{{k_P}}} \qquad \qquad(4)$
For a uniform grid system, eq. (4) becomes
${k_e} = \frac{{2{k_P}{K_E}}}{{{k_P} + {K_E}}},{\rm{ }}{k_w} = \frac{{2{k_W}{K_P}}}{{{k_W} + {K_P}}} \qquad \qquad(5)$
Substituting the above equations into eq. (3), the following discretized equation is obtained
${a_P}{T_P} = {a_E}{T_E} + {a_W}{T_W} + b \qquad \qquad(6)$
where
${a_E} = \frac{{{k_e}{A_e}}}{{{{(\delta x)}_e}}},{\rm{ }}{a_W} = \frac{{{k_w}{A_w}}}{{{{(\delta x)}_w}}} \qquad \qquad(7)$
${a_P} = {a_W} + {a_E} - {S_P}{A_P}{(\Delta x)_P} \qquad \qquad(8)$
$b = {S_C}{A_P}{(\Delta x)_P} \qquad \qquad(9)$
## Boundary Conditions
(a) Left boundary (b) Right boundary
Figure 2: Control volumes at boundaries for Practice A
(a) Left boundary (b) Right boundary
Figure 3: Control volumes at boundaries for Practice B
If the temperature at the boundary is known (boundary condition of the first kind), no specific treatment on the boundary condition is necessary and the temperature at internal grid points can be obtained by solving the above discretized equations. For boundary conditions of the second and third kind, additional equations are necessary for the boundary temperatures. For Practice A, the width of the control volume at the boundary is equal to half of that of the internal control volumes (see Fig. 2). If the heat flux at the left boundary is known, the energy balance for the control volume at the left boundary is
${q''_B}{A_1} + {k_1}{A_1}\frac{{{T_2} - {T_1}}}{{{{(\delta x)}_1}}} + ({S_c} + {S_P}{T_1}){A_1}{(\Delta x)_1} = 0$
which can be expressed as
${a_1}{T_1} = {a_E}{T_2} + b \qquad \qquad(10)$
where
${a_E} = \frac{{{k_1}{A_1}}}{{{{(\delta x)}_1}}} \qquad \qquad(11)$
${a_P} = {a_E} - {S_P}{A_1}{(\Delta x)_1} \qquad \qquad(12)$
$b = {q''_B}{A_1} + {S_C}{A_1}{(\Delta x)_1} \qquad \qquad(13)$
If the left boundary is subject to a convective condition, we have
${q''_B} = h({T_f} - {T_1}) \qquad \qquad(14)$
It can be shown that eqs. (10) and (11) are still valid but eqs. (12) and (13) should be modified to
${a_P} = {a_E} - {S_P}{A_1}{(\Delta x)_1} + h{A_1} \qquad \qquad(15)$
$b = h{T_f}{A_1} + {S_C}{A_1}{(\Delta x)_1} \qquad \qquad(16)$
The boundary condition on the right hand side can be obtained by following a similar procedure (see Problem 3.46). If the computational domain is discretized by using Practice B, the size of the control volume for the grid at the boundary is zero (see Fig. 3), thus the discretized equation for the boundary condition at the left side can be obtained by setting the control volume size x)1 to zero in eqs. (12) – (13) or eqs. (15) – (16). Similarly, the discretized equation for the boundary condition on the right side for Practice B can be obtained by setting x)M in the discretized equation for Practice A.
## Solution of Discretized Equations
When the matrix for the coefficient for discretized algebraic equations of one-dimensional conduction is written, all of the nonzero components are aligned along three diagonals of the matrix. While the discretized algebraic equations can be solved by using Gaussian elimination method, a simple version – referred to as TDMA (TriDiagonal Matrix Algorithm) – can be used (Patankar, 1980). The discretized equations for every point, eq. (6), can be rewritten as
${a_i}{T_i} = {b_i}{T_{i + 1}} + {c_i}{T_{i - 1}} + {d_i},{\rm{ }}i = 1,2,...M \qquad \qquad(17)$
which indicates that the temperature at any point is related to temperatures of its immediate neighbor points only. Obviously, for the left and boundaries, we have
c1 = 0,bM = 0
To get the temperature, we hope that our temperature at the ith grid is related to the neighbor on the right, i.e.,
${T_i} = {P_i}{T_{i + 1}} + {Q_i} \qquad \qquad(18)$
For the (i − 1)th grid point, eq. (18) becomes
${T_{i - 1}} = {P_{i - 1}}{T_i} + {Q_{i - 1}} \qquad \qquad(19)$
Substituting eq. (19) into eq. (17), we get
${a_i}{T_i} = {b_i}{T_{i + 1}} + {c_i}({P_{i - 1}}{T_i} + {Q_{i - 1}}) + {b_i} \qquad \qquad(20)$
which can be rearranged to
${T_i} = \frac{{{b_i}}}{{{a_i} - {c_i}{P_{i - 1}}}}{T_{i + 1}} + \frac{{{d_i} + {c_i}{Q_{i - 1}}}}{{{a_i} - {c_i}{P_{i - 1}}}} \qquad \qquad(21)$
Comparing eq. (18) and (21) yields
${P_i} = \frac{{{b_i}}}{{{a_i} - {c_i}{P_{i - 1}}}},{\rm{ }}{Q_i} = \frac{{{d_i} + {c_i}{Q_{i - 1}}}}{{{a_i} - {c_i}{P_{i - 1}}}} \qquad \qquad(22)$
For the first grid point at the left (i = 1), c1 = 0 and eq. (22) becomes
${P_1} = \frac{{{b_1}}}{{{a_1}}},{\rm{ }}{Q_1} = \frac{{{d_1}}}{{{a_1}}} \qquad \qquad(23)$
For the last point at the right (i = M), bM = 0 and it follows that PM = 0. Equation (18) for i = M becomes
${T_N} = {Q_N} \qquad \qquad(24)$
The solution of the algebraic equations can thus be completed in the following steps: (1) obtain P1 and Q1 from eq. (23), (2) calculate Pi and Qi from eq. (22), (3) obtain temperature at the right side by eq. (24), and (4) calculate temperatures for all grid points by eq. (18).
It should be noted that if the thermal conductivity is independent from temperature, the temperatures at different grid points can be obtained by solving the discretized equations using TDMA once. If the thermal conductivity depends on temperature, the problem needs to be solved by an iteration procedure: (1) assume tentative thermal conductivity at all grid points, (2) obtain coefficients for discretized equation based on the tentative conductivity, (3) solve the discretized equation using TDMA, and (4) update the thermal conductivities at all points and solve for the temperatures. This procedure needs to be repeated until there is no notable change of temperature distributions between two consecutive iterations.
## References
Patankar, S.V., 1980, Numerical Heat Transfer and Fluid Flow, Hemisphere, Washington, DC.
Tao, W.Q., 2001, Numerical Heat Transfer, 2nd Ed., Xi’an Jiaotong University Press, Xi’an, China (in Chinese). | 2,861 | 8,999 | {"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": 29, "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} | 3.640625 | 4 | CC-MAIN-2023-40 | longest | en | 0.850826 |
https://www.nagwa.com/en/videos/298174197808/ | 1,726,154,984,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00735.warc.gz | 848,542,765 | 35,266 | Question Video: Identifying the Power of an 𝑛th Root Irrational Number That Is a Rational Number | Nagwa Question Video: Identifying the Power of an 𝑛th Root Irrational Number That Is a Rational Number | Nagwa
# Question Video: Identifying the Power of an πth Root Irrational Number That Is a Rational Number Mathematics
Let π₯ = the fifth root of (β(7)/β(2)). Which of the following numbers is rational? [A] π₯Β³ [B] π₯β΄ [C] π₯ΒΉΒ² [D] π₯ΒΉβ΅ [E] π₯βΆβ°
04:27
### Video Transcript
Let π₯ be equal to the fifth root of the cubed root of seven over the fourth root of two. Which of the following numbers is rational? (A) π₯ cubed, (B) π₯ to the fourth power, (C) π₯ to the 12th power, (D) π₯ to the 15th power, or (E) π₯ to the 60th power.
Before we start, there are two things we should think about, the first of them being what is a rational number. A rational number is a number that can be written in the form π over π, where π and π are integers and π is not equal to zero. Another way to say this is that a rational number can be made by dividing two integers.
Now, the second thing we might want to think about is seeing if we can simplify what π₯ is equal to here. To do that, letβs think about our exponent rule that tells us that the πth root of π₯ is equal to π₯ to the one over π power. And that means we can rewrite the fifth root as the one-fifth power. And we wanna repeat this process with the two roots inside the parentheses, which means weβll have seven to the one-third power over two to the one-fourth power, all taken to the one-fifth power.
And one further step we can do to simplify is remember the power of a power rule that tells us π₯ to the π power to the π power is equal to π₯ to the π times π power. And this means we can multiply one-third by one-fifth. When we do that, we multiply the numerators, one times one is one, then multiply the denominators, three times five is 15. Weβll have seven to the 15th power.
And then weβll need to multiply one-fourth times one-fifth to find the power of our denominator. One-fourth times one-fifth is one twentieth. And so weβre saying that π₯ is equal to seven to the one fifteenth power over two to the one twentieth power. To find a rational value, we need the numerator and the denominator here to be an integer. So letβs consider some of our options.
If we cube π₯, weβll have seven to the one fifteenth power over two to the one twentieth power cubed. And if we were going to take a power of a power, we would multiply these powers together. We would have π₯ cubed being equal to seven to the three fifteenth power over two to the three twentieths power.
Now, we could plug this value into our calculator to see what would happen. When we do that, we get 1.3304 continuing. This is an irrational value. And so maybe we should consider what kind of powers here would give us integers. If our exponents were whole numbers, if we were taking seven to an integer power, the outcome would be an integer. And that means we want to take both two and seven to some power of an integer. We need to multiply one fifteenth and one twentieth by some number that produces an integer for both of these values.
For example, if we took π₯ to the 15th power, then we would have seven to the first power. But we would still have a fraction in our denominator. But that does get us a bit closer. From our list, we can take π₯ to the 60th power, which would be seven to the 60 over 15 power and two to the 60 over 20 power. 60 divided by 15 is four. 60 divided by 20 is three. π₯ to the fourth power will be an integer, and two cubed will also be an integer. An integer divided by an integer is a rational number. And so we can say that π₯ to the 60th power will be rational.
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• Realistic Exam Questions | 1,125 | 4,054 | {"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} | 4.84375 | 5 | CC-MAIN-2024-38 | latest | en | 0.93421 |
www.schoolproject.in | 1,642,643,624,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301670.75/warc/CC-MAIN-20220120005715-20220120035715-00412.warc.gz | 1,050,586,144 | 23,287 | ### Introduction of Maths working model of Quadrilateral
A quadrilateral is a polygon with four sides and
four vertices or corners. Sometimes, the term quadrangle is used, by analogy with triangle,and sometimes tetragon for consistency with pentagon (5-sided), hexagon (6-sided) and so on.
Quadrilaterals can be classified into Parallelograms, Squares, Rectangles and Rhombuses, Square, Rectangle and Rhombus are also Parallelograms.
Rectangle: A quadrilateral with four right angles, a rectangle is a type of parallelogram.
Square: A quadrilateral with four congruent sides and four right angles, a square is both a rhombus and a rectangle.
Trapezoid: A quadrilateral with exactly one pair of parallel sides (the parallel sides are called bases).
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https://skia.googlesource.com/skia/+/7e20e667978516fabb3a28af60d981f2fd3bdb45/src/core/SkMatrixInvert.cpp | 1,713,707,716,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817780.88/warc/CC-MAIN-20240421132819-20240421162819-00605.warc.gz | 477,853,880 | 5,067 | blob: ea8d36702c1b11976f179b0165afb26c51b48ea7 [file] [log] [blame]
/* * Copyright 2021 Google LLC * * Use of this source code is governed by a BSD-style license that can be * found in the LICENSE file. */ #include "src/core/SkMatrixInvert.h" #include "include/private/base/SkFloatingPoint.h" SkScalar SkInvert2x2Matrix(const SkScalar inMatrix[4], SkScalar outMatrix[4]) { double a00 = inMatrix[0]; double a01 = inMatrix[1]; double a10 = inMatrix[2]; double a11 = inMatrix[3]; // Calculate the determinant double determinant = a00 * a11 - a01 * a10; if (outMatrix) { double invdet = sk_ieee_double_divide(1.0, determinant); outMatrix[0] = a11 * invdet; outMatrix[1] = -a01 * invdet; outMatrix[2] = -a10 * invdet; outMatrix[3] = a00 * invdet; // If 1/det overflows to infinity (i.e. det is denormalized) or any of the inverted matrix // values is non-finite, return zero to indicate a non-invertible matrix. if (!SkScalarsAreFinite(outMatrix, 4)) { determinant = 0.0f; } } return determinant; } SkScalar SkInvert3x3Matrix(const SkScalar inMatrix[9], SkScalar outMatrix[9]) { double a00 = inMatrix[0]; double a01 = inMatrix[1]; double a02 = inMatrix[2]; double a10 = inMatrix[3]; double a11 = inMatrix[4]; double a12 = inMatrix[5]; double a20 = inMatrix[6]; double a21 = inMatrix[7]; double a22 = inMatrix[8]; double b01 = a22 * a11 - a12 * a21; double b11 = -a22 * a10 + a12 * a20; double b21 = a21 * a10 - a11 * a20; // Calculate the determinant double determinant = a00 * b01 + a01 * b11 + a02 * b21; if (outMatrix) { double invdet = sk_ieee_double_divide(1.0, determinant); outMatrix[0] = b01 * invdet; outMatrix[1] = (-a22 * a01 + a02 * a21) * invdet; outMatrix[2] = ( a12 * a01 - a02 * a11) * invdet; outMatrix[3] = b11 * invdet; outMatrix[4] = ( a22 * a00 - a02 * a20) * invdet; outMatrix[5] = (-a12 * a00 + a02 * a10) * invdet; outMatrix[6] = b21 * invdet; outMatrix[7] = (-a21 * a00 + a01 * a20) * invdet; outMatrix[8] = ( a11 * a00 - a01 * a10) * invdet; // If 1/det overflows to infinity (i.e. det is denormalized) or any of the inverted matrix // values is non-finite, return zero to indicate a non-invertible matrix. if (!SkScalarsAreFinite(outMatrix, 9)) { determinant = 0.0f; } } return determinant; } SkScalar SkInvert4x4Matrix(const SkScalar inMatrix[16], SkScalar outMatrix[16]) { double a00 = inMatrix[0]; double a01 = inMatrix[1]; double a02 = inMatrix[2]; double a03 = inMatrix[3]; double a10 = inMatrix[4]; double a11 = inMatrix[5]; double a12 = inMatrix[6]; double a13 = inMatrix[7]; double a20 = inMatrix[8]; double a21 = inMatrix[9]; double a22 = inMatrix[10]; double a23 = inMatrix[11]; double a30 = inMatrix[12]; double a31 = inMatrix[13]; double a32 = inMatrix[14]; double a33 = inMatrix[15]; double b00 = a00 * a11 - a01 * a10; double b01 = a00 * a12 - a02 * a10; double b02 = a00 * a13 - a03 * a10; double b03 = a01 * a12 - a02 * a11; double b04 = a01 * a13 - a03 * a11; double b05 = a02 * a13 - a03 * a12; double b06 = a20 * a31 - a21 * a30; double b07 = a20 * a32 - a22 * a30; double b08 = a20 * a33 - a23 * a30; double b09 = a21 * a32 - a22 * a31; double b10 = a21 * a33 - a23 * a31; double b11 = a22 * a33 - a23 * a32; // Calculate the determinant double determinant = b00 * b11 - b01 * b10 + b02 * b09 + b03 * b08 - b04 * b07 + b05 * b06; if (outMatrix) { double invdet = sk_ieee_double_divide(1.0, determinant); b00 *= invdet; b01 *= invdet; b02 *= invdet; b03 *= invdet; b04 *= invdet; b05 *= invdet; b06 *= invdet; b07 *= invdet; b08 *= invdet; b09 *= invdet; b10 *= invdet; b11 *= invdet; outMatrix[0] = a11 * b11 - a12 * b10 + a13 * b09; outMatrix[1] = a02 * b10 - a01 * b11 - a03 * b09; outMatrix[2] = a31 * b05 - a32 * b04 + a33 * b03; outMatrix[3] = a22 * b04 - a21 * b05 - a23 * b03; outMatrix[4] = a12 * b08 - a10 * b11 - a13 * b07; outMatrix[5] = a00 * b11 - a02 * b08 + a03 * b07; outMatrix[6] = a32 * b02 - a30 * b05 - a33 * b01; outMatrix[7] = a20 * b05 - a22 * b02 + a23 * b01; outMatrix[8] = a10 * b10 - a11 * b08 + a13 * b06; outMatrix[9] = a01 * b08 - a00 * b10 - a03 * b06; outMatrix[10] = a30 * b04 - a31 * b02 + a33 * b00; outMatrix[11] = a21 * b02 - a20 * b04 - a23 * b00; outMatrix[12] = a11 * b07 - a10 * b09 - a12 * b06; outMatrix[13] = a00 * b09 - a01 * b07 + a02 * b06; outMatrix[14] = a31 * b01 - a30 * b03 - a32 * b00; outMatrix[15] = a20 * b03 - a21 * b01 + a22 * b00; // If 1/det overflows to infinity (i.e. det is denormalized) or any of the inverted matrix // values is non-finite, return zero to indicate a non-invertible matrix. if (!SkScalarsAreFinite(outMatrix, 16)) { determinant = 0.0f; } } return determinant; } | 1,719 | 4,585 | {"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} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.329245 |
https://www.aqua-calc.com/one-to-all/temperature/preset/kelvin/918.15 | 1,611,769,866,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704828358.86/warc/CC-MAIN-20210127152334-20210127182334-00159.warc.gz | 620,781,770 | 5,939 | # Convert Kelvins [K] to other units of temperature
## Kelvins [K] temperature conversions
918.15 K = 645 degrees Celsius K to °C 918.15 K = 1 193 degrees Fahrenheit K to °F 918.15 K = 1 652.67 degrees Rankine K to °R 918.15 K = 346.13 degrees Romer K to °Rø
Convert entered temperature to units of energy.
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Manganese(II) hydroxide [Mn(OH)2] weighs 3 260 kg/m³ (203.51515 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-11, liquid (R11) with temperature in the range of -1.12°C (29.984°F) to 115.56°C (240.008°F)
#### Weights and Measurements
terajoule per minute (TJ/min) is the SI multiple of the derived power measurement unit joule per second (J/sec) or watt (W).
The kinematic viscosity (ν) is the dynamic viscosity (μ) divided by the density of the fluid (ρ)
troy/µm² to centner/ha conversion table, troy/µm² to centner/ha unit converter or convert between all units of surface density measurement.
#### Calculators
Calculate Maximum Heart Rate (MHR) based on age and gender | 470 | 1,674 | {"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} | 2.921875 | 3 | CC-MAIN-2021-04 | latest | en | 0.718569 |
http://math.stackexchange.com/questions/219649/curvature-torsion-and-change-of-coordinates/220397 | 1,469,266,032,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257821671.5/warc/CC-MAIN-20160723071021-00294-ip-10-185-27-174.ec2.internal.warc.gz | 160,649,412 | 17,390 | Curvature, Torsion and Change of coordinates
Is there a general way so we can find curvature and torsion of a curve $$\gamma : \mathbb{R} \rightarrow \mathbb{R}^n$$ where $$n=2,3$$ from a coordinate system to a new coordinate system?
-
Do you mean a rigid transformation or a complete change of variables? – Pragabhava Oct 24 '12 at 2:14
No rigid transformation, I mean coordinate transformation, for example from cartesian coordinate system to polar coordinate system or from bipolar coordinates to cartesian coordinates. – stavros kassaras Oct 24 '12 at 6:45
Case $n = 2$ and curvature $\kappa(t)$.
The curve $\gamma(t) = \big(x(t),y(t)\big)^T$, has curvature $$\kappa(t) = \frac{x'y''-x''y'}{(x'^2 + y'^2)^{3/2}} = \frac{\det(\gamma',\gamma'')}{\|\gamma'\|^3}.$$ Now, suppose $x(t) = x\big(\xi(t),\eta(t)\big)$, $y(t) = y\big(\xi(t),\eta(t)\big)$, then \begin{align} \frac{d x}{d t} &= x_\xi \xi' + x_\eta \eta'\\ \frac{d y}{d t} &= y_\xi \xi' + y_\eta \eta' \end{align} and let $\alpha(t) = \big(\xi(t),\eta(t)\big)^T$, we can write the above relation as $$\gamma' = \textbf{J} \, \alpha'$$ where $$\textbf{J} = \begin{pmatrix}x_\xi & x_\eta \\ y_\xi & y_\eta\end{pmatrix}$$ is the Jacobian matrix. Then $$\gamma'' = \frac{d \textbf{J}}{dt} \alpha' + \textbf{J} \alpha''$$ and $$\kappa(t) = \frac{\det(\textbf{J}\,\alpha',\frac{d \textbf{J}}{dt} \alpha' + \textbf{J} \alpha'')}{\|\textbf{J}\,\alpha'\|^3}.$$
Example.
Let $x(t) = r(t) \cos \theta(t)$ and $y(t) = r(t) \sin \theta(t)$, then \begin{align} \det(\textbf{J}\alpha',\textbf{J}'\alpha') &= 2 r'^2 \theta' + r^2 \theta'^3\\ \det(\textbf{J}\alpha',\textbf{J}\alpha'') &= -r r'' \theta' + r r' \theta''\\ \|\textbf{J}\alpha'\|^3 &= (r'^2 + r^2 \theta'^2)^{3/2} \end{align}
and
$$\kappa(t) = \frac{r^2 \theta'^3 + 2 r'^2 \theta' - r r'' \theta' + r r' \theta''}{(r'^2 + r^2 \theta'^2)^{3/2}}.$$
In the special case $\theta = t$ $$\kappa(t) = \frac{r^2 + 2r'^2 - r r''}{(r^2 + r'^2)^{3/2}}$$
which, according to eq. (15) of Wolfram Mathworld, is correct.
Now you can do the rest.
- | 799 | 2,051 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-30 | latest | en | 0.510918 |
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### What is 75% of \$5000?
```
Date: 2 Jan 1995 18:37:01 -0500
From: Anonymous
Subject: math
What is 75% of \$5,000?
```
```
Date: 3 Jan 1995 02:57:23 -0500
From: Dr. Sydney
Subject: Re: math
Hello there!
Thanks for writing Dr. Math! There are couple of different ways to
think about figuring out what is 75% of \$5000.
First, you could think of 75% as a decimal -- taking 75% of \$5000 is the
same as multiplying \$5000 by .75.
Or, instead, you could think of 75% as a fraction -- 75% is the same as
3/4, right? This is because 75% = 75/100 = 3/4. So, 75% of \$5000
would be the same as 3/4 of \$5000.
I hope this helps. If you have any more questions or if you want to see
--Sydney
```
Associated Topics:
Elementary Multiplication
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 350 | 1,180 | {"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} | 3.109375 | 3 | CC-MAIN-2017-26 | longest | en | 0.935719 |
https://forum.allaboutcircuits.com/threads/mic-preamp-ltspice.177338/ | 1,695,669,348,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510085.26/warc/CC-MAIN-20230925183615-20230925213615-00481.warc.gz | 284,145,274 | 25,794 | Mic preamp, LTSpice
upand_at_them
Joined May 15, 2010
940
I made this microphone preamp in LTSpice (haven't used LTS much before) and I noticed that the output signal has a fatter top than bottom....Or maybe it's the bottom that is skinnier than the top.
Did I do something wrong in the design? Or is this just the nature of the simulation? The signal isn't near either power rail.
Attachments
• 1.7 KB Views: 26
Papabravo
Joined Feb 24, 2006
20,379
You want Vb to be in the middle of the supply range and don't bypass the collector resistor.
Try something like this:
Boost the input AC signal to get a bigger output. The .png picture was take from a previous run where the input was 10 mV P-P so the RMS values are not correct. They should read:
virms: RMS(v(vin))=0.00353148 FROM 0 TO 0.001
vorms: RMS(v(vout))=0.361432 FROM 0 TO 0.001
av: vorms/virms=102.346
The actual design procedure for a CE amplifier is a bit more complicated than is practical for a forum post. What does the online textbook for this site have to say?
Sorry. It doesn't look like there is much practical design help there.
Attachments
• 2.5 KB Views: 16
Last edited:
Bordodynov
Joined May 20, 2015
3,087
With an input voltage source signal with an amplitude of 10mV the distortion is about 10%. The main harmonic distortion is second. And this is even before the voltage limits are reached. Do a Fourier analysis (.Four 2.5kHz).
crutschow
Joined Mar 14, 2008
32,852
Did I do something wrong in the design? Or is this just the nature of the simulation?
Not wrong just bad, and no.
And that distortion is not specifically related to the bias point.
You are seeing the inherent non-linearity of the large-signal Vbe voltage versus the collector current.
(The collector current is a logarithmic function of the base-emitter voltage.)
You can see that, since if you drive the amp with a current source instead of a voltage source, that distortion will be significantly reduced (as the collector current versus base current is fairly linear).
Alternately, if you add a small un-bypassed emitter degeneration resistor (R5, see below, which gives negative feedback) you can reduce that distortion with a voltage input.
The tradeoff is gain versus the reduction in distortion.
The higher the feedback resistor the lower the gain and distortion.
You can check the distortion using the .four command as shown by looking at the View/Spice Error Log.
Without R5 the harmonic distortion is over 8%.
With R5 the distortion is reduced to <3% (shown).
You can vary the value of R5 to see how the distortion changes.
To maintain the same DC bias conditions, the sum of R4 and R5 should stay constant.
Last edited:
upand_at_them
Joined May 15, 2010
940
Great. Thanks guys! Looking forward to making changes and actually building it.
upand_at_them
Joined May 15, 2010
940
So this design is bad? What is bad about it? Is there a better design I should be looking at? I don't mind getting this one the best I can get it and then building a different one to compare.
upand_at_them
Joined May 15, 2010
940
I changed R3 to 1.2K, C3 to 220uF, and the new R5 to 20R. Distortion is now just under 1%.
upand_at_them
Joined May 15, 2010
940
You want Vb to be in the middle of the supply range and don't bypass the collector resistor.
I have the collector capacitor there because of someone else's design. It was said that it limits the bandwidth of the amp, since this is for a microphone and doesn't need beyond 20kHz. I haven't run the circuit through different frequencies yet, though.
crutschow
Joined Mar 14, 2008
32,852
I was referring to it not having any negative feedback to reduce the distortion.
I changed R3 to 1.2K, C3 to 220uF, and the new R5 to 20R. Distortion is now just under 1%.
Does it still have sufficient gain for your mic?
upand_at_them
Joined May 15, 2010
940
Does it still have sufficient gain for your mic?
Gain is less. But I figure that I can add a second stage.
Audioguru again
Joined Oct 21, 2019
6,152
An opamp has been used as a low distortion preamp for many years. Before that, a preamp with two transistors using lots of negative feedback for low distortion was used.
Attachments
• 302.9 KB Views: 23
upand_at_them
Joined May 15, 2010
940
I'm going to try an opamp after this. Thanks.
Audioguru again
Joined Oct 21, 2019
6,152
Do you have a common electret mic or an old dynamic (coil and magnet) mic?
upand_at_them
Joined May 15, 2010
940
Electret.
Audioguru again
Joined Oct 21, 2019
6,152
I have used this circuit many times. Texas Instruments shows an inverting opamp for a little less hiss noise.
Attachments
• 85.2 KB Views: 27
• 29.3 KB Views: 25
DickCappels
Joined Aug 21, 2008
10,067
Your original problem was caused by C2. I would leave that out.
upand_at_them
Joined May 15, 2010
940
@Audioguru again Thanks for the circuits. I do have some opamps around here somewhere.
@DickCappels I did test removing C2, though. Without it I get no noticeable change that I could see, but slightly more distortion (8.7% v. 8.4%). Even with my latest circuit changes I have 0.887% distortion with C2 and 0.996% without it. I don't know enough to comment on it, but if it has slightly less distortion and also limits bandwidth to 20kHz I'm in favor of it.
crutschow
Joined Mar 14, 2008
32,852
Your original problem was caused by C2.
No, the distortion was caused by the nonlinearity of the BJT as I stated in post #4.
Audioguru again
Joined Oct 21, 2019
6,152
A transistor with a high voltage gain (no negative feedback) and at a fairly high output level 9but not clipping) has severe even-order harmonics distortion. The top part of the waveform is squashed so much that the voltage gain cannot be measured.
Attachments
• 24.6 KB Views: 16 | 1,560 | 5,776 | {"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} | 2.734375 | 3 | CC-MAIN-2023-40 | latest | en | 0.924748 |
https://numberworld.info/root-of-802 | 1,597,380,062,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739177.25/warc/CC-MAIN-20200814040920-20200814070920-00227.warc.gz | 420,174,811 | 2,947 | # Root of 802
#### [Root of eight hundred two]
square root
28.3196
cube root
9.2909
fourth root
5.3216
fifth root
3.8092
In mathematics extracting a root is declared as the determination of the unknown "x" in the equation $y=x^n$ The outcome of the extraction of the root is declared as a so-called root. In the case of "n equals 2", one talks about a square root or sometimes a second root also, another possibility could be that n is equivalent to 3 at that time one would call it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 802 is represented as this: $$\sqrt[]{802}=28.319604517013$$
Furthermore it is legit to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 802 is 28.319604517013. The cube root of 802 is 9.2909072109025. The fourth root of 802 is 5.3216167202282 and the fifth root is 3.809209630567.
Look Up | 281 | 972 | {"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} | 3.734375 | 4 | CC-MAIN-2020-34 | latest | en | 0.916049 |
https://www.actucation.com/grade-6-maths/transformation-of-expression-into-phrase-(multiple-operations) | 1,627,718,508,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154085.58/warc/CC-MAIN-20210731074335-20210731104335-00543.warc.gz | 618,335,763 | 27,223 | Informative line
# Transformation of Expression into Phrase involving Addition and Subtraction
• Transformation of an expression into a phrase means writing the algebraic expression into words.
• Here, we will transform the expression in which two operations i.e., addition and subtraction are used.
For example: $$(x+7) - 2$$
According to PEMDAS rule, the addition or subtraction should be performed in the order from left to right.
Here, first, a variable $$x$$ is added to $$7$$ and then $$2$$ is subtracted from the sum of $$x$$ and $$7$$.
Thus, this can be written as:
$$2$$ is subtracted from the sum of $$x$$ and $$7$$.
#### Which statement correctly represents the expression: $$(y-6) + 10$$?
A $$10$$ is added to the subtraction of $$6$$ from $$y$$
B $$6$$ is added to the subtraction of $$10$$ from $$y$$
C $$6$$ is added to the subtraction of $$y$$ from $$10$$
D $$y$$ is added to the subtraction of $$10$$ from $$6$$
×
Given expression:
$$(y-6)+10$$
According to PEMDAS rule, the addition or subtraction should be performed in the order from left to right.
First, $$6$$ is subtracted from $$y$$.
Then, $$10$$ is added.
Thus, $$(y-6)+10$$ can be written as:
$$10$$ is added to the subtraction of $$6$$ from $$y$$.
Hence, option (A) is correct.
### Which statement correctly represents the expression: $$(y-6) + 10$$?
A
$$10$$ is added to the subtraction of $$6$$ from $$y$$
.
B
$$6$$ is added to the subtraction of $$10$$ from $$y$$
C
$$6$$ is added to the subtraction of $$y$$ from $$10$$
D
$$y$$ is added to the subtraction of $$10$$ from $$6$$
Option A is Correct
# Transformation of Expression into Phrase involving Addition and Multiplication
• Transformation of an expression into a phrase means writing the algebraic expression into words.
• Here, we will transform the expression in which two operations i.e., addition and multiplication are used.
For example: $$2y+4$$
According to PEMDAS rule, the multiplication is performed first and then the addition.
First, a variable $$y$$ is multiplied by $$2$$ and then $$4$$ is added to the product of $$2$$ and $$y$$.
Hence, this can be written as:
$$4$$ is added to the product of $$2$$ and $$y$$.
#### Which statement correctly represents the expression: $$7 + 3x$$?
A $$3$$ is added to the product of $$7$$ and $$x$$
B $$7$$ is added to the product of $$3$$ and $$x$$
C $$x$$ is added to the product of $$7$$ and $$3$$
D $$x$$ is added to the sum of $$3$$ and $$7$$
×
Given expression:
$$7+3x$$
According to PEMDAS rule, the multiplication is performed first and then the addition.
First, $$3$$ is multiplied by $$x$$.
Then, $$7$$ is added.
Thus, $$(7+3x)$$ can be written as:
$$7$$ is added to the product of $$3$$ and $$x$$.
Hence, option (B) is correct.
### Which statement correctly represents the expression: $$7 + 3x$$?
A
$$3$$ is added to the product of $$7$$ and $$x$$
.
B
$$7$$ is added to the product of $$3$$ and $$x$$
C
$$x$$ is added to the product of $$7$$ and $$3$$
D
$$x$$ is added to the sum of $$3$$ and $$7$$
Option B is Correct
# Transformation of Expression into Phrase involving Subtraction and Multiplication
• Transformation of an expression into a phrase means writing the algebraic expression into words.
• Here, we will transform the expression in which two operations i.e., subtraction and multiplication are used.
For example: $$24-5a$$
According to PEMDAS rule, the multiplication is performed first and then the subtraction.
Thus, first, $$5$$ is multiplied by $$a$$, then the product of $$5$$ and $$a$$ is subtracted from $$24$$.
Hence, this can be written as:
The product of $$5$$ and $$a$$ is subtracted from $$24$$
#### Which statement correctly represents the expression: $$2c-16$$?
A The product of $$16$$ and $$c$$ is subtracted from $$2$$
B The product of $$2$$ and $$c$$ is subtracted from $$16$$
C $$2$$ is subtracted from the product of $$c$$ and $$16$$
D $$16$$ is subtracted from the product of $$2$$ and $$c$$
×
Given expression:
$$2c-16$$
According to PEMDAS rule, the multiplication is performed first and then the subtraction.
First, $$2$$ is multiplied with $$c$$.
Then, $$16$$ is subtracted.
Thus, $$2c-16$$ can be written as:
$$16$$ is subtracted from the product of $$2$$ and $$c$$.
Hence, option (D) is correct.
### Which statement correctly represents the expression: $$2c-16$$?
A
The product of $$16$$ and $$c$$ is subtracted from $$2$$
.
B
The product of $$2$$ and $$c$$ is subtracted from $$16$$
C
$$2$$ is subtracted from the product of $$c$$ and $$16$$
D
$$16$$ is subtracted from the product of $$2$$ and $$c$$
Option D is Correct
# Transformation of Expression into Phrase involving Addition and Division
• Transformation of an expression into a phrase means writing the algebraic expression into words.
• Here, we will transform the expression in which two operations i.e., addition and division are used.
For example: $$\dfrac{6}{a} + 3$$
According to PEMDAS rule, the division is to be performed first and then the addition.
Thus, first, $$6$$ is divided by '$$a$$' and then $$3$$ is added to the quotient of $$6$$ by '$$a$$ '.
Hence, this can be written as:
$$3$$ is added to the quotient of $$6$$ by '$$a$$'.
#### Which statement correctly represents the expression: $$\dfrac{b}{4} + 6$$?
A The sum of $$b$$ by $$6$$ is divided by $$4$$
B The quotient of $$6$$ by $$4$$ is added to $$b$$
C The quotient of $$b$$ by $$4$$ is added to $$6$$
D The quotient of $$b$$ by $$6$$ is added to $$4$$
×
Given expression:
$$\dfrac{b}{4} + 6$$
According to PEMDAS rule, the division is to be performed first and then the addition.
First, find the quotient of $$b$$ by $$4$$.
Then, $$6$$ is added.
Thus, $$\dfrac{b}{4} +6$$ can be written as:
The quotient of $$b$$ by $$4$$ is added to $$6$$.
Hence, option (C) is correct.
### Which statement correctly represents the expression: $$\dfrac{b}{4} + 6$$?
A
The sum of $$b$$ by $$6$$ is divided by $$4$$
.
B
The quotient of $$6$$ by $$4$$ is added to $$b$$
C
The quotient of $$b$$ by $$4$$ is added to $$6$$
D
The quotient of $$b$$ by $$6$$ is added to $$4$$
Option C is Correct
# Transformation of Expression into Phrase involving Subtraction and Division
• Transformation of an expression into a phrase means writing the algebraic expression into words.
• Here, we will transform the expression in which two operations i.e., subtraction and division are used.
For example: $$\dfrac{4}{c}- 2$$
According to PEMDAS rule, the division is to be performed first and then the subtraction.
Thus, first, $$4$$ is divided by c and then $$2$$ is subtracted from the quotient of $$4$$ by c.
Hence, this can be written as:
$$2$$ is subtracted from the quotient of $$4$$ by c.
#### Which statement correctly represents the expression: $$5-\dfrac{b}{10}$$?
A $$10$$ is subtracted from the quotient of $$b$$ by $$5$$
B $$5$$ is subtracted from the quotient of $$b$$ by $$10$$
C The quotient of $$b$$ by $$5$$ is subtracted from $$10$$
D The quotient of $$b$$ by $$10$$ is subtracted from $$5$$
×
Given expression:
$$5-\dfrac {b}{10}$$
According to PEMDAS rule, the division is to be performed first and then the subtraction.
First, find the quotient of $$b$$ by $$10$$.
Then, subtract the quotient from $$5$$.
Thus, $$5-\dfrac{b}{10}$$ can be written as:
The quotient of $$b$$ by $$10$$ is subtracted from $$5$$.
Hence, option (D) is correct.
### Which statement correctly represents the expression: $$5-\dfrac{b}{10}$$?
A
$$10$$ is subtracted from the quotient of $$b$$ by $$5$$
.
B
$$5$$ is subtracted from the quotient of $$b$$ by $$10$$
C
The quotient of $$b$$ by $$5$$ is subtracted from $$10$$
D
The quotient of $$b$$ by $$10$$ is subtracted from $$5$$
Option D is Correct
# Transformation of Expression into Phrase involving Multiplication and Division
• Transformation of an expression into a phrase means writing the algebraic expression into words.
• Here, we will transform the expression in which two operations i.e., multiplication and division are used.
For example: $$\dfrac{2c}{3}$$
According to PEMDAS rule, the multiplication and division should be operated in the order from left to right.
Here, first, $$2$$ is multiplied by $$c$$ and then the product of $$2$$ and $$c$$ is divided by $$3$$.
Thus, this can be written as:
The product of $$2$$ and $$c$$ is divided by $$3$$.
#### Which statement correctly represents the expression: $$\dfrac{3a}{2}$$?
A The product of $$2$$ and $$a$$ is divided by $$3$$
B The product of $$3$$ and $$a$$ is divided by $$2$$
C $$2$$ is divided by the product of $$3$$ and $$a$$
D $$3$$ is divided by the product of $$2$$ and $$a$$
×
Given expression:
$$\dfrac{3a}{2}$$
According to PEMDAS rule, the multiplication and division should be operated in the order from left to right.
Here, first, $$3$$ is multiplied by $$a$$ and then the product is divided by $$2$$.
Thus, $$\dfrac{3a}{2}$$ can be written as:
The product of $$3$$ and $$a$$ is divided by $$2$$.
Hence, option (B) is correct.
### Which statement correctly represents the expression: $$\dfrac{3a}{2}$$?
A
The product of $$2$$ and $$a$$ is divided by $$3$$
.
B
The product of $$3$$ and $$a$$ is divided by $$2$$
C
$$2$$ is divided by the product of $$3$$ and $$a$$
D
$$3$$ is divided by the product of $$2$$ and $$a$$
Option B is Correct | 3,021 | 9,393 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-31 | latest | en | 0.790345 |
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2 Fixed author name
Nowadays we can associate to a topological space $X$ a category called the fundamental (or Poincare) $\infty$-groupoid given by taking $Sing(X)$.
There are many different categories that one can associate to a space $X$. For example, one could build the small category whose object set is the set of points with only the identity morphisms from a point to itself. It is claimed that the classifying space of this category returns the space: $BX=X$
The inspiration for these examples comes from three primary sources: Graeme Segal's famous 1968 paper Classifying Spaces and Spectral Sequences, Raoul Bott's Mexico notes (taken by Lawrence Conlon) Lectures on characteristic classes and foliations, and a 1995 pre-print called Morse Theory and Classifying Spaces by Ralph Cohen, G. Segal and Vaughan John Jones.
In each of these papers there is a notion of a topological category. It is not just a category enriched in Top, since the set of objects can have non-discrete topology. Here is the definition that I can gleam from these articles:
A topological category consists of a pair of spaces $(Obj,Mor)$ with four continuous structure maps:
• $i:Obj\to Mor$, which sends an object to the identity morphism
• $s:Mor\to Obj$, which gives the source of an arrow
• $t:Mor\to Obj$, which gives the target of an arrow
• $\circ:Mor\times_{t,s}Mor\to Mor$, which is composition.
Were $i$ is a section of both $s$ and $t$, and all the axioms of a small category hold.
Is the appropriate modern terminology to describe this a Segal Space? What would Lurie call it? Based on reading Chris Schommer-Pries MO post and elsewhere this seems to be true. Would the modern definition of the above be a Segal Space where the Segal maps are identities? Also, why do we demand that the topology on objects be discrete for Segal Categories? Is there something wrong with allowing the object sets to have topologies?
1
# Segal's Original Definition of a Topological Category
Nowadays we can associate to a topological space $X$ a category called the fundamental (or Poincare) $\infty$-groupoid given by taking $Sing(X)$.
There are many different categories that one can associate to a space $X$. For example, one could build the small category whose object set is the set of points with only the identity morphisms from a point to itself. It is claimed that the classifying space of this category returns the space: $BX=X$
The inspiration for these examples comes from three primary sources: Graeme Segal's famous 1968 paper Classifying Spaces and Spectral Sequences, Raoul Bott's Mexico notes (taken by Lawrence Conlon) Lectures on characteristic classes and foliations, and a 1995 pre-print called Morse Theory and Classifying Spaces by Ralph Cohen, G. Segal and Vaughan Jones.
In each of these papers there is a notion of a topological category. It is not just a category enriched in Top, since the set of objects can have non-discrete topology. Here is the definition that I can gleam from these articles:
A topological category consists of a pair of spaces $(Obj,Mor)$ with four continuous structure maps:
• $i:Obj\to Mor$, which sends an object to the identity morphism
• $s:Mor\to Obj$, which gives the source of an arrow
• $t:Mor\to Obj$, which gives the target of an arrow
• $\circ:Mor\times_{t,s}Mor\to Mor$, which is composition.
Were $i$ is a section of both $s$ and $t$, and all the axioms of a small category hold.
Is the appropriate modern terminology to describe this a Segal Space? What would Lurie call it? Based on reading Chris Schommer-Pries MO post and elsewhere this seems to be true. Would the modern definition of the above be a Segal Space where the Segal maps are identities? Also, why do we demand that the topology on objects be discrete for Segal Categories? Is there something wrong with allowing the object sets to have topologies? | 945 | 4,032 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2013-20 | latest | en | 0.899633 |
https://www.studypug.com/ca/grade7/patterns | 1,675,220,252,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00189.warc.gz | 1,032,877,004 | 40,362 | # Patterns
#### All in One Place
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#### Learn with Ease
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0/8
##### Examples
###### Lessons
1. A bedroom wall is painted with blue polka dots and green flowers. The basic design is shown.
Below are two patterns made when 2 and 3 of the basic designs are put together.
1. Create a table that shows the pattern of blue polka dots and green flowers for the first 6 figures. Describe the pattern.
2. Use b to represent the number of green flowers. Write an expression to describe the number of blue polka dots.
3. How many polka dots will there be if there are 32 flowers?
2. Bianca was changing fractions into decimal numbers and noticed a pattern.
$\frac{1}{11}=0.0909...$
$\frac{2}{11}=0.1818...$
$\frac{3}{11}=0.2727...$
1. Describe the pattern.
2. Write $\frac{4}{11}$ as a decimal number, using bar notation.
3. Write $0.54545454...$ as a fraction.
3. Bella always sleeps for 8 hours on school days and 2 extra hours on holidays.
1. Write an expression to show the hours of sleep that Bella gets in a week.
2. If Bella only has 2 days of school this week, how many hours of sleep does she have for the whole week? Use the expression in the previous part to answer this question.
0%
##### Practice
###### Topic Notes
In this section, we are asked to describe given patterns and copy and complete tables of values for these patterns. Also, we use variables to write expressions for the patterns and then solve for these variables. In this section, we will gain fundamental knowledge about expressions and variables that we will build on in other sections. | 430 | 1,819 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "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} | 4.5 | 4 | CC-MAIN-2023-06 | latest | en | 0.885814 |
https://betterlesson.com/lesson/532050/interactive-anchor-chart-4?from=breadcrumb_lesson | 1,623,751,791,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00035.warc.gz | 132,001,094 | 22,154 | # Interactive Anchor Chart - 4
4 teachers like this lesson
Print Lesson
## Objective
SWBAT exemplify key elements of a given number.
#### Big Idea
Students help BUILD an anchor chart for the number 4, showing different ways to "make" 4.
## Attention Grabber/Introduction
5 minutes
“It’s Friday, you know what that means…” I say as kindergartners walk in after lunch.
“It’s Saturday tomorrow! Sleep in!” a student annouces.
“Yes, that is true, but we have some important work to do before we are ready to go off for the weekend. You will notice a big number by the big screen. What are we going to do with it?” I ask.
Students respond that we are going to color and make ways to show 4. They catch on quickly, my little cherubs!
“Yes—for sure!” I say. This time, I say that our 4 things are going to be apples. A few students say, “Yeah!” in response to our beloved book that currently holds such an important place in our instruction.
## Independent Practice
10 minutes
The students begin working on their illustrations, and again, I take requests first, encouraging kiddos to use their math vocabulary and really stretch as we are learning these first numbers.
Talking to students during the illustration process is critical. This is a one-of-a-kind opportunity to use language and really see where students are “at” in terms of concept development. Sometimes, a kid will want to color a coconut tree because he likes coconut trees, but that’s when I get him talking about the specific number of coconut trees and why that number is so important to what we’re doing. Little things like that make the difference, I’m finding.
## Closing
15 minutes
As we gather to attach the examples to the 4, a kiddo says to a friend, “It’s time to put the 4 together!”
“We are starting to get the idea here!” I declare. “Let’s talk about what we’re putting on this anchor chart as we attach each thing!.”
I won’t lie—I kind of “work it” to get the kiddos talking as much as possible during this assembly process. When one student gets “stuck,” I offer support or get other kiddos talking on the topic, never letting on that someone has “limited” knowledge. Being conversational and upbeat is critical, I’m discovering.
“So what number did we “put together?” I ask.
“You are so very smart!” I say with a smile. | 539 | 2,332 | {"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} | 3.078125 | 3 | CC-MAIN-2021-25 | latest | en | 0.937822 |
https://cxc.cfa.harvard.edu/sherpa/faq/user_stat.html | 1,716,665,439,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00853.warc.gz | 159,091,696 | 5,150 | How can I define my own fit statistic in Sherpa?
The load_user_stat function provides an interface for adding user-defined statistic functions in Sherpa. It accommodates user-defined functions for a statistic and statistical errors, in addition to defining a list of model parameters and hyperparameters for prior distributions (if prior desired).
For example, a simple user-defined statistic in Python Sherpa would look like this:
```sherpa> def my_stat_func(data, model, staterror, syserror=None, weight=None, bkg=None):
# A simple function to replicate χ2
fvec = ((data - model) / staterror)**2
stat = fvec.sum()
return (stat, fvec)
sherpa> def my_staterr_func(data):
# A simple staterror function
return numpy.sqrt(data)
sherpa> load_user_stat("mystat", my_stat_func, my_staterr_func)
sherpa> set_stat(mystat)
```
And a more complex user-defined statistic, with prior distributions, would look like this:
```sherpa> def my_stat_func(data, model, staterror=None, syserror=None, weight=None, bkg=None):
...
return (stat, fvec)
sherpa> load_user_stat("mystat", my_stat_func, priors=dict(mugamma=0.017, ..., gamma=abs1.nh, ... ))
sherpa> set_stat(mystat)
```
In these examples, 'stat' is a scalar statistic value and 'fvec' is an array of the statistic contributions of each bin. This method caters to both types of optimization methods in Sherpa: levmar expects the statistic contribution per bin, whereas simplex and moncar expect a single scalar statistic value. | 358 | 1,476 | {"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-2024-22 | latest | en | 0.601989 |
http://css-tricks.com/forums/topic/how-do-i-calulate-two-values/ | 1,419,187,802,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802772134.89/warc/CC-MAIN-20141217075252-00025-ip-10-231-17-201.ec2.internal.warc.gz | 66,911,752 | 22,709 | # How Do I Calulate two values….
• # September 3, 2009 at 3:53 pm
I want to take the price of an item and multiply it by the quantity (as entered by shopper) to display what the total would be …
Price Per Inch * # of inches = Total
The idea is this…
So the inch price will be displayed with the minimum number of inches in a text box…
then the overall total for this product is calculated as the customer changes the value in the number of inches box…
This way they know the price of the necklace before they hit the add to cart button.
I have found a few scripts that deal with multiple items with different values totaling then totaling the grand total but nothing basic like what I am looking for. Plus I really want to use jQuery.
Can anyone help me learn this? :?:
# September 4, 2009 at 8:34 am
have a variable for your price A a variable for your size B
var a = 10; //used an integer you can set manually
var b = 12; //used an integer, but you can get what was input by user if needed
var c = a * b; //calculation
# September 5, 2009 at 4:14 pm
ikthius has given the answer needed to achieve this. I would just like to elaborate on what he said :)
Code:
\$(“#inches-select”).change( //inches selected with value changed
function(){
var inches= // Some value which you get from user on value being changed
var pricePerInch = 50; //The value setting for the necklace
//Attaching a \$ sign to the total price
var priceUpdate = ‘\$’+(inches*pricePerExtraPage);
\$(“.price”).text(“”+priceUpdate); //Here you just update the price in particular header or paragraph with the class ‘price’
}
);
Performing the price calculation on input change would show instant price change to the user. It would really look cool if you could take the input of the number of inches using a slider ;)
I have implemented a similar thing in the order form of my home page. In my form, the user can select total number of pages from a drop down menu and the price is being shown instantly on selection. The site is still under construction, but the order form is in working condition. :)
P.S.: The above code is incomplete and just a framework of how the final version will look. You can look up in the jQuery documentation on how to access values from an input field and in the process learn as well :)
Viewing 3 posts - 1 through 3 (of 3 total)
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*May or may not contain any actual "CSS" or "Tricks". | 592 | 2,444 | {"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-2014-52 | latest | en | 0.911445 |
https://riunitedformarriage.org/what-is-isheisenberg-uncertainty-principle/ | 1,716,207,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058278.91/warc/CC-MAIN-20240520111411-20240520141411-00147.warc.gz | 434,252,090 | 11,765 | ## What is Isheisenberg uncertainty principle?
uncertainty principle, also called Heisenberg uncertainty principle or indeterminacy principle, statement, articulated (1927) by the German physicist Werner Heisenberg, that the position and the velocity of an object cannot both be measured exactly, at the same time, even in theory.
## What violates Heisenberg uncertainty?
(Phys.org)—If an object traveling through spacetime can loop back in time in a certain way, then its trajectory can allow a pair of its components to be measured with perfect accuracy, violating Heisenberg’s uncertainty principle.
What is the theory of uncertainty?
Introduced first in 1927 by the German physicist Werner Heisenberg, the uncertainty principle states that the more precisely the position of some particle is determined, the less precisely its momentum can be predicted from initial conditions, and vice versa.
What is the Heisenberg Uncertainty Principle and why is it important?
The Heisenberg uncertainty principle is a law in quantum mechanics that limits how accurately you can measure two related variables. Specifically, it says that the more accurately you measure the momentum (or velocity) of a particle, the less accurately you can know its position, and vice versa.
### Is Heisenberg Uncertainty Principle true?
At the foundation of quantum mechanics is the Heisenberg uncertainty principle. Physics students are still taught this measurement-disturbance version of the uncertainty principle in introductory classes, but it turns out that it’s not always true.
### Can the uncertainty principle be violated?
A violation of the uncertainty principle implies a violation of the second law of thermodynamics. Uncertainty relations state that there exist certain incompatible measurements, to which the outcomes cannot be simultaneously predicted.
Is the uncertainty principle used today?
The one that physicists use in everyday research and call Heisenberg’s uncertainty principle is in fact Kennard’s formulation. It is universally applicable and securely grounded in quantum theory. Heisenberg’s formulation, however, was proposed as conjecture, so quantum mechanics is not shaken by its violation.
Is Heisenberg uncertainty principle true?
#### What is the uncertainty principle used for?
Roughly speaking, the uncertainty principle (for position and momentum) states that one cannot assign exact simultaneous values to the position and momentum of a physical system. Rather, these quantities can only be determined with some characteristic “uncertainties” that cannot become arbitrarily small simultaneously.
#### What is the purpose of Schrodinger’s cat?
Schrödinger’s cat is a famous hypothetical experiment designed to point out a flaw in the Copenhagen interpretation of superposition as it applies to quantum theory.
Why did Walter White call himself Heisenberg?
Despite the alias, Walt didn’t fully transform into Heisenberg until season 4. The name of Walt’s alter ego came from Werner Heisenberg, a German physicist known as a pioneer of quantum mechanics. More than likely, he used the name and the altered look as something to hide behind as a way to cope with his own actions.
How is the Heisenberg uncertainty principle related to quantum mechanics?
– Heisenberg uncertainty principle is a principle of quantum mechanics. And so if we take a particle, let’s say we have a particle here of Mass M, moving with Velocity V, the momentum of that particle, the linear momentum is equal to the Mass times the Velocity.
## How is the uncertainty principle different from classical physics?
This should not suggest that the uncertainty principle is the only aspect of the conceptual difference between classical and quantum physics: the implications of quantum mechanics for notions as (non)-locality, entanglement and identity play no less havoc with classical intuitions. 1. Introduction 2. Heisenberg
## Is the energy-time uncertainty principle a relation?
The energy-time uncertainty principle does not result from a relation of the type expressed by (Figure) for technical reasons beyond this discussion. Nevertheless, the general meaning of the energy-time principle is that a quantum state that exists for only a short time cannot have a definite energy.
How is the Bohr model different from the uncertainty principle?
So the uncertainty in the position would be greater than the diameter of the hydrogen atom, using the Bohr model. So the Bohr model is wrong. It’s telling us we know the electron is orbiting the nucleus at a certain radius, and it’s moving at a certain velocity. The uncertainty principle says this isn’t true. | 893 | 4,668 | {"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-2024-22 | latest | en | 0.911508 |
https://academia.stackexchange.com/questions/111242/how-to-choose-a-good-grading-curve-for-yes-no-exams/111255 | 1,603,453,247,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00104.warc.gz | 188,077,084 | 47,173 | How to choose a good grading curve for yes/no exams?
I teach a certain course where I think the best way to test knowledge is using yes/no questions. However, with a simple yes/no exam, a student can just answer randomly to all questions and get, in expectation, a score of 50%, which is not very fair. I thought of several solutions, but each of them has a disadvantage:
1. Give a negative score to a wrong answer, such that the expected value of a student who answers randomly is 0. However, this rule made some students highly anxious, and I realized that this anxiety could unjustly harm their performance .
2. Ask the students to explain each answer. Award points only to correct answers with correct explanations. The problem is that it takes much more time to write and to grade, negating one of the main advantages of a multi-choice exam.
3. Give a grade of 0 to any student with less than 50% correct answers, and give twice the points for each answer above 50% (e.g, a student with 90% correct answers will receive 2·(90−50) = 80% score). However, I fear this might not be accepted by the university, which encourages additive scoring. E.g, students with 60% correct answers will complain that it is unfair to give them only 20% grade.
What is a good way to write an exam based on yes/no questions?
• The conversation on the pros and cons of multiple choice has been moved to chat. Please use comments only to suggest improvements to the question and similar. – Wrzlprmft Jun 16 '18 at 20:14
• No grade scheme will improve the situation that yes/no questions suffer from not much useful signal range (if you account for the occasional error) above background (50%). – Trilarion Jun 16 '18 at 21:26
• Did you consider extending the number of possible answers to a question from (1:yes; 2:no) to (1:yes, because XYZ; 2:yes, because ABC; 3:no, because PQR; 4:it may be true depending on IJL), so that the expected mark of a random guesser will be 100%*1/n? Does distinguishing the right reasoning(beyond answer) shows deeper knowledge in your course? – fixerlt Jun 18 '18 at 13:08
• Mathematically, 0 for wrong, 1 for right (and 0.5 for blank) is entirely equivalent to -1 for wrong, 1 for right (and 0 for blank). So the exact points you give for each option doesn't really matter. With that in mind, giving students the additional worry that they feel penalized for giving a wrong answer isn't a good idea in my opinion. Thus a wrong answer should give 0, not negative. – Arthur Jun 18 '18 at 14:51
• @FourOhFour students with exam anxiety told me that, in such a case, they will always select only the "unconfident" options, since they are afraid to lose points. So, their maximum possible score will be quite low. – Erel Segal-Halevi Jun 20 '18 at 12:16
For this situation, I provide three answer options:
• Yes.
• No.
• I don't know.
• I don't know: +0.5
That way, if the student really doesn't know, they are guaranteed the expectation of a random guess and you get feedback on holes in student knowledge, which can be used to take remedial action. If your aim is more effective student learning, this can be a good strategy.
I've tried to avoid negative marks for a couple of reasons:
Imagine two candidates who achieve the same final score in a negatively marked exam. One candidate answers only part of the paper, losing a few marks for wrong answers. The other answers almost all of the paper, getting far more correct but also getting several wrong and having marks deducted. Which one knows more?
The second problem is that the use of negative marking extends the theoretical range of marks for the exam. For example, if one mark is awarded for every correct answer and one deducted for each wrong answer, the theoretical range of the exam is - 100 to +100 per cent.
From the comments moved to chat: someone asked about a slightly different value for the “I don’t know” option. This is my response:
I tend not to use Yes/No questions, but multiple choice for some assessment. In those cases, I make the "I don't know" option worth 1/(N-1) where N is the number of other options. That way, if there are 3 other options, a random guess would expect to get 1/3 but the I don't know option gains 1/2 --- the expected value if the student can knock out one of the three options..
• Comments are not for extended discussion; this conversation has been moved to chat. – Wrzlprmft Jun 17 '18 at 8:45
• Students who don't answer at all (check neither option) are treated the same as "I don't know"? – Bergi Jun 17 '18 at 21:25
• @Bergi In this case, I'm partial to saying that checking no option is a 0. If you don't know the answer, then you don't know the answer, and should be able to own up to that fact. Incentivising students to actively give an answer (and therefore at least to some degree actually consider the question) would be a good thing, I think. – Arthur Jun 18 '18 at 14:53
• A true academic should admit when they don't know. There's a lot more that I don't know than I do know. – corsiKa Jun 18 '18 at 16:46
You are, essentially, evaluating a binary predictor. The usual score is known as accuracy (i.e., probability of giving the correct answer). For a random (coin toss) predictor you will get 50% (if the correct answers are evenly distributed between yes and no).
Note that the condition that the correct answers are distributed %50:%50 between yes and no is itself a huge hint to the students if they know that in advance.
I suggest that you consider two alternative metrics:
Matthews correlation coefficient
1. Range: from -1 to 1
2. All correct: 1
3. All wrong: -1
4. Random: 0
Proficiency
Measures the share of information (percentage of bits) contained in the true answers which is captured by the submitted answers (information here means entropy of the distribution).
1. Range: from 0 to 1
2. All correct: 1
3. All wrong: 1 (sic!)
4. Random: 0
Note that the scores of "all correct" and "all opposite" are the same. This is a feature, not a bug: the person who got all answers wrong has probably confused the labels ("yes" and "no") but knows the subject.
It also gives a lower score to "close to random" results than MCC. IOW, it discriminates good results better.
Caveats
If an answer is missing, it should be replace with a random yes/no answer.
This is correct statistically, but wrong pedagogically because the honest "I don't know" answer shows second order knowledge that should be encouraged.
The metrics behave the same regardless of the distribution of the correct answers, except when all the correct answers are the same (in which case neither metric is defined). This is addressed by the Bernoulli shuffle (step 2 in the protocol).
These are statistical metrics
These metrics make little sense when there are very few questions. Specifically, a mistake in one question may have a very different cost depending on the question (this can be addressed by repeating the Bernoulli shuffle many times and taking the average score - see protocol step 2 below).
These metrics are not additive: you cannot score two halves of the test separately and somehow combine them to get the total test score (let alone add the scores for the two halves).
This can make these metrics a very hard sell to both students and administrators.
Protocol
Given the above, the correct protocol for these metrics is:
1. Prepare hundreds of questions (it is okay for all questions to have the same correct answer).
2. Using a Bernoulli rng with p=1/2, for each question, map yes/no to A/B (i.e., for about 50% questions, yes will be A and for 50% it will be B, and for half of questions the correct answer is A) - this is done completely behind the screen. Neither the students nor the professor need to be aware what the actual map is.
3. Score the resulting binary predictors (now both the true base rate and the predicted base rates are the same - 50%).
The dependency on the Bernoulli rng does introduce some noise (order 1/N where N is the number of test questions). I.e., the standard deviation of mcc and proficiency on a test with 100 questions will be under 1%. Since the steps 2 & 3 are automatic, they can be repeated many times and the mean or median score used.
• As a stats nerd, I love this answer, but I suspect some students might not understand it and conclude you are grading unfairly. – Superbest Jun 16 '18 at 1:45
• Looks good. I wonder if I can convince the university to accept such a scale. – Erel Segal-Halevi Jun 16 '18 at 18:32
• How would leaving an answer blank be handled with these systems? – John K Jun 16 '18 at 19:09
• @JohnK: I would treat blank as submitting a random answer. – sds Jun 17 '18 at 18:18
• The Matthews coefficient has some (presumably) undesired behaviour in not treating questions equally. For example, if the test has five statements, three of which are actually true, then a student who gets only one wrong will score 0.61 if the mistake was thinking a false statement was true, but will score 0.67 if the mistake was thinking a true statement was false. – Especially Lime Jun 18 '18 at 14:07
Sure, a student recording random guesses on a yes/no or true/false exam will have an expected score of 50%. So? Under most grading schemes, this is a solid F.
A more pertinent question to ask is "What is the probability that a randomly guessing student will receive a passing grade / a B or higher / an A?"
For this problem, exam length is your friend! Let's assume that each question is an independent trial and that correct and incorrect answers are equally likely (p=0.5). This places us in the context of the Binomial Distribution. I'll consider a passing grade to be a score of 70% or higher and an A to be a 90% or higher, with a correct answer getting 1pt and an incorrect answer getting 0pts on each question (which meets your institution's preference for additive grading).
• For a short 10 question exam, a random guesser has a 17.2% chance of passing and a 1.07% chance of getting an A.
• For a 20 question exam, a random guesser has a 5.77% chance of passing and a 0.0201% chance of getting an A.
• For a 50 question exam, a random guesser has only a 0.330% chance of passing and a 0.000000210% chance of getting an A.
• For a 100 question exam, a random guesser has only a 0.00393% chance of passing and a 1.52*10^{-15}% chance of getting an A.
While there are many caveats to this rough analysis, this should highlight that random guessing is not a viable strategy for an exam of reasonable length. Keep in mind that the probability of passing is also the same probability that they get less than a 30% on the exam! Along the lines of this statistical analysis, there are many similar takeaways: a student who can make an educated guess (say p=0.7) at each question will almost always outperform a random guesser; a student who knows the answer to the first n-1 questions and then randomly guesses on the last question will do even better. Another takeaway would be that a multiple choice exam has a better likelihood of punishing a random guesser than a yes/no exam.
Personally, I don't find a multiple choice or yes/no exam all that great at evaluating student learning, as it generally hews towards regurgitation of facts and technicalities of wording, as opposed to demonstrating critical analysis and synthesis. If you do persist with grading yes/no or multiple choice, here's a few personal opinions:
• Use a random (or pseudo-random) process to determine the answers. Exams where every correct answer is choice (B) are cute, but are not a good evaluation tool. Students will also pick up on patterns in the order of correct answers, leading to meta-gaming.
• Use a simple metric (0pts incorrect/ 1pt correct or similar) and be sure to explain whether or not they will be penalized for guessing---exams are stressful enough without needing to consider game theory when recording each answer.
• Make the exam sufficiently long. This will help even out the noise from random guessing and make each individual question less stressful. Consider the extreme case of what it would feel like to take an exam with a single T/F question.
• On the flip side, make sure that there is enough time to give a reasoned response to each question.
Edit
I was unaware of the different correlations between raw percentage score and typical grades, as pointed out in the comments. My thanks to @curiousdanii and @cfr for pointing this out to me and my sincere apologies for the ethnocentrism! I greatly appreciate the information from academia.stackexchange that can help us peer beyond our institutional bubbles.
Having only graded within the American system, I don't have a good qualitative feel for the various grades in other systems. However, I feel that a central theme of my answer is still worth considering: random guessing is almost always an inferior strategy to studying and answering questions correctly. As such, I would give serious thought to whether it is truly a problem that you need to control for when designing a rubric.
If pressed, I think the most natural way to account for an unacceptably high expected value (relative to a given grade scheme) would be to switch to a multiple choice format. As with a good T/F question, strive to make all the answers outwardly plausible. Having a "joke" response that can immediately be ruled out accomplishes little.
• In Australia 50% is a pass. An individual lecturer wouldn't be allowed to change this. Is there any way to adapt your strategy for such a situation? – curiousdannii Jun 16 '18 at 22:51
• In the UK, 40% is a pass; 50% is a 2.2. – cfr Jun 18 '18 at 2:18
• Where is 70% passing? Genuinely curious – Azor Ahai -- he him Jun 18 '18 at 14:47
• I think this answer missing the point a bit. If a student only knows half the answers, and guesses on the other half, they'd get a 75% (instead of a 50% like they would for a fill-in-the-blank style test.) The problem isn't "Guessing on everything is a valid strategy" - it's "Guessing on everything you don't know" is, and will give you a much higher percent than you deserve. – Kevin Jun 18 '18 at 17:21
• @erfink - On just a T/F sheet? Yeah, there's not really any way of distinguishing which one had wrong answers they were certain about, and which ones that it was just a guess (and on the flip side, no way of determining a right answer that's just a guess vs know-for-a-fact.) My personal opinion is that I like "Full points for correct, subtract some portion of points for a wrong answer." Because I think "Knowing What You Know And Don't Know" is an underappreciated ability. I'm a programmer, and I hate programmers that don't know stuff but aren't self aware enough to realize it. – Kevin Jun 18 '18 at 20:50
One possibility that solves all of the listed problems: Grading on a curve.
Granted, Curve Grading is something that gets a lot of peoples' hackles up (either for or against) but this is a situation that it's practically built for: you have a range of results that don't line up with fair letter grades but that are sorted/ranked by degree of knowledge.
Before, a student's 75% would be a 'C', even though they only actually knew the answers to half the questions (and guessed the other half.) On a curve, though, that student's 75% would likely give them a low or failing grade since most other people in the class would hopefully have a higher success rate.
Edit: For clarification, when I say "Grading on a Curve," I mean that X% of the class will get an 'A', Y% will get a 'B', etc. Heck, to be honest, if you explained up front, "I'll be grading this on a curve because, well, if I didn't, people could get 50% for just guessing everything randomly!" and made the curve semi-generous, then you would engender very little ill-will amongst the students.
• Similarly -- you can accomplish the same and avoid the controversial curve by assigning pre-determined letter grades: 94% will be an A, 85% a B, etc. These are pretty high standards (unusual in STEM classes), but the students will be familiar with such a scheme, and it's appropriate since 50% is random chance. The only disadvantage is that since your worst students should get 60-70% and the best should get 90-100%, you need to make sure these relatively small differences will propagate to the final grade (rather than being drowned out by homework, etc., where the variation might be greater) – cag51 Jun 15 '18 at 21:00
• Doesn't this lead to hostility among the students? Since a student doing well on the exam, makes the scores of the others lower... – Erel Segal-Halevi Jun 16 '18 at 18:36
• Nope, you will always engender ill-will for grading on a curve as it is inherently unfair, and IMO, unethical. – curiousdannii Jun 16 '18 at 22:52
• @curiousdannii But if you don't grade on a curve what you end up evaluating is the professors ability to write a fair test which is drowning out the actual students abilities. – Keatinge Jun 17 '18 at 2:45
• Eh, if I was a student and you told me, "For this next quiz, we'll be grading on a curve. On average, my tests have 15% A's, 50% B's, and 35% C's - but for this test, I'll be curving it so that 30% are A's, 60% are B's, and only 10% are C's" - I'd be ecstatic. Students typically hate the curve when it's harsh or when they perceive that they're getting a lower letter grade than they'd get ordinarily. That's why I suggest making the curve semi-generous - make it clear up front that the students are getting a higher grade than they'd likely get without it. – Kevin Jun 18 '18 at 13:10
There are a number of papers published on the over the years. A table of options is shown below from Bandaranayake, et. al., Using Multiple Response True-False Multiple Choice Questions, Royal Australian College of Surgeons, 1999. Some of these consider the construction of "multiple true/false" questions, that is, questions that have one "stem" and several (weakly or strongly) linked true-false statements:
Another option is suggested by Frank Reid, An Alternative Scoring Formula for Multiple-Choice and True False Tests, Journal of Educational Research, 2001:
Some other papers I found include the following:
• Multiple True-False Questions; Hill, G. C.; Woods, G. T. Education in Chemistry, 11, 3, 86-87, May 74
• Scoring Multiple True/False Tests: Some Considerations; Gross, Leon J., Evaluation and the Health Professions, v5 n4 p459-68 Dec 1982
• Burton, Multiple choice and true/false tests: reliability measures and some implications of negative marking, Assessment & Evaluation in Higher Education, 2004.
• Burton R. Misinformation, partial knowledge and guessing in true/false tests. Medical Education [serial online], 2002.
• Tsai F, Suen H. A brief report on a comparison of six scoring methods for multiple true-false items. Educational & Psychological Measurement, 1993.
• Muijtjens, Mameren H, Hoogenboom, Evers, Vleuten C, Muijtjens. The effect of a ‘don't know’ option on test scores: number-right and formula scoring compared. Medical Education, 1999.
• Interesting. I am also interested to know how such questions are graded. – Erel Segal-Halevi Jun 16 '18 at 18:31
• Very cool! Thanks for all the nice references. – Peter K. Jun 18 '18 at 22:08
Another way to help lower the guessable score while still rewarding the students who have prepared themselves for the exams is to group like questions into one, where answering all the questions about one topic is showing mastery of that topic and earns full credit, whereas only knowing some of the answers is worth less credit.
This is easier to explain with an example. Let's say I'm preparing an exam for a US History class for the unit on the events leading up to the revolutionary war. Some of the events that I might want to be sure my students have studied and understood could be The Establishment of Jamestown, Virginia as the first permanent English settlement, Bacon's Rebellion, The Boston Massacre, The Boston Tea Party, and some others.
When preparing the question on Bacon's Rebellion, I might want to check that the students know when and where it was. So I could set up the question like this:
Bacon's Rebellion (4 points. Possible scores: 4, 1, 0):
• A) Bacon's Rebellion was in 1676. T F
• B) Nathaniel Bacon led the rebellion in Jackson, Mississippi. T F
(Answers: A) True, B) False, it was in Jamestown, Virginia)
If the student gets both questions right, they get all 4 points. If they get only one right, they still get one point. If they get both questions wrong, they get 0 points.
This scoring system would have to be thoroughly explained to the students before hand and written at the top of the test, but to give you an idea of how this distributes the exam scores, if the exam is made up of questions each with two parts like this and a student simply guesses on every question, they have an expected average score of 1.5 points per question, or 37.5% on the whole exam. Yet for a student that knows most of the material and knows 95% of the answers is expected to get about a 92% on the exam.
By awarding one point instead of 0 for the partially correct answer, this helps mitigate the students complaints of "But I knew that the rebellion was in 1676, I just forgot where it was!" because they are still getting partial credit, just not as much as they would get if they showed a mastery of the material by being able to answer both questions.
By never giving any negative points either, this helps with the anxiety you were mentioning your students face with that prospect.
Two tips if you decide to implement this idea:
1. Stick to the same format for the whole exam, and make sure to explain to the students what they are going to see. Either make them all questions with two parts, or do something else, but don't mix them up. Students may become caught up in trying to figure out how the questions are scored, and will lose valuable time that should be used showing their knowledge of the material.
2. Don't try to group more than two questions together in this way. The math might look nice how the expected score of the person guessing drops greatly with each question you add about Bacon's Rebellion, but from the student perspective, if he or she has studied and knows most of the material but just can't remember one little detail that happens to be something you ask, that student immediately loses the majority of the credit for a question that they might know a lot about.
• I like your grading scheme but I think you would have less of a fight with students if you awarded points on a logarithmic-like scale rather than exponential-like scale. In other words, if there are 10 problems in a set, the gap in points awarded between 9 correct and 10 correct is smaller than 5 correct and 6 correct. – JWH2006 Jun 15 '18 at 20:16
• Very interesting. In effect, if a group has $n$ questions, then a student with $k$ correct answers receives $k^2/n$ points on that group (e.g. if $n=2$ then $k=2$ gives a score of $2$ and $k=1$ gives a score of $1/2$). – Erel Segal-Halevi Jun 16 '18 at 18:14
One easy scheme that has been used by some mathematical olympiads is to make every question's answer to be an integer from 1 to 999. If you are careful in setting the question, you usually can make the probability of guessing the correct answer (even after some common sense elimination) to be no bigger than 10%. For example, if the question asks to find the length of some line segment in some geometric construction, engineer the construction so that there are no obvious inequalities that can bound the length to less than 100.
The advantages of this scheme is that the exam is easy to grade and yet difficult for students to get undeserved credit. I am strongly of the opinion that mathematics students should be graded based on proofs, but if you are considering multiple choice questions at least this scheme is a far better option.
a student can just answer randomly to all questions and get, in expectation, a score of 50%, which is not very fair
Not really, unless your grading scheme also gives a high grade to scores above 50%. If you assume the scores follow a normal-like distribution (they almost always will) then the mean will come out somewhere well above 50% and the guessers who got 50% will end up with a pretty negative z-score, and thus a low grade (possibly F, depending on how you set it up).
I think the real issue with true/false questions is that you actually have pretty good odds of winning the lottery, and getting a good grade without knowing the material. Because of this, at least multiple choice or fill in the blank questions are much better. Essay questions, as you mention, are of course the best but much more laborious to actually grade.
Note that you can always convert T/F questions to multiple choice, by using "which of the three above statements are true?". This way it is impossible to guess independently, so the expected score can drop from 50% down to 12% (but in reality will be 20% or 25% because of the number of choices).
After reviewing all of the previously posted answers, I can't see any way in which the inherent problems of a yes-no exam can be overcome. It is simply not a good exam format. You can't extract a useful quantification of your students' knowledge and skill from it.
You can't, but your students can. So if you want to set yes-no exams, do so with the purpose of giving your students a regular purpose for reviewing what they have learnt. Give these exams a nominal 1-2% contribution to the final grade, with a max of 10% for all the yes-no exams combined, and then set one or two more involved assessments to make up the rest of the final grade.
Each yes or no question can be answered in three ways: yes, no, (blank).
State at the start that each correct answer has a +1 score, each blank has a 0 score, and each incorrect answer has, say, a -0.5 score. This discourages guessing but has a lesser penalty than a -1 score. The range of possible grades would then be from -50% to 100%.
• I quite like this scheme in theory, but in practice it is likely to be gender biased since there is evidence that women are less likely to guess than men if there is a penalty for guessing and less likely to fill in answers they are unsure of: e.g. mpra.ub.uni-muenchen.de/39987/1/MPRA_paper_39987.pdf – Jack Aidley Jun 18 '18 at 13:10
• I think that one of the papers in my answer suggested upping the incorrect penalty to deal with that general issue. – Daniel R. Collins Jun 19 '18 at 13:56
You can also make a combination of a linear and a non-linear scales. Say, you give 5 problems with 4 yes/no questions in each one. To determine the problem score from, say, 0 to 20 you can use any reasonable (=monotone and explicable to the students) function on ${0,1}^4$ you wish (you can even assign different values to different questions) but then you just add the scores for the problems. This eliminates the threshold effect that is the main disadvantage of any "cutoff amplification" of just one final score but preserves the general idea that correct random guessing isn't worth much.
As to the common smooth amplification techniques, the ones I've seen in action are $x^2/100$ (so the lucky random guy with nominal 60% gets only 36) and $10\sqrt x$ (if everyone has really low score, this stretches the bottom part and boosts the morale a bit).
If there is a reason why you are considering 50% to be of some benefit, you could restructure the course's grading structure to not be based on the Score (percentage), but instead be based on the Grade (A-F), using the same scale as for GPA calculation (A=4,F=0).
This would mean that scoring 0-59% on any test or assignment would result in 0 points contributing to the final course Grade.
This will make simply guessing on a True/False or Yes/No test to be statistically useless.
I will state that I believe that worrying over a student improperly "gaming" the system to score a failing grade that might be higher than another failing grade he might have received if he had properly taken the test, is stressing yourself out for no useful purpose.
The more I think about it, the more I have a problem with using GPA within a class. For one, it would be almost impossible to get an A, as that would be a perfect score.
Perhaps Scoring for assignments could be
• F=0
• D=2
• C=4
• B=6
• A=8
With the final course grade being
• '>=7 gives A
• '>=5 gives B
• '>=3 gives C
• '>=1 gives D
• '<1 gives F
I see three main issues:
1. Someone without any domain knowledge can still get points. You say this is unfair, but unfair to whom? People without any knowledge are getting points, but those with knowledge are getting more points. And the grade thresholds tend to take this into account; the thresholds tend to run from about 50% (F) to 90% (A), so when you take the free points into account, that's effectively 0% knowledge is an F, while 80% knowledge is an A.
2. Negative points make people feel bad. The importance you should give this depends on the level. For elementary school, psychological issues are prominent, but by college, it should be less of an issue (although not entirely eliminated). This can be reduced by presenting students will mathematically equivalent situations with different formulations. For instance, with a 100 point test, you could tell students that they start the test with -50 points, but they get .5 points back every time they skip a question.
3. People who guess do better than those who leave questions they don't know blank. This does create a certain sort of "busy work" for students, having them choose between randomly filling in answers they don't know, or missing out on "free points". Note that this isn't addressed by including a "don't know" option, since they'll still have the busy work of marking "don't know" for all those questions, instead of just skipping them. There is, however, an advantage here where you're basically asking students "Okay, you don't know what the answer is here. But which do you think is more likely?"
There's no way you can fully address all three of these issues, so you're going to have to choose at least one to ignore. You can give half a point to each question left blank, which completely takes care of (3) and mostly takes care of (2) (students aren't getting negative points for getting questions wrong, but they are missing out on the points they could have gotten if they left them blank). You can leave things as they are, which fully addresses (2), but doesn't address (1) or (3). Or you can give negative points, which addresses (1) and (3), but not (2). | 7,242 | 30,666 | {"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} | 3.078125 | 3 | CC-MAIN-2020-45 | latest | en | 0.939576 |
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You are here: HomeBlue Series → Add & Subtract 2-B
# Math Mammoth Add & Subtract 2-B
95 pages
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Math Mammoth Add & Subtract 2-B is a continuation to the book Math Mammoth Add & Subtract 2-A. The goal of this book is to study addition and subtraction within 0-100, both mentally and in columns, especially concentrating on regrouping in addition (carrying) and in subtraction (borrowing).
The PDF version of this book is enabled for annotation. This means that if you prefer, your student can fill it in on the computer, using the typewriter and drawing tools in Adobe Reader version 9 or greater.
### Mental math
Mental math is important because it builds number sense. This book includes many lessons that practice mental math. For example, the child practices adding and subtracting two-digit numbers when one of the numbers is a whole ten (problems such as 30 + 14, or 66 − 20).
Also studied are problems such as 36 + 8 or 45 + 9. These problems use the idea of going over ten as in problems 6 + 8 and 5 + 9. The child knows that 6 + 8 fills the first ten and is four more than the ten. He/she will learn to use that fact when adding 36 + 8. The sum 36 + 8 fills the next whole ten (40), and is four more than that, or 44.
### Regrouping in tens
Simultaneously with this, the child learns adding two-digit numbers in columns, and regrouping with tens, or "carrying," which is illustrated and explained in detail with the help of visual models. These visual models take the place of base-ten blocks or other manipulatives. You are welcome to use manipulatives to get the idea across, if you prefer. The main concept to understand is that 10 ones are regrouped to form a new ten, and this new ten is written using a little "1" in the tens column.
### Regrouping in subtraction
The next lessons teach subtracting in columns. First we only deal with the easy problems where you don't need to regroup (borrow). Then the following lessons practice in detail the process of regrouping or borrowing. You can use either term with your child, or even choose not to use either if you feel it is confusing. You can alternatively use the phrase "breaking a ten into ten ones."
First, the lesson Regrouping practices breaking down a ten into ten ones because we cannot subtract from the ones. It is crucial that the child understands what happens here. Otherwise, he/she might end up only memorizing the procedure, and will probably at some point misremember how it was done. That is why this lesson deals with the regrouping process in detail with plenty of visual exercises.
If you notice that the child does not understand the concept of regrouping, he/she may need more practice with concrete manipulatives or visual exercises before proceeding.
### More mental math
After learning regrouping, we practice mental subtraction in three separate lessons. One of them expounds on several methods for mental subtracting. Another is about Euclid's game—a fun game that also practices subtraction of two-digit numbers.
The video below shows a similar idea as to what is in the book about regrouping (borrowing).
## Feedback
Thank you SO much for the free download of Data & Graphs!! I had purchased several blue series books for my son but hadn't gotten that yet and it was exactly what I needed next.
I've been so impressed with your curriculum. My son is autistic and non-speaking and your curriculum has been accessible to him. He does not do well with manipulatives like some other curricula use because he seems to think in numbers like I do. My son is 6 and using Place Value 1&2, Add/Subtract 2A and 2B right now.
Lynn
June 2018
You might also be interested in:
• Math Mammoth Add & Subtract 2-A—a worktext about "easy" additions and subtractions within 0-100, memorizing the basic addition facts of single-digit numbers, and word problems (grades 1-2).
• Math Mammoth Add & Subtract 3—a worktext about adding and subtracting 3-digit numbers with regrouping and with mental math, estimation, and word problems (grades 2-3).
• Math Mammoth Place Value 2—a worktext about three-digit numbers (place value), comparing, ordering, rounding, bar graphs, and pictographs (grade 2).
• Math Mammoth Place Value 3—a worktext about four-digit place value, adding & subtracting 4-digit numbers mentally, rounding, estimating, and order of operations (grade 3).
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You can buy Math Mammoth books at:
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GAMEMASTERX40 May 7, 2019 | 455 | 1,180 | {"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} | 3.171875 | 3 | CC-MAIN-2020-05 | latest | en | 0.766936 |
https://www.hse.ru/en/ba/data/courses/301408361.html | 1,627,204,914,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00532.warc.gz | 828,231,098 | 16,185 | • A
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Regular version of the site
# Probability Theory and Mathematical Statistics
2019/2020
ENG
Instruction in English
10
ECTS credits
Course type:
Compulsory course
When:
2 year, 1-4 module
### Course Syllabus
#### Abstract
This course is designed to introduce students to the basic ideas and methods of statistics as well as the application of statistical methods in econometrics, data science and the social sciences. This course provides some of the analytical tools that are required by advanced courses of data science and machine learning. This course provides students with experience in the methods and applications of statistics to a wide range of theoretical and practical situations. The course is taught in English. Prerequisites are Calculus (functions of several variables, partial derivatives, integrals, maximum of functions), and elements of Linear algebra (vectors, matrices, linear equations).
#### Learning Objectives
• This course introduces some of the basic ideas of theoretical statistics, emphasizing the applications of these methods and the interpretation of tables and results.
• We will introduce concepts and methods that provide the foundation for more specialised courses in statistics.
#### Expected Learning Outcomes
• Students will be able to routinely apply a variety of methods for explaining, summarising and presenting data and interpreting results clearly using appropriate diagrams, titles, and labels when required.
• Students will be able to summarise the ideas of randomness and variability and the way in which these link to probability theory to allow the systematic and logical collection of statistical techniques of great practical importance in many applied areas.
• Students will have a grounding in probability theory and some grasp of the most common statistical methods.
• Students will be able to recall a large number of distributions and be a competent user of their mass/density and distribution functions and moment generating functions.
• Students will be able to perform inference to test the significance of common measures such as means and proportions and conduct chi-squared tests of contingency tables.
• Students will be able to use simple linear regression and correlation analysis and know when it is appropriate to do so.
• Students will be able to apply and be competent users of standard statistical operators and be able to recall a variety of well-known distributions and their respective moments.
• Students will be able to explain the fundamentals of statistical inference and apply these principles to justify the use of an appropriate model and perform hypothesis tests in a number of different settings.
• Students will demonstrate an understanding that statistical techniques are based on assumptions and the plausibility of such assumptions must be investigated when analysing real problems.
• Students will be able to explain the principles of data reduction.
• Students will be able to choose appropriate methods of inference to tackle real problems.
#### Course Contents
• Data presentation.
• Elements of probability theory.
• Discrete random variables.
• Continuous random variables.
• Multivariate random variables.
• Conditional distributions.
• Limit theorems.
• The normal distribution and ideas of sampling.
• Populations and samples.
• Point estimation of parameters.
• Confidence intervals.
• Testing of statistical hypotheses.
• Linear regression.
• ANOVA.
• Experiment design.
#### Assessment Elements
• First semester Home assignments and Quizzes
• Second semester Home assignments and Quizzes
• FallMock (October Midterm)
At the end of each module the students sit a written exam.
• SpringMock (Spring Midterm)
At the end of each module the students sit a written exam.
• WinterExam (December Exam)
At the end of each module the students sit a written exam.
• University of London exams (May Exam)
• FinalExam (June Exam)
At the end of each module the students sit a written exam. Экзамен - письменный Прокторинг не требуется. Будет проходить в формате домашнего проекта. При не соблюдении сроков сдачи экзамена, оценка ставится ноль. В случае отсутствия студента по уважительной причине, вес экзамена перераспределяется на зимний экзамен в Декабре. В случае отсутствия по неуважительной причине или нарушение срока сдачи, оценка ноль.
#### Interim Assessment
• Interim assessment (2 module)
0.3 * FallMock (October Midterm) + 0.1 * First semester Home assignments and Quizzes + 0.6 * WinterExam (December Exam)
• Interim assessment (4 module)
0.4 * FinalExam (June Exam) + 0.05 * Second semester Home assignments and Quizzes + 0.15 * SpringMock (Spring Midterm) + 0.4 * University of London exams (May Exam)
#### Recommended Core Bibliography
• Statistics for business and economics, Newbold, P., Carlson, W. L., 2007
#### Recommended Additional Bibliography
• Bartoszyński, R., & Niewiadomska-Bugaj, M. (2008). Probability and Statistical Inference (Vol. 2nd ed). Hoboken, N.J.: Wiley-Interscience. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=219782
• Freund, J. E., Miller, I., & Miller, M. (2014). John E. Freund’s Mathematical Statistics with Applications: Pearson New International Edition (Vol. Eighth edition, Pearson new international edition). Essex, England: Pearson. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=1418305
• Hogg, R. V., McKean, J. W., & Craig, A. T. (2014). Introduction to Mathematical Statistics: Pearson New International Edition. Harlow: Pearson. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=1418145
• Hogg, R. V., Zimmerman, D. L., & Tanis, E. A. (2015). Probability and Statistical Inference, Global Edition (Vol. Ninth edition. Global edition). Boston: Pearson. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=1419274
• Larsen, R. J., & Marx, M. L. (2015). An introduction to mathematical statistics and its applications. Slovenia, Europe: Prentice Hall. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsbas&AN=edsbas.19D77756
• Lindgren, B. W. (1993). Statistical Theory (Vol. Fourth edition). Boca Raton, Florida: Routledge. Retrieved from http://search.ebscohost.com/login.aspx?direct=true&site=eds-live&db=edsebk&AN=1683924 | 1,480 | 6,451 | {"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-2021-31 | latest | en | 0.884069 |
https://www.thestudentroom.co.uk/showthread.php?t=6306370 | 1,659,947,193,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570767.11/warc/CC-MAIN-20220808061828-20220808091828-00395.warc.gz | 866,660,537 | 32,385 | # What have I done wrong in this Q? Congruent triangles GCSE
Announcements
#1
-We are given that triangle PQR is an isosceles triangle in which PQ = PR -Since the base angles of an isosceles triangle are equal, angle PQR = angle PRQ
-And we are given that angle MRQ = angle NQR
-And we know that QR = QR - shared side
-triangles QNR is congruent to triangles RMQ - ASA - congruent angle side angle
I got 2/4, what's wrong?
0
2 years ago
#2
(Original post by Obolinda)
-We are given that triangle PQR is an isosceles triangle in which PQ = PR -Since the base angles of an isosceles triangle are equal, angle PQR = angle PRQ
-And we are given that angle MRQ = angle NQR
-And we know that QR = QR - shared side
-triangles QNR is congruent to triangles RMQ - ASA - congruent angle side angle
I got 2/4, what's wrong?
It looks fine to me. Is this an online working checker that’s giving you 2/4?
1
#3
(Original post by Sir Cumference)
It looks fine to me. Is this an online working checker that’s giving you 2/4?
Yes, it was Mathswatch. But it's cool now, I finally got it to mark me correctly
0
2 years ago
#4
(Original post by Obolinda)
Yes, it was Mathswatch. But it's cool now, I finally got it to mark me correctly
I thought so. Often with Mathswatch getting it to understand that your working is correct is harder than the actual maths
Last edited by Notnek; 2 years ago
0
1 year ago
#5
how did you get the right answer?
0
1 year ago
#6
(Original post by Obolinda)
Yes, it was Mathswatch. But it's cool now, I finally got it to mark me correct
how did you get the correct answer?
0
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The Car That Passes
Age 7 to 11 Challenge Level:
What statements can you make about the car that passes the school gates at 11am on Monday? How will you come up with statements and test your ideas?
In the Playground
Age 5 to 11 Challenge Level:
What can you say about the child who will be first on the playground tomorrow morning at breaktime in your school?
Birds in the Garden
Age 5 to 11 Challenge Level:
This activity asks you to collect information about the birds you see in the garden. Are there patterns in the data or do the birds seem to visit randomly?
Inspector Morse
Age 11 to 14 Challenge Level:
You may like to read the article on Morse code before attempting this question. Morse's letter analysis was done over 150 years ago, so might there be a better allocation of symbols today?
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Age 5 to 11
Written for teachers, this article discusses mathematical representations and takes, in the second part of the article, examples of reception children's own representations.
Enriching Data Handling
Age 5 to 11
This article for teachers looks at some suggestions taken from the NRICH website that offer a broad view of data and ask some more probing questions about it.
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In this problem you will do your own poll to find out whether your friends think two squares on a board are the same colour or not.
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Terrariums
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Build a mini eco-system, and collect and interpret data on how well the plants grow under different conditions.
Observing the Sun and the Moon
Age 7 to 14 Challenge Level:
How does the time of dawn and dusk vary? What about the Moon, how does that change from night to night? Is the Sun always the same? Gather data to help you explore these questions. | 586 | 2,650 | {"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-2020-40 | latest | en | 0.905299 |
http://www.chegg.com/homework-help/questions-and-answers/figure-water-density-1000-flowing-pipeline-point-1-water-velocity-47-point-2-68above-point-q2441935 | 1,472,435,890,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982950764.62/warc/CC-MAIN-20160823200910-00028-ip-10-153-172-175.ec2.internal.warc.gz | 361,272,833 | 14,396 | In the figure, water (density 1000) is flowing in a pipeline. At point 1 the water velocity is 4.7. Point 2 is 6.8above point 1. The cross -sectional area of the pipe is 0.06at point 1 and 0.020at point 2. What is the pressure difference-between points 1 and 2?
## Answers
### Get this answer with Chegg Study
Practice with similar questions
Q:
In the figure, water (density 1000 kg/m^3 ) is flowing in a pipeline. At point 1 the water velocity is 4.7 m/s . Point 2 is 6.8 m above point 1. The cross -sectional area of the pipe is 0.06 m^2 at point 1 and 0.020 m^2 at point 2. What is the pressure difference p2 -p1 between points 1 and 2?
A: See answer
Q:
In the figure, water density 1000 kg/m^3 is flowing in a pipeline. At point 1 the water velocity is 4.7 m/s Point 2 is 6.8 m above point 1. The cross -sectional area of the pipe is 0.06 m^2 at point 1 and 0.020 m^2 at point 2. What is the pressure difference p1 - p2 between points 1 and 2?
A: See answer
Show more questions [+] | 322 | 988 | {"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-2016-36 | latest | en | 0.891438 |
https://fightinggg.github.io/hexonext/PX3W2O.html | 1,696,439,979,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511386.54/warc/CC-MAIN-20231004152134-20231004182134-00390.warc.gz | 275,963,149 | 10,642 | ###name path
###descripption You have a directed weighted graph with n vertexes and m edges. The value of a path is the sum of the weight of the edges you passed. Note that you can pass any edge any times and every time you pass it you will gain the weight.
Now there are q queries that you need to answer. Each of the queries is about the k-th minimum value of all the paths.
###input The input consists of multiple test cases, starting with an integer t (1≤t≤100), denoting the number of the test cases. The first line of each test case contains three positive integers n,m,q. $$(1≤n,m,q≤5∗10^4)$$
Each of the next m lines contains three integers ui,vi,wi, indicating that the i−th edge is from ui to vi and weighted wi.$$(1≤u_i,v_i≤n,1≤w_i≤10^9)$$
Each of the next q lines contains one integer k as mentioned above.$$(1≤k≤5∗10^4)$$
It's guaranteed that$Σn ,Σm, Σq,Σmax(k)≤2.5∗10^5$ and max(k) won't exceed the number of paths in the graph.
###output For each query, print one integer indicates the answer in line.
###sample input 1 2 2 2 1 2 1 2 1 2 3 4
###sample output 3 3
###hint 1->2 value :1
2->1 value: 2
1-> 2-> 1 value: 3
2-> 1-> 2 value: 3
###toturial 拓展一条边有两种方式，第一终点往外走 ， 第二相同起点的下一条边，这样做的前提是边要有序，从小到大排好
###code
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INFERENSI DATA UJI HIDUP TERSENSOR TIPE II BERDISTRIBUSI RAYLEIGH Widiharih, Tatik; Mardjiyati, Wiwin
MEDIA STATISTIKA Vol 1, No 2 (2008): Media Statistika
Publisher : Department of Statistics, Faculty of Science and Mathematics, Universitas Diponegoro
Abstract
Abstrak Analisis data uji tahan hidup merupakan salah satu teknik analisis statistika yang banyak digunakan di bidang industri dan kesehatan. Data waktu hidup dapat berupa data lengkap atau data tersensor, dan merupakan variabel random nonnegatif. Estimator titik untuk parameter q digunakan MLE, kemudian MLE tersebut digunakan untuk uji kecocokan distribusi Rayleig dengan metode Anderson Darling. Estimator titik uji tahan hidup meliputi rata-rata waktu kegagalan (mean time to failure / MTTF), fungsi kegagalan h(t), dan fungsi ketahanan S(t). Estimasi interval dilakukan dengan metode besaran pivot.  Kata kunci: data tersensor, MLE, rata-rata waktu kegagalan, fungsi                     kegagalan, fungsi ketahanan
MODEL PENILAIAN KREDIT MENGGUNAKAN ANALISIS DISKRIMINAN DENGAN VARIABEL BEBAS CAMPURAN BINER DAN KONTINU Mukid, Moch. Abdul; Widiharih, Tatik
MEDIA STATISTIKA Vol 9, No 2 (2016): Media Statistika
Publisher : Department of Statistics, Faculty of Science and Mathematics, Universitas Diponegoro
Abstract
Credit scoring models is an important tools in the credit granting process. These models measure the credit risk of a prospective client. This study aims to applied a discriminant model with mixed predictor variables (binary and continuous) for credit assesment. Implementation of the model use debitur characteristics data from a bank in Lampung Province which the used binary variables involve sex and marital status. Whereas, the continuous variables that was considered appropriate in the model are age, net income, and length of work. By using the data training, it was known that the misclassification of the model is 0.1970 and the misclassification of the testing data reach to 0.3753. Keywords: discriminant analysis, mixed variables, credit scoring
ANALISIS EFISIENSI BANK PERKREDITAN RAKYAT DI KOTA SEMARANG DENGAN PENDEKATAN DATA ENVOLEPMENT ANALYSIS Septianto, Hendi; Widiharih, Tatik
MEDIA STATISTIKA Vol 3, No 1 (2010): Media Statistika
Publisher : Department of Statistics, Faculty of Science and Mathematics, Universitas Diponegoro
Abstract
The research was conducted to measure rural banks (Bank Perkreditan Rakyat / BPR) efficiency level in Semarang city. The measurement was done using non parametric approach with Data Envolepment Analysis (DEA) method constant return to scale assumption (CCR model). The research was using all rural banks in Semarang (16 rural banks). The result indicated that 6 rural banks were efficient and 10 rurals banks were inefficient.  Keywords: CCR Model, Efficient, Rural Bank
METODE TAGUCHI UNTUK OPTIMALISASI PRODUK PADA RANCANGAN FAKTORIAL Wuryandari, Triastuti; Widiharih, Tatik; Anggraini, Sayekti Dewi
MEDIA STATISTIKA Vol 2, No 2 (2009): Media Statistika
Publisher : Department of Statistics, Faculty of Science and Mathematics, Universitas Diponegoro
Abstract
Taguchi methods represent the effort quality improvement which known as off-line quality control  method because the method design quality into every appropriate process and product. Taguchi methods is represent quality repair with attempt “new†methods, its meaning do dissimilar approach giving same belief storey by SPC (Statistical Proces Control), very effective in quality improvement as well as lessening expense of same. Fractional factorial design represent base from Taguchi method by fraction from factorial design. Fractional factorial with 4 factors and defining relations p = 2 is or 81 run become or 9 blocks with each blocks there are 9 run just eligible one block. The block name that is Orthogonal Array which lessen time and attemp fare. Orthogonal Array used to device of factorial attemp 3 level by 4 factors that is Orthogonal Array L9. Optimalitation product of factorial design  can be determinate with tables of anova, table of response and tables of Signal to Noise Ratio.  Keywords: Taguchi Methods, Signal to Noise Ratio, Orthogonal Array
ANALISIS KLASTER UNTUK SEGMENTASI PEMIRSA PROGRAM BERITA SORE STASIUN TV SWASTA Rosiatun, Aan; Widiharih, Tatik; Safitri, Diah
MEDIA STATISTIKA Vol 3, No 2 (2010): Media Statistika
Publisher : Department of Statistics, Faculty of Science and Mathematics, Universitas Diponegoro
Abstract
A procedure market segmentation is designing the market segmentation use the method of cluster k-means analyze which applied in process designing the market evening news audiences on tv chanels. The process of grouping audiences into each segment which formed, based on likeness of characteristic owned and it formed 3 market segment evening news audiences, that is audiences group who give low evaluation, audiences group who give enough evaluation, and audiences group who give high evaluation. Result from the market segmentation with case study at Pangkah district Tegal regency got first cluster is 25.2 %, second cluster is 46 %, and third cluster is 28.8 %. Marketing strategy can target be old > 20 years because it has members total of cluster is biggest. The result can be used by a television company to determine marketing strategy.  Keywords: Characteristic, Market Segmentation, Cluster K-Means Analysis
ANALISIS CLUSTER PADA KABUPATEN/KOTA DI JAWA TENGAH BERDASARKAN PRODUKSI PALAWIJA Safitri, Diah; Widiharih, Tatik; Wilandari, Yuciana; Saputra, Arsyil Hendra
MEDIA STATISTIKA Vol 5, No 1 (2012): Media Statistika
Publisher : Department of Statistics, Faculty of Science and Mathematics, Universitas Diponegoro
Abstract
Production of palawija, namely maize, cassava, sweet potato, peanut, soybean, and green bean is an important food crop in Central Java. In this article, districts/cities in Central Java are grouped into three groups based on the production of palawija so as to know which group have high potential the production of maize, cassava, sweet potato, peanut, soybean or green bean by using k-means cluster analysis. Cluster 1 consists of District Cilacap, Wonosobo, Magelang, Karanganyar, Semarang, Temanggung, Kendal, and Batang that have a high potential in maize production. Cluster 2 consists of District Banyumas, Purbalingga, Banjarnegara, Kebumen, Purworejo, Boyolali, Klaten, Sukoharjo, Sragen, Blora, Rembang, Pati, Kudus, Jepara, Demak, Pekalongan, Pemalang, Tegal, Brebes, Magelang City, Surakarta City, Salatiga City, Semarang City, Pekalongan City, and Tegal City that have a high potential in peanut production. Cluster 3 consist of District Wonogiri and Grobogan that have a high potential in soybean production, green bean production, cassava production, and sweet potato production
Rancangan D-Optimal Model Gompertz dengan Maple Widiharih, Tatik; Warsito, Budi
MEDIA STATISTIKA Vol 10, No 1 (2017): Media Statistika
Publisher : Department of Statistics, Faculty of Science and Mathematics, Universitas Diponegoro
Abstract
Gompertz model is used in many areas including biological growth studies, animal and husbandry, chemistry, and agricultural. Locally D-optimal designs for Gompertz models with three parameters is investigated. We used the Generalized Equivalence Theorem of Kiefer and Wolvowitz to determine D-optimality criteria. Tchebysheff system is used to decide that the D-optimal design is minimally supported design or nonminimally supported design. The result, D-optimal design for Gompertz model is minimally supported design with uniform weight on its support.Keywords:D-optimal, Generalized Equivalence Theorem, Tchebysheff System, Minimally Supported, Uniform Weight.
PENANGANAN MULTIKOLINEARITAS (KEKOLINEARAN GANDA) DENGAN ANALISIS REGRESI KOMPONEN UTAMA Widiharih, Tatik
MATEMATIKA Vol 4, No 2 (2001): JURNAL MATEMATIKA
Publisher : MATEMATIKA
Abstract
Multikolinearitas yang tinggi diantara peubah-peubah bebas, mengakibatkan pendugaan dengan metode kuadrat terkecil tidak dapat diandalkan, ditandai dengan koefisien regresi tidak nyata dan adanya multikolinieritas. Pendeteksian multikolinearitas dapat dilakukan secara informal salah satunya dengan koefisien korelasi lenear antar peubah bebas maupun dengan cara formal dengan faktor inflasi ragam. Analisis regresi komponen utama digunakan untuk menghilangkan multikolinieritas dan semua peubah bebas masuk dalam model, analisis regresi ini merupakan teknik analisis regresi yang dikombinasikan dengan teknik analisis komponen utama . Analisis komponen utama bertujuan menyederhanakan peubah yang diamati dengan mereduksi dimensinya , hal ini dilakukan dengan jalan menghilangkan korelasi diantara peubah-peubah melalui transformasi . Teknik analisis komponen utama dijadikan sebagai tahap analisis antara untuk memperoleh hasil akhir dalam analisis regresi.
MODEL LOGIT KUMULATIF UNTUK RESPON ORDINAL Raharjanti, Robiah Peni Raharjanti; Widiharih, Tatik
MATEMATIKA Vol 8, No 3 (2005): JURNAL MATEMATIKA
Publisher : MATEMATIKA
Abstract
Logit cumulative model is used to discribe the relationship between a response variable and one or more explanatory variables which response variable is of ordinal scale. To estimate the parameters, use maximum likelihood method with Newton Raphson iteration. Testing for the significance of the coefficients is done to fit the model. Test for overall significance of the variables in the model is performed by likelihood ratio test and test on individual coefficient is done using Wald’s test
ESTIMASI DATA HILANG PADA RANCANGAN ACAK KELOMPOK LENGKAP widiharih, Tatik
MATEMATIKA Vol 10, No 2 (2007): JURNAL MATEMATIKA
Publisher : MATEMATIKA
Abstract
Randomized complete block design is a design to reduce the residual error in an experiment by removing variability due to a known and controllable nuisance variable. Missing observations introduce a new problem into the analysis since treatments are no longer orthogonal to blocks, that is, every treatment does not occur in every block, There are two general approaches to the missing values problem. The first is an exact analysis, the second is an approaximate analysis in which the missing observations are estimated and usual analysis of variance is performed just as if the estimated observations were real data, with the error degrees of freedom reduced by the number of missing observations. In this paper was discussed the second approach with completely analysis. Bigger’s method is a simple method for estimating missing observations by using matrix approximation.  | 2,768 | 10,891 | {"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-2024-33 | latest | en | 0.283635 |
https://www.traditionaloven.com/tutorials/surface-area/convert-mm-diam-circle-to-sq-km-square-kilometer.html | 1,643,411,965,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00219.warc.gz | 1,094,393,529 | 17,352 | Convert ∅ 1mm to km2, sq km | circle one millimeter diam. to square kilometers
# area surface units conversion
## Amount: 1 circle one millimeter diam. (∅ 1mm) of area Equals: 0.00000000000079 square kilometers (km2, sq km) in area
Converting circle one millimeter diam. to square kilometers value in the area surface units scale.
TOGGLE : from square kilometers into ∅ one millimeter circles in the other way around.
## area surface from circle one millimeter diam. to square kilometer conversion results
### Enter a new circle one millimeter diam. number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other area surface measuring units - complete list.
How many square kilometers are in 1 circle one millimeter diam.? The answer is: 1 ∅ 1mm equals 0.00000000000079 km2, sq km
## 0.00000000000079 km2, sq km is converted to 1 of what?
The square kilometers unit number 0.00000000000079 km2, sq km converts to 1 ∅ 1mm, one circle one millimeter diam.. It is the EQUAL area value of 1 circle one millimeter diam. but in the square kilometers area unit alternative.
∅ 1mm/km2, sq km area surface conversion result From Symbol Equals Result Symbol 1 ∅ 1mm = 0.00000000000079 km2, sq km
## Conversion chart - ∅ one millimeter circles to square kilometers
1 circle one millimeter diam. to square kilometers = 0.00000000000079 km2, sq km
2 ∅ one millimeter circles to square kilometers = 0.0000000000016 km2, sq km
3 ∅ one millimeter circles to square kilometers = 0.0000000000024 km2, sq km
4 ∅ one millimeter circles to square kilometers = 0.0000000000031 km2, sq km
5 ∅ one millimeter circles to square kilometers = 0.0000000000039 km2, sq km
6 ∅ one millimeter circles to square kilometers = 0.0000000000047 km2, sq km
7 ∅ one millimeter circles to square kilometers = 0.0000000000055 km2, sq km
8 ∅ one millimeter circles to square kilometers = 0.0000000000063 km2, sq km
9 ∅ one millimeter circles to square kilometers = 0.0000000000071 km2, sq km
10 ∅ one millimeter circles to square kilometers = 0.0000000000079 km2, sq km
11 ∅ one millimeter circles to square kilometers = 0.0000000000086 km2, sq km
12 ∅ one millimeter circles to square kilometers = 0.0000000000094 km2, sq km
13 ∅ one millimeter circles to square kilometers = 0.000000000010 km2, sq km
14 ∅ one millimeter circles to square kilometers = 0.000000000011 km2, sq km
15 ∅ one millimeter circles to square kilometers = 0.000000000012 km2, sq km
Convert area surface of circle one millimeter diam. (∅ 1mm) and square kilometers (km2, sq km) units in reverse from square kilometers into ∅ one millimeter circles.
## Area units calculator
Main area or surface units converter page.
# Converter type: area surface units
First unit: circle one millimeter diam. (∅ 1mm) is used for measuring area.
Second: square kilometer (km2, sq km) is unit of area.
QUESTION:
15 ∅ 1mm = ? km2, sq km
15 ∅ 1mm = 0.000000000012 km2, sq km
Abbreviation, or prefix, for circle one millimeter diam. is:
∅ 1mm
Abbreviation for square kilometer is:
km2, sq km
## Other applications for this area surface calculator ...
With the above mentioned two-units calculating service it provides, this area surface converter proved to be useful also as a teaching tool:
1. in practicing ∅ one millimeter circles and square kilometers ( ∅ 1mm vs. km2, sq km ) measures exchange.
2. for conversion factors between unit pairs.
3. work with area surface's values and properties.
To link to this area surface circle one millimeter diam. to square kilometers online converter simply cut and paste the following.
The link to this tool will appear as: area surface from circle one millimeter diam. (∅ 1mm) to square kilometers (km2, sq km) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,091 | 3,969 | {"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-2022-05 | latest | en | 0.681303 |
https://www.magesblog.com/tags/graphics/ | 1,721,788,667,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00606.warc.gz | 750,382,127 | 6,384 | # graphics
## Using SVG graphics in blog posts
My traditional work flow for embedding R graphics into a blog post has been via a PNG files that I upload online. However, when I created a ‘simple’ graphic with only basic curves and triangles for a recent post, I noticed that the PNG output didn’t look as crisp as I expected it to be. So, eventually I used a SVG (scalable vector graphic) instead. Creating a SVG file with R could’t be easier; e.
## Visualising the predictive distribution of a log-transformed linear model
Last week I presented visualisations of theoretical distributions that predict ice cream sales statistics based on linear and generalised linear models, which I introduced in an earlier post. Theoretical distributions Today I will take a closer look at the log-transformed linear model and use Stan/rstan, not only to model the sales statistics, but also to generate samples from the posterior predictive distribution.
## Visualising theoretical distributions of GLMs
Two weeks ago I discussed various linear and generalised linear models in R using ice cream sales statistics. The data showed not surprisingly that more ice cream was sold at higher temperatures. icecream <- data.frame( temp=c(11.9, 14.2, 15.2, 16.4, 17.2, 18.1, 18.5, 19.4, 22.1, 22.6, 23.4, 25.1), units=c(185L, 215L, 332L, 325L, 408L, 421L, 406L, 412L, 522L, 445L, 544L, 614L) ) I used a linear model, a log-transformed linear model, a Poisson and Binomial generalised linear model to predict sales within and outside the range of data available.
## Adding mathematical notations to R plots
I have to admit that I find the plotmath expressions in R a little fiddly to annotate plots with mathematical notation. Apparently I am not the only one, but Stefano Meschiari did actually something about it. A few days ago his package latex2exp appeared on CRAN. The package provides the wonderful function latex2exp that translates LaTeX code into plotmath expressions. Brillant! All I have to remember is to escape the “"
## Unknown pleasures
Have I missed unknown pleasures in Python by focusing on R? A comment on my blog post of last week suggested just that. Reason enough to explore Python a little. Learning another computer language is like learning another human language - it takes time. Often it is helpful to start by translating from the new language back into the old one. I found a Python script by Ludwig Schwardt that creates a plot like this:
## GrapheR: A GUI for base graphics in R
How did I miss the GrapheR package? The author, Maxime Hervé, published an article about the package [1] in the same issue of the R Journal as we did on googleVis. Yet, it took me a package update notification on CRANbeeries to look into GrapheR in more detail - 3 years later! And what a wonderful gem GrapheR is. The package provides a graphical user interface for creating base charts in R.
## Review: Kölner R Meeting 13 December 2013
Last week’s Cologne R user group meeting was the best attended so far. Well, we had a great line up indeed. Matt Dowle came over from London to give an introduction to the data.table package. He was joined by his collaborator Arun Srinivasan, who is based in Cologne. Their talk was followed by Thomas Rahlf on Datendesign mit R (Data design with R). data.table Download slides Matt’s goal with the data.
## High resolution graphics with R
For most purposes PDF or other vector graphic formats such as windows metafile and SVG work just fine. However, if I plot lots of points, say 100k, then those files can get quite large and bitmap formats like PNG can be the better option. I just have to be mindful of the resolution. As an example I create the following plot: x <- rnorm(100000) plot(x, main="100,000 points", col=adjustcolor("black", alpha=0.2)) Saving the plot as a PDF creates a 5. | 916 | 3,836 | {"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 | 0.891596 |
https://www.opposingviews.com/category/frustrated-father-sends-sons-school-message-common-core-check-photo | 1,632,802,936,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00369.warc.gz | 931,116,887 | 42,582 | # Frustrated Father Sends Son's School A Message With This 'Common Core' Check (Photo)
Author:
Updated:
Original:
A parent’s recent prank at the expense of his child’s elementary school is bringing new laughs to an old string of jabs against the Common Core State Standards Initiative.
The clever Ohio parent decided to put his money where his mouth was when it came time for him to write his child’s school a check. The payment, made out to Millridge Elementary, replaces the dollar amounts with “common core math,” which has been criticized for overcomplicating simple equations, according to the International Journal Review.
Now, if the school tries to cash the check, someone will need to explain the common core style of solving math equations to the bank clerk. Since elementary math teachers are generally not expected to cash checks made out to the school, the person cashing the check may find the process as confusing as the bank cashier.
The Ohio parent’s action is only the most recent in a string of jabs made by parents in protest of Common Core.
Rare reports that model Britney Lace posted a math problem on Instagram and Twitter, writing: “This is the new common core first grade math... what happened to the old way?? This is stupid.”
Other models agreed. One wrote back: “It really is! It makes it almost impossible for kids to get help with their homework because no one understands it!”
Even parents who deal with mathematics as a part of their career have raised concerns. One became resentful when trying to help his child with his math homework. He vented his frustration by letting his child’s teacher know exactly how he felt about the method by which his child was told to solve the equation 427 – 316.
The problem asked the child to use the “number line” method to solve the equation and to write a short letter, reported International Journal Review.
“I have a Bachelor of Science Degree in Electronics Engineering which included extensive study in differential equations and other higher math applications. Even I cannot explain the Common Core Mathematics approach, nor get the answer correct,” the parent wrote. “In the real world, simplification is valued over complication.”
The parent then used the “stacking” method to solve the problem. Afterward, he wrote, “The answer is solved in under 5 seconds – 111. The process used is ridiculous and would result in termination if used.”
Sources: International Journal Review (2), Rare / Photo Credit: Twitter
undefined | 520 | 2,510 | {"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-2021-39 | latest | en | 0.970028 |
cclccorp.com | 1,526,907,152,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864186.38/warc/CC-MAIN-20180521122245-20180521142245-00192.warc.gz | 512,313,363 | 13,203 | # Understanding Dim Factors
Parcel carriers utilize dim factors to determine the dimensional weight of a shipment. The dim factor is the divisor utilized in determining the dimensional weight. Dimensional weight is determined by multiplying the length x width x height of the package, or pallet, and dividing by the dim factor. Parcel carriers invoice based on the greater of the actual weight or the dimensional weight. (Freight forwarders also use dimensional weight for air and time definite services.)
There are different dim factors for domestic and international shipments. In 2011, both FedEx and UPS reduced the standard dim factors. Domestic shipments used to have a dim factor of 194, but it is now 166. International shipments used to have a dim factor of 166, but it is now 139.
Not sure how the dim factor plays a role? Let’s evaluate. Let’s say you ship a 20x20x20 domestic package weighing 25 pounds; the dimensional weight based on a 194 dim factor was 42 pounds. Since the shipment is invoiced at the greater of the actual weight or the dimensional weight, and the dimensional weight was larger, you would have been invoiced based on a package weight of 42 pounds. However, now that the dim factor is 166, the dimensional weight would calculate at 49 pounds, which is almost double the actual weight of the package.
So, what does this mean to you? Do you have any options to reduce the application or effect of a dim factor? You will need to read a future post to find out! | 314 | 1,494 | {"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-2018-22 | latest | en | 0.949368 |
https://www.enotes.com/homework-help/what-57th-derivative-y-cos7x-588197 | 1,521,298,576,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645177.12/warc/CC-MAIN-20180317135816-20180317155816-00673.warc.gz | 807,265,494 | 9,775 | # What is the 57th derivative of y=cos7x?
Expert Answers
ishpiro | Certified Educator
Let's take a few derivatives of y = cos(7x) in order to observe the pattern they will create:
`y' = -7sin(7x)`
`y'' = -7*7cos(7x) = -7^2cos(7x)`
`y''' = -7^2*(-7)sin(7x) = 7^3sin(7x)`
`y^((4)) = 7^3*7cos(7x) = 7^4cos(7x)`
It is apparent that the power of 7 in front of the trigonometric function will be the same as the order of the derivative. The sign (plus or minus) and whether the function is sine or cosine will have the following order:
-sin
-cos
+sin
+ cos
and this pattern will then repeat all over. Since it will repeat every four derivatives, we can determine how derivative of nth order will look like by dividing n by 4 :
if the remainder is 1 (such as first derivative), the derivative will contain - sin
if the remainder is 2, it will contain -cos
if the remainder is 3, it will contain + sin
if there is no remainder(such as original function, or 4th derivative), it will contain +cos.
So, for the 57th derivative, 57 divides by 4 with the remainder 1. Therefore,
`y^((57)) = -7^57sin(7x)` .
The 57th derivative is `-7^57sin(7x)` .
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Popular Questions | 384 | 1,250 | {"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} | 4.1875 | 4 | CC-MAIN-2018-13 | latest | en | 0.877206 |
https://programmer.group/5eb9f7b82cf92.html | 1,708,798,997,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00665.warc.gz | 484,032,585 | 6,099 | # Machine Learning Basics - Neural Network (Debugging Optimization) (Random Initialization, Gradient Detection)
Keywords: network encoding
# 1: Random Initialization
When we use gradient descent or other advanced optimization algorithms, we need to select some initial values for the parameter theta.For advanced optimization algorithms, we assume by default that we have set the initial value for the variable theta:
Similarly, for the gradient descent method, we also need to initialize theta.Then we can minimize the cost function J by descending the gradient step by step, so how do we initialize theta?
## (1) Setting theta all to 0--not applicable in neural networks
Although this can be used in logistic regression.But it does not play a role in the actual training of neural networks.
If we initialize all theta zeros (all equal):
The inputs for each hidden unit are essentially the same:
So the partial derivatives are the same.
When we just set a step, after each parameter update.The weight of each hidden unit is also consistent.
This means that this neural network can't compute a good function. When we have many hidden cells, all the hidden cells are calculating the same characteristics and taking the same function as input.- High redundancy.So no matter how many output units follow, you end up with only one feature, which prevents the neural network from learning something interesting
## (3) Code implementation
``````def rand_initiation(l_in,l_out): #For adjacent layers
w = np.zeros((l_out,l_in+1)) #Need to add the paranoid unit weight of the previous layer
eps_init = 0.12 #Near 0, keep at-εreachεBetween
w = np.random.rand(l_out,l_in+1)*2*eps_init-eps_init
return w``````
``````w = rand_initiation(3,5)
print(w.shape)
print(w)``````
Or use the following method for random initialization:
``````def debug_initialize_weights(fan_in,fan_out):
w = np.zeros((fan_out,1+fan_in))
w = np.sin(np.arange(w.size)).reshape(w.shape) / 10 #Using sin guarantees that values are between (-1, 1) and divided by 10, the values are initialized between (-0.1, 0.1)
return w``````
``````w = debug_initialize_weights(3,5)
print(w.shape)
print(w)``````
# 2: Gradient Test--Ensure the correctness of the reverse propagation code we implemented
## (1) Problems Existing in Reverse Propagation
Previously, we learned how to use the forward and reverse propagation algorithms in the neural network to calculate derivatives, but the back propagation algorithm has many details, is complex to implement, and has a bad feature:
When reverse propagation works with gradient descent or other algorithms, there may be some undetectable errors, meaning that although the cost function J appears to be decreasing, the final result may not be the optimal solution.The error will be one magnitude higher than if there were no bugs.And we probably don't know that the problem we're having is caused by Bug.
Almost all of these problems can be solved.Gradient tests are performed almost at the same time as all reverse propagation or other similar gradient descent algorithms are used.He will ensure that forward propagation is completely correct and that backward propagation is already complete.
Can be used to verify that the code you write actually correctly calculates the derivative of the cost function J
## (2) Definition method for solving derivatives
When theta is a vector, we need to check the partial derivatives:
## (3) algorithm implementation
n represents the theta parameter dimension, and we perform the derivation above for each theta_i.We then test whether the derivatives we use to define them are close (several decimal differences) or equal to those we use reverse propagation.Then we can be sure that the implementation of reverse propagation is correct.J( theta) can be optimized very well
## (4) Code implementation
``````def compute_numerial_gradient(cost_func,theta): #Used to calculate gradients
perturb = np.zeros(theta.size) #Keeping every calculated gradient, theta_j of only one dimension subtracts epsilon
e = 1e-4
for i in range(theta.size):
perturb[i] = e
J_1 = cost_func(theta-perturb)
J_2 = cost_func(theta+perturb)
perturb[i] = 0 #Modified location data needs to be restored
``````def check_gradients(lamda):
#Because a real training set has too much data, here we randomly generate some small-scale data for statistical testing.
input_layer_size = 3 #input layer
hidden_layer_size = 5 #Hidden Layer
num_labels = 3 #output layer---Number of Classifications
m = 5 #Number of data sets
#Initialization Weight Parameter
theta1 = debug_initialize_weights(input_layer_size,hidden_layer_size)
theta2 = debug_initialize_weights(hidden_layer_size,num_labels)
#Similarly: Initialize some training set X and tag value y
X = debug_initialize_weights(input_layer_size - 1,m) #The training set is m rows, input_layer_size column
y = 1 + np.mod(np.arange(1,m+1),num_labels) #Remaining Category 1->num_labels
encoder = OneHotEncoder(sparse=False) #sparse=True Represents the format of the encoding, defaulting to True，Is the sparse format, specified False Then you don't need to toarray() Yes
y_onehot = encoder.fit_transform(np.array([y]).T) #y needs to be a column vector
#Expand parameters
theta_param = np.concatenate([np.ravel(theta1),np.ravel(theta2)])
#1.Using derivative definition to derive
def cost_func(theta_p):
return cost(theta_p,input_layer_size,hidden_layer_size,num_labels,X,y_onehot,lamda) #Cost Function Partial Derivative J( theta)
``check_gradients(0) #Modifying lamda is still close`` | 1,254 | 5,585 | {"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-2024-10 | latest | en | 0.799125 |
https://www.gnu.org/software/mit-scheme/documentation/stable/mit-scheme-ref/Advanced-Operations-on-Weight_002dBalanced-Trees.html | 1,713,004,063,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816587.89/warc/CC-MAIN-20240413083102-20240413113102-00301.warc.gz | 739,000,814 | 4,388 | #### 11.7.3 Advanced Operations on Weight-Balanced Trees
In the following the size of a tree is the number of associations that the tree contains, and a smaller tree contains fewer associations.
procedure: wt-tree/split< wt-tree bound
Returns a new tree containing all and only the associations in wt-tree that have a key that is less than bound in the ordering relation of the tree type of wt-tree. The average and worst-case times required by this operation are proportional to the logarithm of the size of wt-tree.
procedure: wt-tree/split> wt-tree bound
Returns a new tree containing all and only the associations in wt-tree that have a key that is greater than bound in the ordering relation of the tree type of wt-tree. The average and worst-case times required by this operation are proportional to the logarithm of the size of wt-tree.
procedure: wt-tree/union wt-tree-1 wt-tree-2
Returns a new tree containing all the associations from both trees. This operation is asymmetric: when both trees have an association for the same key, the returned tree associates the datum from wt-tree-2 with the key. Thus if the trees are viewed as discrete maps then `wt-tree/union` computes the map override of wt-tree-1 by wt-tree-2. If the trees are viewed as sets the result is the set union of the arguments. The worst-case time required by this operation is proportional to the sum of the sizes of both trees. If the minimum key of one tree is greater than the maximum key of the other tree then the worst-case time required is proportional to the logarithm of the size of the larger tree.
procedure: wt-tree/intersection wt-tree-1 wt-tree-2
Returns a new tree containing all and only those associations from wt-tree-1 that have keys appearing as the key of an association in wt-tree-2. Thus the associated data in the result are those from wt-tree-1. If the trees are being used as sets the result is the set intersection of the arguments. As a discrete map operation, `wt-tree/intersection` computes the domain restriction of wt-tree-1 to (the domain of) wt-tree-2. The worst-case time required by this operation is proportional to the sum of the sizes of the trees.
procedure: wt-tree/difference wt-tree-1 wt-tree-2
Returns a new tree containing all and only those associations from wt-tree-1 that have keys that do not appear as the key of an association in wt-tree-2. If the trees are viewed as sets the result is the asymmetric set difference of the arguments. As a discrete map operation, it computes the domain restriction of wt-tree-1 to the complement of (the domain of) wt-tree-2. The worst-case time required by this operation is proportional to the sum of the sizes of the trees.
procedure: wt-tree/subset? wt-tree-1 wt-tree-2
Returns `#t` iff the key of each association in wt-tree-1 is the key of some association in wt-tree-2, otherwise returns `#f`. Viewed as a set operation, `wt-tree/subset?` is the improper subset predicate. A proper subset predicate can be constructed:
```(define (proper-subset? s1 s2)
(and (wt-tree/subset? s1 s2)
(< (wt-tree/size s1) (wt-tree/size s2))))
```
As a discrete map operation, `wt-tree/subset?` is the subset test on the domain(s) of the map(s). In the worst-case the time required by this operation is proportional to the size of wt-tree-1.
procedure: wt-tree/set-equal? wt-tree-1 wt-tree-2
Returns `#t` iff for every association in wt-tree-1 there is an association in wt-tree-2 that has the same key, and vice versa.
Viewing the arguments as sets, `wt-tree/set-equal?` is the set equality predicate. As a map operation it determines if two maps are defined on the same domain.
This procedure is equivalent to
```(lambda (wt-tree-1 wt-tree-2)
(and (wt-tree/subset? wt-tree-1 wt-tree-2
(wt-tree/subset? wt-tree-2 wt-tree-1)))
```
In the worst case the time required by this operation is proportional to the size of the smaller tree.
procedure: wt-tree/fold combiner initial wt-tree
This procedure reduces wt-tree by combining all the associations, using an reverse in-order traversal, so the associations are visited in reverse order. Combiner is a procedure of three arguments: a key, a datum and the accumulated result so far. Provided combiner takes time bounded by a constant, `wt-tree/fold` takes time proportional to the size of wt-tree.
A sorted association list can be derived simply:
```(wt-tree/fold (lambda (key datum list)
(cons (cons key datum) list))
'()
wt-tree))
```
The data in the associations can be summed like this:
```(wt-tree/fold (lambda (key datum sum) (+ sum datum))
0
wt-tree)
```
procedure: wt-tree/for-each action wt-tree
This procedure traverses wt-tree in order, applying action to each association. The associations are processed in increasing order of their keys. Action is a procedure of two arguments that takes the key and datum respectively of the association. Provided action takes time bounded by a constant, `wt-tree/for-each` takes time proportional to the size of wt-tree. The example prints the tree:
```(wt-tree/for-each (lambda (key value)
(display (list key value)))
wt-tree))
```
procedure: wt-tree/union-merge wt-tree-1 wt-tree-2 merge
Returns a new tree containing all the associations from both trees. If both trees have an association for the same key, the datum associated with that key in the result tree is computed by applying the procedure merge to the key, the value from wt-tree-1 and the value from wt-tree-2. Merge is of the form
```(lambda (key datum-1 datum-2) …)
```
If some key occurs only in one tree, that association will appear in the result tree without being processed by merge, so for this operation to make sense, either merge must have both a right and left identity that correspond to the association being absent in one of the trees, or some guarantee must be made, for example, all the keys in one tree are known to occur in the other.
These are all reasonable procedures for merge
```(lambda (key val1 val2) (+ val1 val2))
(lambda (key val1 val2) (append val1 val2))
(lambda (key val1 val2) (wt-tree/union val1 val2))
```
However, a procedure like
```(lambda (key val1 val2) (- val1 val2))
```
would result in a subtraction of the data for all associations with keys occuring in both trees but associations with keys occuring in only the second tree would be copied, not negated, as is presumably be intent. The programmer might ensure that this never happens.
This procedure has the same time behavior as `wt-tree/union` but with a slightly worse constant factor. Indeed, `wt-tree/union` might have been defined like this:
```(define (wt-tree/union tree1 tree2)
(wt-tree/union-merge tree1 tree2
(lambda (key val1 val2) val2)))
```
The merge procedure takes the key as a parameter in case the data are not independent of the key. | 1,601 | 6,799 | {"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-2024-18 | latest | en | 0.939205 |
https://www.freecodecamp.org/forum/t/sum-all-primes-right-answers-wont-pass/142167 | 1,553,162,973,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202510.47/warc/CC-MAIN-20190321092320-20190321114320-00517.warc.gz | 778,671,019 | 5,187 | # Sum All Primes, right answers, won't pass
Sum All Primes, right answers, won't pass
0
#1
code outputs answer, but won’t register as a pass
function isPrime(num) {
var all = [];
for (var count = 0; count <= num; count++){
all.push(count);}
all = all.filter(function(number) {
for (var i = 2; i <= Math.sqrt(number); i++) {
if (number % i === 0) return false; }
return true; });
return all.slice(2, all.length -0).reduce(function(a, b){
return a + b;
});
}
isPrime(10);```
firefox
Your Browser User Agent is: ```Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:54.0) Gecko/20100101 Firefox/54.0```.
https://www.freecodecamp.org/challenges/sum-all-primes
#2
The original challenge function name was sumPrimes, so that is what the tests are looking for to test against. Just rename your isPrime function to sumPrimes and all should be good.
#3
Wow! Thank you so much!
I never would have guessed that you can’t change the function name. I rewrote that code so many times, I can’t even remember why I renamed it…
Thank you Thank you Thank you!
#4
You probably had a second function at one point called isPrime that was used in your sumPrimes, but then you combined the two and forgot to rename it. I have done something like that before.
#5
I think that is exactly what I did.
#6
I looked at your solution again and noticed a tiny change you could make to the following for loop declaration which would simplify your return statement at the end of the function.
for (var count = 0; count <= num; count++){
If you change the above 0 to a 2, then your return statement could simplify to the following without the slice.
return all.reduce(function(a, b){
return a + b;
});
If you were to use an isPrime function, then you could avoid creating an array of all the numbers 2 thru num and then having to filter out the primes, by just looping through the numbers 2 thru num with the following code. It avoids the extra step up creating an all numbers array and just creates a primes array.
function sumPrimes(num) {
function isPrime(number) {
for (var i = 2; i <= Math.sqrt(number); i++) {
if (number % i === 0) return false;
}
return true;
}
var primes = [];
for (var count = 2; count <= num; count++){
if (isPrime(count)) primes.push(count);
}
return primes.reduce(function(a, b) {return a + b});
}
#7
Thank you! That does simplify things. | 612 | 2,356 | {"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-2019-13 | latest | en | 0.86478 |
https://jamesshieh.wordpress.com/2012/10/ | 1,619,191,480,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594808.94/warc/CC-MAIN-20210423131042-20210423161042-00078.warc.gz | 418,800,579 | 13,727 | # Monthly Archives: October 2012
Often people shouldn’t reinvent the wheel, however in my learning at Hacker School I feel that for certain areas it is likely important to try and think up a solution for a well known (and most likely solved) problem and see what you come up with.
Today, I was thinking of a problem that would appear in a possible project that Sitong suggested. The project involved creating a type of “heat” map which would represent the minutes to a destination given walking and public transportation options. In mulling over this in my head it presented one of the classic problems in algorithms… the shortest path problem. I have little experience with algorithms so I thought this would be a great problem to come up with a personal solution before checking for the optimal solution. Hopefully my solution will turn out to be the optimal.
In this problem, you are given a graph with nodes and connections with values, in this case the values for the connections represent minutes of travel between each node. The nodes will represent travel options between two points on a map and the goal is to find the shortest path from an initial starting node and a chosen end node.
In my haze of sleep I began to think of solutions for a quick optimization of finding the shortest path and began to envision the graph node network as a net that is malleable. If one picks the starting node and ending node in each hand and stretches, the shortest path will appear as it will be the first path that can no longer stretch between the two hands (ignoring of course all physical tangling of the net). The problem I face is to quickly calculate the distance from every node on a map to another node, and if this act of “lifting” up the node from the net instantly reveals the shortest node it may be somehow translatable in code.
My currently train of thought brings me to trees. Given a graph network, it may be possible to translate that graph into a tree and stretch it out until you find the shortest path between the two nodes…
~ Hacking in Progress 😛 ~
After doing some reading and implementing Dijkstra’s algorithm in Ruby, I see the parallels of my visual thoughts and the mathematical explanations. Creating a even set of sub-nodes and traveling down all parallel paths until you hit nodes is analogous to lifting a net and having all connections lift off a table until node by node the solution appears. I find this connection between my thoughts and the derivation fascinating and really do wonder if there are other things out there that I could find novel solutions to. Looking forward to more late-night epiphanies and other sparks of inspiration!
Getting started with this book isn’t quite easy for someone who hasn’t worked too much in setting up a machine in these databases. I’m going to log my steps so that in the future people (or myself) can quickly reference my steps and see what worked for me!
Using Ubuntu 12.04
First step is to install the postgresql packages which comes in two pieces, the client and the dev package with the additional add-ons that the book requires:
``````sudo apt-get install postgresql-client
sudo apt-get install postgresql-contrib``````
Postgresql creates a postgres user which manages the server if you run it locally. Ideally you should create a user for yourself:
``````# Opens psql as the postgres superuser
sudo -u postgres psql
# Creates a user for yourself
createuser --superuser [user]
``````
Exit the psql dashboard and return to terminal. Now I can create and log into my server by using the psql command and install the extensions:
``````# Create DB
createdb book
# Open the newly created DB
psql book
# Install the extensions needed for the exercises in 7 Databases in 7 Days
create extension tablefunc
create extension dict_xsyn
create extension fuzzystrmatch
create extension pg_trgm
create extension tablefunc``````
Another link which was quite helpful for getting psql setup and running:
https://help.ubuntu.com/community/PostgreSQL | 839 | 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} | 3.359375 | 3 | CC-MAIN-2021-17 | latest | en | 0.952299 |
https://www.thestudentroom.co.uk/showthread.php?t=6446408 | 1,596,710,331,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736902.24/warc/CC-MAIN-20200806091418-20200806121418-00182.warc.gz | 873,834,826 | 36,299 | # Finding the Determinants for Matrix
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#1
The question: https://gyazo.com/005997e8069185fb56235fe15ef10e0e
I did manage to successfully find the answers for AB, |A| and |B| and |AB| but I'm having some difficulty in properly re-arranging the answer for |AB| to equal |A||B|.
For |A||B| I got (ad-bc)(eh-fg) and for |AB| it was (ae+bg)(cf+ch) - (af+bh)(ce+dg)
Not exactly sure how I'm supposed to completely simplify it to get |A||B|
0
3 months ago
#2
(Original post by OJ Emporium)
The question: https://gyazo.com/005997e8069185fb56235fe15ef10e0e
I did manage to successfully find the answers for AB, |A| and |B| and |AB| but I'm having some difficulty in properly re-arranging the answer for |AB| to equal |A||B|.
For |A||B| I got (ad-bc)(eh-fg) and for |AB| it was (ae+bg)(cf+ch) - (af+bh)(ce+dg)
Not exactly sure how I'm supposed to completely simplify it to get |A||B|
The only thing you can realistically do is to expand and simplify both versions and show that they are the same. However, you have a small error in you expression for |AB|. Check the term (cf+ch) - it should be slightly different.
1
#3
(Original post by Pangol)
The only thing you can realistically do is to expand and simplify both versions and show that they are the same. However, you have a small error in you expression for |AB|. Check the term (cf+ch) - it should be slightly different.
Most certainly did not see that error, thank you for the heads up
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## Income is an example of what variable? Yahoo Answers
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Variables and Scales of Measurement The Oxford Math Center | 1,386 | 6,950 | {"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-2020-45 | latest | en | 0.885039 |
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1. Feb 22, 2016
chrispypatt
My homework problem is as follows:
Consider the Laplace transform shown below.
(4s3+15s2+s+30)/(s2+5s+6)
a. What is the value of f(t=0) and f(t=∞)? Use the initial and final value theorems.
b. Find the inverse transform f(t). Use this expression to find f(t=0) and f(t=∞) and compare with the result of part a).
I know to find the final value, using FVT, limt->inff(t)=lims->0sF(s), but I am given the stipulation that the poles must be in the left side of the domain. My book words the definition of IVT and FVT by saying IVT is only valid if f(t) has no impulse functions (the function must be rational) and that for FVT, we must add the rule about the poles. I am confused because everywhere online, including how my professor explained it, is that the impulse rule applies to ONLY IVT and the poles ONLY applies to FVT.
If I went with how my prof taught, Neither theorem can be applied to the problem but if I go by the book's wording, IVT does not apply but FVT shows the final value of f(t) will go to 0.
If someone more knowledgeable with these two theorems could clarify this, that would be helpful! Thanks in advance.
Last edited by a moderator: Feb 28, 2016
2. Feb 28, 2016
Greg Bernhardt
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
3. Feb 28, 2016
Staff: Mentor
In future posts, please don't delete the homework template -- its use is required.
Presumably, the above is F(s); i.e., $\mathcal{L}[f(t)]$.
I'm not familiar with either of these theorems, and I don't understand how this stipulation fits in.
What I do know is that $\mathcal{L}[f'(t)] = sF(s) - f(0)$, where F(s) is as above.
For the b) part, I would carry out long division, and for the remainder, use partial fractions to write it as a sum of two fractions.
4. Feb 28, 2016
Ray Vickson
A lot of the on-line literature on these matters is confusing, and sometimes even self-contradictory (proving the result in one section, then showing a couner-example in another section). For the FVT, a nice article that sets it out properly and clearly is
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LREtools
divconq
find solutions of "divide and conquer" recurrence equations
Calling Sequence divconq(problem)
Parameters
problem - problem statement or RESol
Description
• Solves divide and conquer recurrences, meaning those of the form $Af\left({R}^{n}k\right)+Bf\left({R}^{m}k\right)=0$, where R is either numeric or a name, m and n are integers, and where A and B are independent of k.
• See the help page for LREtools[REcreate] for the definition of the format of a problem.
Examples
> $\mathrm{with}\left(\mathrm{LREtools}\right):$
> $\mathrm{prob}≔\mathrm{REcreate}\left(\left\{y\left(nk\right)=2y\left(k\right)\right\},y\left(k\right),\varnothing \right)$
${\mathrm{prob}}{≔}{\mathrm{RESol}}{}\left(\left\{{y}{}\left({n}{}{k}\right){-}{2}{}{y}{}\left({k}\right){=}{0}\right\}{,}\left\{{y}{}\left({k}\right)\right\}{,}{\varnothing }{,}{\mathrm{INFO}}\right)$ (1)
> $\mathrm{divconq}\left(\mathrm{prob}\right)$
${y}{}\left({1}\right){}{{k}}^{\frac{{\mathrm{ln}}{}\left({2}\right)}{{\mathrm{ln}}{}\left({\mathrm{n~}}\right)}}$ (2)
> $\mathrm{divconq}\left(f\left({r}^{3}k\right)=2f\left({r}^{2}k\right),f\left(k\right),\varnothing \right)$
${f}{}\left({1}\right){}{{k}}^{\frac{{\mathrm{ln}}{}\left({2}\right)}{{\mathrm{ln}}{}\left({r}\right)}}$ (3) | 503 | 1,421 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "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-2021-21 | longest | en | 0.333781 |
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2004.SU/Hall/Assignment11/Assignment11.html | 1,544,866,344,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826842.56/warc/CC-MAIN-20181215083318-20181215105318-00157.warc.gz | 154,009,890 | 3,375 | Assignment 11
Polar Equations
by Jeff Hall
Problem 1
Investigate the following equation:
It produces this graph:
When a and b are equal, and k is an integer, this equation
generates one textbook version of the "n-leaf rose."
In this example, a, b, and k all equal 1, and t goes from -pi to pi.
How does this equation compare to a more basic equation:
When k=1, we get this graph.
When k=2, we get this graph:
At this point, it looks like the number of blue petals is determined by k-squared.
Let's try k=3 just to be sure:
Hmm.... Now the number of blue petals equals k.
Let's try k=4:
The number of petals now equals k times 2.
Let's try 5:
Okay, now we are starting to see a more defined pattern.
When k is odd, the number of petals equals k.
Now let's try one more even number...try k=6:
Since there are 12 petals, we can safely safe that k*2 works when k is even.
Let's change the original equation's k value, as well. Let k=2 for both:
Now try k=3 for both:
This time, the number of red petals equals k every time.
Let's try k=4 just to be sure:
That's it!
What would happen if we changed these equations from cos to sin?
We get this when k=1:
It rotated the image 90 degrees counterclockwise.
Problem 2
Investigate the following equations varying a,b,c, and k:
When a,b,c, and k=1, we get this graph:
Let's manipulate k first. Let k=2:
Now let k=3:
And k=4:
Now let's put k back to 1 and try a=2:
The variable a seems to determine the size of each circle.
Let's try a=3 to be sure:
Yes, that's what is happening.
Notice that the red and blue circles have a as their radius.
Also note that the black line is getting a steeper slope.
Now put a back to 1 and change b to equal 2:
Now try b=3:
And finally, try b=4:
Changing b obviously enlarges the green and purple shapes,
but notice the distortion in their shapes, as well.
They are becoming more circular.
Also note that the black line's slope is becoming more horizontal.
Now let's manipulate c. Change b back to 1 and let c=2:
Now try c=3:
Finally, try c=4:
As you can see, changing the c only moves the black line.
Notice the x- and y- intercepts equal c.
Problem 3
Investigate the following equations for different values of p:
The parameter k is called the "eccentricity" of these conics.
When p=1 and k=1, we get this:
When p=2, we get this:
When p=3, we get this:
As you can see, each parabola passes through -p and p on either the x- or y-axis.
Now let k=2 and p=1:
Now try p=2:
It seems that when k>1, the parabolas become hyperbolic figures.
Increasing the value of p just widens these figures apart. Notice that some of
these figures now cross the x- and y-axes at -2p and +2p.
Now let's try k<1. Change p back to 1 and let k=0.5
Now change p=2:
Finally, change p=3:
As you can see, when k<1, the parabolic figures begin to look
like circles. Increases in p make the figures grow larger.
What happens when k=0? Absolutely nothing, since each equation is r=0.
What happens when k<0? The graphs look identical to the respective positive k graph.
The figures may be inverted, but you can't tell by the way they look on a graph.
Problem 4
Investigate the following equations:
These equations generate the following graph:
Notice that the blue figure is rotated in comparison to the red figure.
The purple equation generated a hyperbola.
Now let's see what happens when you place a 2 in front of each theta:
Changing the cos to sin in the original first two equations gives us this:
Again, the red and blue figures appear to be similar, just rotated differently.
Problem 5
Try some interesting graphs.
Try this one when a is small, such as a=0.1:
We get this graph:
Now put a=1 and add these equations:
Since this graph has become cluttered (a hazard of geometry), let's try
graphing these equations on a brand new graph: | 1,013 | 3,867 | {"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} | 4.75 | 5 | CC-MAIN-2018-51 | latest | en | 0.932612 |
https://stackcodereview.com/finding-the-minimum-number-of-moves-to-achieve-equal-array-elements-2/ | 1,686,297,439,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655446.86/warc/CC-MAIN-20230609064417-20230609094417-00211.warc.gz | 581,051,042 | 13,548 | Finding the minimum number of moves to achieve equal array elements
Posted on
Problem
Long story short, the other day I discovered the site LeetCode and started solving some of its problems until I stumbled onto this one.
I’ll paste the description here:
Given a non-empty integer array of size n, find the minimum number of
moves required to make all array elements equal, where a move is
incrementing n – 1 elements by 1.
Input:
``````[1,2,3]
``````
Output:
``````3
``````
Explanation:
Only three moves are needed (remember each move increments two
elements): [1,2,3] => [2,3,3] => [3,4,3] => [4,4,4]
The problem is, the solution I am trying to submit always exceeds the allowed processing time limit and I feel like I’ve reached a dead end. Can anyone suggest a way to improve my code?
`````` static void Main()
{
int res = MinMoves(new int[] { 1, 2, 3 });
}
public static int MinMoves(int[] nums)
{
if (nums.Length < 2)
return 0;
int moves = 0;
int diff = 0;
int min = 0;
int max = 0;
int maxIX = 0;
do
{
min = nums.Min();
max = nums.Max();
maxIX = Array.IndexOf(nums, max);
diff = Math.Abs(max - min);
moves += diff;
for (int i = 0; i < nums.Length; i++)
if (i != maxIX)
nums[i] += diff;
} while (min != max);
return moves;
}
``````
Solution
The problem can be re-formulated into a much easier-to-conceptualize form.
Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n – 1 elements by 1.
where a move is decrementing a single element by 1
Do you see why these are the same? Suppose we start with `1, 2, 3`. Adding `1, 1, 0` produces `2, 3, 3`. Adding `0, 0, -1` produces `1, 2, 2`, which is effectively the same thing. However many moves it takes to make `2, 3, 3` all equal is surely the same number of moves as it takes to make `1, 2, 2`, all equal, or `0, 1, 1`, or `1000, 1001, 1001`, or whatever. Where you start is irrelevant.
Once you know that the problem is actually “how many single decrements does it take to get all the elements equal?” the answer is easily seen. There is never any point in decrementing the smallest element, and every element will have to eventually equal the smallest element. Therefore the answer is: it’s the sum of the differences of every element from the minimal element.
That’s a trivial program in C#:
``````var min = nums.Min();
return nums.Select(x => x - min).Sum();
``````
Or alternatively, in a single line:
``````return nums.Sum() - nums.Length * nums.Min();
``````
(Though as noted in the comments, this second version can overflow. Consider doing all the math in longs or BigIntegers if you’re worried about overflow cases.)
Firstly, identify hidden loops, such as `Min`, `Max` and `IndexOf`, and merge those together – as per Paparazzi’s answer. There’s no need to count to `n` 3 times in a row.
The question asks you to find the number of moves, not to actually execute the number of moves. You’ve already optimized this by working with `diff` rather than adding just `1` each time. However, this is an optimization for the best-case scenario (e.g., when diff > 1), the worst case performance remains `n^2` with respect to the length of `nums`.
Further optimization could be achieved by removing the lowest duplicated numbers – though this once again is a best-case optimization. (e.g., when your input has few unique numbers).
Finally, rethink the whole thing. Below is what I came up with. It’s based on permutations and triangular numbers. I had drafted a proof of sorts, but my lunch break was too short to finish it. Initial unit tests show the output to be identical to your implementation, but requiring just `O(n log(n))` time complexity.
``````public static int GerardMinMoves(int[] nums) {
// Sort in place. Assumed O(n log(n)) complexity
Array.Sort(nums);
int moves = 0;
int sum = 0;
for(int i = 1; i < nums.Length; ++i)
{
// Difference beween current and previous.
int delta = (nums[i] - nums[i - 1]);
// Add current delta to previous deltas
sum += delta;
// Triangular number, accumulate.
moves += sum;
}
return moves;
}
``````
You don’t need `Math.Abs`. You are making 3 passes to get `min`, `max`, and `maxIX` – do it in one pass and stop when `min == max`.
``````min = max = num[0];
maxIX = 0;
for (int i = 1; i < nums.Length; i++)
{
if(nums[i] > max)
{
max = nums[i];
maxIX = i;
}
else if (nums[i] < min)
min = nums[i];
}
if (min != max)
{
....
}
``````
The answer may be calculated without actually processing. | 1,221 | 4,521 | {"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} | 3.875 | 4 | CC-MAIN-2023-23 | latest | en | 0.848479 |
http://mathcurve.com/courbes2d.gb/spiricdeperseus/spiricdeperseus.shtml | 1,563,653,034,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526670.1/warc/CC-MAIN-20190720194009-20190720220009-00512.warc.gz | 98,296,870 | 2,678 | SPIRIC OF PERSEUS
Perseus, 2nd century BC: Greek scholar.
Reduced Cartesian equation: with A > B, i.e. . Bicircular quartic, rational if C = 0, having real points only if . Polar equation: .
The spirics of Perseus are the bicircular quartics with a symmetry centre. Therefore, they are the cyclic curves with respect to a centred conic, the deferent (corresponding to the "initial curve" in the link) of which is this conic. In other words, they are the envelopes of the circles the centres of which describe a centred conic, and such that the centre of the conic has a constant power with respect to these circles.
Case where 0 < B < A are fixed, and C is variable. the curves inside the two circles, in the left and right parts, are not planar sections of a real torus. Case where B < 0 < A are fixed, and C is variable.
Historically, these curves were defined as the sections of a torus by a plane parallel to its axis; but to obtain all the real curves given above, one has to consider complex tori.
For a torus with centre O, axis Oz, with major and minor radii a and b, cut by the plane parallel to Oz located at distance d from O, we get, in a frame with origin the projection of O on this plane, the Cartesian equation above with: .
This comes from the equation: of these curves.
Note that the torus above is real only if .
Spirics of Perseus of a ring torus Spirics of Perseus of a spindle torus
When , i.e. db (distance of the plane to the axis equal to the minor radius), we get the Cassinian ovals, which reduce to the lemniscate of Bernoulli when C = 0, i.e. a = 2 b.
When C = 0, i.e. (plane tangent inside the torus), we get the Booth curves (or Hippopedes of Proclus), which also reduce to the lemniscate of Bernoulli when a = 2 b.
The limit case A = B (case where the torus is reduced to a sphere) gives circles.
The spirics of Perseus are also the isoptic curves of the centred conics.
Picture by Mr Konieczka and Ms Gautherin, Laboratory of Mechanics of the Department of Mechanical Engineering of École normale supérieure (ENS) de Cachan Disk composed of a birefringent diametrically charged material observed under polarised monochromatic light. The black lines - the isochromatic lines - are the loci of the points for which the difference of the 2 principal constraints is constant. | 604 | 2,325 | {"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} | 3.6875 | 4 | CC-MAIN-2019-30 | latest | en | 0.918825 |
https://convertoctopus.com/638-cubic-inches-to-quarts | 1,643,231,607,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304961.89/warc/CC-MAIN-20220126192506-20220126222506-00470.warc.gz | 238,824,466 | 7,939 | ## Conversion formula
The conversion factor from cubic inches to quarts is 0.017316017316055, which means that 1 cubic inch is equal to 0.017316017316055 quarts:
1 in3 = 0.017316017316055 qt
To convert 638 cubic inches into quarts we have to multiply 638 by the conversion factor in order to get the volume amount from cubic inches to quarts. We can also form a simple proportion to calculate the result:
1 in3 → 0.017316017316055 qt
638 in3 → V(qt)
Solve the above proportion to obtain the volume V in quarts:
V(qt) = 638 in3 × 0.017316017316055 qt
V(qt) = 11.047619047643 qt
The final result is:
638 in3 → 11.047619047643 qt
We conclude that 638 cubic inches is equivalent to 11.047619047643 quarts:
638 cubic inches = 11.047619047643 quarts
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 quart is equal to 0.090517241379112 × 638 cubic inches.
Another way is saying that 638 cubic inches is equal to 1 ÷ 0.090517241379112 quarts.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that six hundred thirty-eight cubic inches is approximately eleven point zero four eight quarts:
638 in3 ≅ 11.048 qt
An alternative is also that one quart is approximately zero point zero nine one times six hundred thirty-eight cubic inches.
## Conversion table
### cubic inches to quarts chart
For quick reference purposes, below is the conversion table you can use to convert from cubic inches to quarts
cubic inches (in3) quarts (qt)
639 cubic inches 11.065 quarts
640 cubic inches 11.082 quarts
641 cubic inches 11.1 quarts
642 cubic inches 11.117 quarts
643 cubic inches 11.134 quarts
644 cubic inches 11.152 quarts
645 cubic inches 11.169 quarts
646 cubic inches 11.186 quarts
647 cubic inches 11.203 quarts
648 cubic inches 11.221 quarts | 512 | 1,892 | {"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-2022-05 | latest | en | 0.764212 |
https://research.chalmers.se/publication/252740 | 1,582,795,146,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146665.7/warc/CC-MAIN-20200227063824-20200227093824-00402.warc.gz | 513,811,132 | 7,824 | # A Novel Approach for Ellipsoidal Outer-Approximation of the Intersection Region of Ellipses in the Plane Artikel i vetenskaplig tidskrift, 2018
In this paper, a novel technique for tight outer-approximation of the intersection region of a finite number of ellipses in 2-dimensional space is proposed. First, the vertices of a tight polygon that contains the convex intersection of the ellipses are found in an efficient manner. To do so, the intersection points of the ellipses that fall on the boundary of the intersection region are determined, and a set of points is generated on the elliptic arcs connecting every two neighbouring intersection points. By finding the tangent lines to the ellipses at the extended set of points, a set of half-planes is obtained, whose intersection forms a polygon. To find the polygon more efficiently, the points are given an order and the intersection of the half-planes corresponding to every two neighbouring points is calculated. If the polygon is convex and bounded, these calculated points together with the initially obtained intersection points will form its vertices. If the polygon is non-convex or unbounded, we can detect this situation and then generate additional discrete points only on the elliptical arc segment causing the issue, and restart the algorithm to obtain a bounded and convex polygon. Finally, the smallest area ellipse that contains the vertices of the polygon is obtained by solving a convex optimization problem. Through numerical experiments, it is illustrated that the proposed technique returns a tighter outer-approximation of the intersection of multiple ellipses, compared to conventional techniques, with only slightly higher computational cost.
Intersection of ellipses
Computational geometry
Intersection of half-planes
Convex optimization
Ellipsoidal outer approximation
Minimum volume enclosing ellipsoid
## Författare
#### Siamak Yousefi
McGill University
#### Xiao-Wen Chang
McGill University
#### Henk Wymeersch
Chalmers, Elektroteknik, Kommunikations- och antennsystem, Kommunikationssystem
#### Benoit Champagne
McGill University
#### Godfried Toussaint
New York University Abu Dhabi
#### Computational Optimization and Applications
0926-6003 (ISSN) 1573-2894 (eISSN)
Vol. 69 2 383-402
#### Styrkeområden
Informations- och kommunikationsteknik
#### Ämneskategorier
Kommunikationssystem
#### DOI
10.1007/s10589-017-9952-3
2019-09-09 | 544 | 2,447 | {"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} | 2.578125 | 3 | CC-MAIN-2020-10 | latest | en | 0.863111 |
https://www.indiabix.com/electronics-and-communication-engineering/digital-electronics/113004 | 1,632,385,127,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00264.warc.gz | 852,329,573 | 6,155 | # Electronics and Communication Engineering - Digital Electronics
16.
The total number of fundamental products of three variables is
A. 3 B. 5 C. 8 D. 16
Explanation:
No answer description available for this question. Let us discuss.
17.
As per Boolean algebra, inputs can be interchanged in
A. OR gates B. AND gates C. both OR and AND gates D. none of the above
Explanation:
No answer description available for this question. Let us discuss.
18.
Medium scale integration refers to ICs with
A. more than 12 but less than 30 gates on the same chip B. more than 50 gates on the same chip C. more than 20 but less than 100 gates on the same chip D. more than 12 but less than 100 gates on the same chip
Explanation:
No answer description available for this question. Let us discuss.
19.
The decimal equivalent of the hexadecimal number (3 E 8)16 is
A. 1000 B. 982 C. 768 D. 323
Explanation:
No answer description available for this question. Let us discuss.
20.
Four memory chips of 16 x 4 size have their address buses connected together. This system will be of size
A. 64 x 4 B. 16 x 16 C. 32 x 8 D. 256 x 1 | 293 | 1,134 | {"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} | 2.578125 | 3 | CC-MAIN-2021-39 | longest | en | 0.832698 |
https://www.nagwa.com/en/worksheets/609137204081/ | 1,529,926,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867666.97/warc/CC-MAIN-20180625111632-20180625131632-00573.warc.gz | 867,763,550 | 26,760 | # Worksheet: Adding and Subtracting Decimals
Q1:
Calculate the following: .
Q2:
Complete .
Q3:
Calculate .
Q4:
Calculate .
Q5:
Calculate the following: .
Q6:
Calculate .
Q7:
Calculate the following: .
Q8:
What is the result of ?
Q9:
Calculate .
Q10:
Calculate .
Q11:
Calculate .
Q12:
.
Q13:
A lion ate 11.4 kg of meat one day and 14.6 kg of meat the next day. How many kilograms of meat did it eat across the two days?
• A2.6 kg
• B26 kg
• C16.64 kg
• D3.2 kg
Q14:
One basket contains 24.1 kg of apples. Another basket contains 21.5 kg of apples. What is the total weight of the apples in the baskets?
• A456 kg
• B51.81 kg
• C45.6 kg
• D518.15 kg
• E2.6 kg
Q15:
On a dive, some oceanographers saw a humpback whale which was 13.7 m long, and a killer whale which was 6.85 m long. How much longer was the humpback whale than the killer whale?
Q16:
Between 2017 and 2050, the world’s population is expected to increase by 2.3 billion. Given that the world’s population in 2017 was 7.5 billion, what is the predicted population in 2050?
Q17:
Karen wanted to buy some snacks and juice. If she had \$25 and bought some fruits for \$14.35 and a gallon of apple juice for \$2.22, find how much she would have after paying.
Q18:
Mason bought a tie for \$14.51 and a shirt for \$23.38. Determine how much he will get back if he pays with a \$50 bill.
Q19:
If Michael scored a total 400 points or more on his math tests for the first five weeks of the quarter, his father would take him to a basketball game. Michael scored 70.9, 85.4, 86.6, 86.3, and 72.4 on the five tests. By how many points did Michael earn the reward?
Q20:
Use the table below to find out how many more people there are per square mile in Iowa than in Colorado.
State Number of People Per Square Mile Colorado Iowa Arkansas Oklahoma 41.5 52.4 51.3 50.3
Q21:
The table shows the top three times for the women’s 100-meter butterfly event at the 2004 Summer Olympics.
SwimmerCountryTime (s)
Petria ThomasAustralia57.47
Otylia JedrzejczakPoland57.84
Inge de BruijnNetherlands57.99
What is the difference between Petria Thomas’ time and Otylia Jedrzejczak’s time?
Q22:
Sophia goes jogging every day. If she jogged for 12 hours this week, complete the given table.
Day Number of Hours Monday Tuesday Wednesday Thursday Friday Saturday Sunday 0.5 1 3 1.5 1 2 | 684 | 2,365 | {"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} | 4.15625 | 4 | CC-MAIN-2018-26 | longest | en | 0.932366 |
https://avia_en_ru.academic.ru/7192/design_algorithm | 1,709,031,150,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00254.warc.gz | 114,253,172 | 11,773 | # design algorithm
design algorithm
алгоритм проектирования; алгоритм синтеза <системы управления>
Авиасловарь. . 2004.
### Смотреть что такое "design algorithm" в других словарях:
• Parks-McClellan filter design algorithm — The Parks McClellan filter design algorithm is a digital signal processing algorithm developed by James H. McClellan (now at Georgia Tech) and Thomas W. Parks (now at Cornell University) while McClellan was a graduate student working with Parks… … Wikipedia
• Algorithm design — is a specific method to create a mathematical process in solving problems. Applied algorithm design is algorithm engineering.Algorithm design is identified and incorporated into many solution theories of operation research, such as dynamic… … Wikipedia
• Algorithm engineering — is a combination of theoretical algorithm design with real world data. By taking an algorithm and combining it with a hardware device connected to the real world, you are able to more accurately verify and validate the algorithm results and… … Wikipedia
• Algorithm — Flow chart of an algorithm (Euclid s algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers a and b in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≤ A yields yes… … Wikipedia
• Design Patterns — Not to be confused with the concept of a Design pattern. Design Patterns: Elements of Reusable Object Oriented Software … Wikipedia
• Algorithm examples — This article Algorithm examples supplements Algorithm and Algorithm characterizations. = An example: Algorithm specification of addition m+n =Choice of machine model:This problem has not been resolved by the mathematics community. There is no… … Wikipedia
• algorithm — algorithmic, adj. /al geuh ridh euhm/, n. a set of rules for solving a problem in a finite number of steps, as for finding the greatest common divisor. [1890 95; var. of ALGORISM, by assoc. with Gk arithmós number. See ARITHMETIC] * * * Procedure … Universalium
• Design Quality Indicator — The Design Quality Indicator (DQI) is a toolkit to measure the design quality of buildings. Development of DQI was started by the Construction Industry Council in 1999[1] and the toolkit was launched as an online resource for the UK construction… … Wikipedia
• Computer-automated design — Design Automation usually refers to electronic design automation. Extending Computer Aided Design (CAD), automated design and Computer Automated Design (CAutoD) [1][2][3] are more concerned with a broader range of applications, such as automotive … Wikipedia
• Memetic algorithm — Memetic algorithms (MA) represent one of the recent growing areas of research in evolutionary computation. The term MA is now widely used as a synergy of evolutionary or any population based approach with separate individual learning or local… … Wikipedia
• Genetic algorithm — A genetic algorithm (GA) is a search heuristic that mimics the process of natural evolution. This heuristic is routinely used to generate useful solutions to optimization and search problems. Genetic algorithms belong to the larger class of… … Wikipedia
### Поделиться ссылкой на выделенное
##### Прямая ссылка:
Нажмите правой клавишей мыши и выберите «Копировать ссылку» | 717 | 3,302 | {"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-2024-10 | latest | en | 0.7971 |
https://mathematica.stackexchange.com/questions/274535/replacing-only-variables-in-specific-locations-with-replace-all | 1,723,147,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00720.warc.gz | 308,707,290 | 42,568 | # Replacing only variables in specific locations with Replace All
Suppose I have an example expression defined like this:
line = R0*f[R0,x] + R0^2*42*D[g[R0,x],x]
I would like to replace R0 with r, but only in the arguments of functions.
In this example I would like to use a replacement rule line /. { ??? } in order to obtain
R0*f[r,x] + R0^2*42*D[g[r,x],x]
What replacement rule ??? do I need to use here?
In my case, line is hundreds of terms long and it would be very cumbersome to not automate this. (If it helps, all functions always take the same list of arguments.)
• By "functions" do you mean "user-defined functions"? Commented Oct 10, 2022 at 16:11
• If you have a list of the relevant functions, it might be easier to work from that direction. For example, you might add DownValues directly to (for example) f & g that handle R0. Or you might build a helper function, say CleanR0 that replaces f[R0,...] with f[r,...] (and similarly with all relevant functions). Commented Oct 10, 2022 at 16:14
• @lericr Yes, the functions are user defined, but it is a long list and I would prefer not having to think about it. Commented Oct 10, 2022 at 17:05
Clear["Global*"]
line = R0*f[R0, x] + R0^2*42*D[g[R0, x], x]
line /. {p_Symbol[R0, a_] /; Context[p] == "Global" -> p[r, a],
Derivative[k__][p_Symbol][R0, b_] /; Context[p] == "Global" ->
Derivative[k][p][r, b]}
$$\text{R0} f(r,x)+42 \text{R0}^2 g^{(0,1)}(r,x)$$
• This gives warning/message. Screen shot !Mathematica graphics V 13.1. What is the purpose for checking that context is global in your method? Commented Oct 10, 2022 at 15:56
• Add parentheses, i.e., line /. {(p_[R0, a_] /; Context[p]) == "Global" :> p[r, a], Derivative[k__][p_][R0, b_] /; Context[p] == "Global" -> Derivative[k][p][r, b]} Commented Oct 10, 2022 at 16:00
• @Nasser, I used context "Global" since Context[Times] and for other built-in functions is "System" and this way I can avoid conversions in other parts. May be there is a more elegant way to do this. The messages box was in the background so I didn't see it, although it computed. Thanks to Bob Hanlon for the correction.
– Syed
Commented Oct 10, 2022 at 16:16
• @Syed Thank you, that is perfect! The trick with Context is great, I hadn't thought of that. Commented Oct 10, 2022 at 17:16
Something like this maybe?
line = R0*f[R0, x] + R0^2*42*D[g[R0, x], x];
MyFunctions = {f, g, Derivative[0, 1][g]};
Clean[fn_][R0, args___] := fn[r, args];
line /. {fn : Alternatives @@ MyFunctions :> Clean[fn]}
This assumes that R0 will have no OwnValues. Also, you might be able to automate the generation of MyFunctions if there will be several different patterns of derivatives (or other functions).
The following uses Syed's idea for checking the context, and it generalizes a bit (you could build out that generalization further).
line = R0*f[R0, x] + R0^2*42*D[g[R0, x], x];
R0 /: Derivative[params___][fun_Symbol][R0, args___] :=
Derivative[params][fun][r, args] /; Context[f] == "Global";
R0 /: f_Symbol[R0, args___] := f[r, args] /; Context[f] == "Global";
line
This assumes that you don't need the original form with the explicit R0s, because the up values for R0 will be applied automatically.
• Thanks, it is a very nice solution, I just find the other one a bit more practical because I do not need to define MyFunctions`. The generalization looks great though! (Although it has the drawback of influencing my entire notebook globally.) Commented Oct 10, 2022 at 17:18 | 1,044 | 3,487 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.879139 |
https://drugtruthaustralia.org/and-pdf/2684-physic-questions-and-answers-pdf-510-949.php | 1,642,765,560,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00283.warc.gz | 286,666,890 | 8,637 | Saturday, June 5, 2021 3:25:08 AM
# Physic Questions And Answers Pdf
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Published: 05.06.2021
Students often face hard-to-solve and mind-numbing physics problems, that cause a lot of distress into the studying process.
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## Physics Quiz Questions & Answers - Set 09
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#### Attachments
Practice with these Physics Multiple Choice Questions with Answers or general knowledge questions and answers for better results in competitive exams and all state-related exams. And it will be very easy for you to revise it. World GK Questions on Animals. Books And Authors Gk Questions. One liner GK Questions. Static GK Questions.
Что? - Сьюзан встала, глаза ее сверкали. Стратмор подавил желание встать с ней. Он многое знал об искусстве ведения переговоров: тот, кто обладает властью, должен спокойно сидеть и не вскакивать с места. Он надеялся, что она сядет. Но она этого не сделала. - Сьюзан, сядь.
Не коснувшись краев, он вытащил из нее ключ Медеко. - Поразительно, - пробурчал он, - что сотрудникам лаборатории систем безопасности ничего об этом не известно. ГЛАВА 47 - Шифр ценой в миллиард долларов? - усмехнулась Мидж, столкнувшись с Бринкерхоффом в коридоре. - Ничего. - Клянусь, - сказал .
Насколько мне известно, ты сотрудник АНБ. - Ненадолго, - буркнул Хейл. - Не зарекайся.
Она подошла к окну, вертя бумагу перед глазами, чтобы найти лучший угол для падения лунного света. - Мидж… пошли. Это личный кабинет директора. - Это где-то здесь, - пробормотала она, вглядываясь в текст. - Стратмор обошел фильтры.
Как и большинство талантливых программистов, Танкада сделался объектом настойчивого внимания со стороны АНБ. От него не ускользнула ирония ситуации: он получал возможность работать в самом сердце правительства страны, которую поклялся ненавидеть до конца своих дней. Энсей решил пойти на собеседование.
Quien es. Кто он. - Понятия не имею.
Вопреки широко распространенному мнению о том, что такой компьютер создать невозможно, АНБ осталось верным своему девизу: возможно все; на невозможное просто требуется больше времени. Через пять лет, истратив полмиллиона рабочих часов и почти два миллиарда долларов, АН Б вновь доказало жизненность своего девиза. Последний из трех миллионов процессоров размером с почтовую марку занял свое место, все программное обеспечение было установлено, и керамическая оболочка наглухо заделана. ТРАНСТЕКСТ появился на свет. Хотя создававшийся в обстановке повышенной секретности ТРАНСТЕКСТ стал плодом усилий многих умов и принцип его работы не был доступен ни одному человеку в отдельности, он, в сущности, был довольно прост: множество рук делают груз легким.
Стратмор был уверен, что предусмотрел . | 1,622 | 6,935 | {"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} | 2.625 | 3 | CC-MAIN-2022-05 | latest | en | 0.917453 |
https://www.flexiprep.com/Expected-Exam-Questions/Physics/Class-11/CBSE-Class-11-Physics-Thermodynamics-Problems.html | 1,611,318,655,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529331.99/warc/CC-MAIN-20210122113332-20210122143332-00098.warc.gz | 785,671,047 | 16,806 | # CBSE Class 11-Physics: Thermodynamics Problem's: Questions
Doorsteptutor material for CBSE is prepared by world's top subject experts: fully solved questions with step-by-step explanation- practice your way to success.
Question:
One mole of an ideal monoatomic gas is taken round the cyclic process ABCDA as shown in figure.
A) Work done by the gas
b. Heat absorbed by the gas in AB and BC
c. Heat in process CD
d. Find the temperature at C and D
e. Maximum temperature attained by the gas during the cycle
f. Net change in the internal energy and the heat
Solution:
For monoatomic gas
a) Work done by the gas Area enclosed by the curve
Solution continued:
Heat absorbed the gas in
Heat absorbed the gas in BC
Solution continued
Heat rejected in DA
Now for the cycle process
So,
Solution continued:
From diagram and
Solution Continued:
Max temperature will be on the slope CD
Equation of Slope CD as Coordinated system
Taking the values for C and D, we get
Now
So,
For max
should be zero
So
Net heat
Max temperature will be on the slope CD | 250 | 1,073 | {"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-2021-04 | longest | en | 0.923095 |
http://www.klaus-boening.de/se%20amp%20math.htm | 1,718,897,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00022.warc.gz | 46,817,482 | 5,609 | Nashville SE 20 - Some basic Math on the SuperTriode Concept Basic calculations require one component to start with. I chose the output transformer as the most crucial component. For the project the single ended version of a Lundahl LL1693 seemed to be a good match regarding quality, pricing, technical data and size (fig. 1). The SE version of the 4.5kg transformer has a core air gap of 450µm. Recommended idle current is 230mA with all primary windings in series. Primary inductance is 16H. With a secondary connection B (4Ohm, see Lundahl datasheet) the reflected primary impedance is 1kOhm. Inherent Output Power Limit The recommended transformer idle current of 230mA allows a theoretical current swing from 0mA to 460mA peak to peak (p-p). According to I 2 * R = P a simplified power calculation neglecting transformer losses reveals a maximum output power of (0.23A / sqrt2) 2 * 1000Ohm = 26.45W ≈ 26.5W (equation 1) which is the inherent output power limit, not too much but fairly above the requested minimum of 20W. Hence I stayed with the LL1693 for further calculations. By the way, the push-pull version of the LL1693 is capable to handle around 300W output power. Output Voltage Swing (fig. 2) Considering a „worst case“ estimation of 30W output power and 1kOhm primary impedance the power output voltage swing p-p at the primary coils according to U 2 / R = P is sqrt 2 * (sqrt(30W * 1000Ohm)) * 2 ≈ 490V p-p (equation 2) With an estimated minimum residual voltage of 100V across the triode the power supply voltage would have to be around 100V + 490V / 2 = 345V ≈ 350V (equation 3) Thus the maximum voltage peak expected to be at the plate is 100V + 490V = 590V (equation 4) Adding a comfortable safety margin (speakers unintentionally disconnected) the circuit components should withstand 900V to 1000V peak voltage, nothing special for a tube but pretty much for a transistor. Power Dissipation From the above equation 3 and a given idle current of 0.23A the total power dissipation is 350V * 0.23A = 80.5W. Assuming an idle current of 30mA for the triode and an idle current of 200mA for the transistor plus a plate / collector voltage of 350 V power dissipation at the plate will be 10.5 W and power dissipation of the transistor will be around 70W. 350V and 200mA idle current must be within the DC safe operating area of the transistor with a comfortable safety margin. This will rule out BJTs because of their inherent secondary breakdown. That leaves power MosFets. Power MosFets meeting the specifications above are available for instance the IXFN32N120P. However, these MosFets are designed for industrial high power switching applications. They are quite non-linear at low drain currents and have large gate capacitances. Below some data on the IXFN32N120P are discussed (fig. 3). Fig. 3: Maximum ratings and transconductance of the IXFN32N120P. At low currents transconductance g m is a linear function of I drain . Estimated transconductance g m at 230mA is around 0.6S. U max of the IXFN32N120P is 1200V, P d is 1000W, gate capacitance C iss is 21nF. With a load resistance of 1kOhm the effective (or Miller) capacitance is C M = C iss * ( 1 + gain) = C iss * (1+ (gm * 1000)) = 0.021µF + (1+(0.6*1000)) = 12.62µF (equation 5) 12µF (!) dynamic input capacitance is difficult to drive without compromising high frequency response even without the essential gate stopper resistor. Thus the IXFN32N120P or similar devices with even lower gate capacitances are hardly suitable for a SE power amplifier according to the SuperTriode concept. Core Magnetization The SuperTriode configuration (fig. 4) suffers from an additional flaw: Even at large signals the plate current remains almost constant. Thus in a single ended design the current through the tube just magnetizes the output transformer‘s iron core but does not contribute to the output power. If 200mA idle current are assumed for the transistor, the inherent power limit calculation from above equation 1 drops to 20W and will be less if transformer losses are taken into consideration. Conclusion With regard to the required minimum specifications of the transistor it may be concluded that the SuperTriode concept as illustrated in figure 4 is not suitable for SE power amplifiers if an output power above a few watts is requested despite its beauty and simplicity. On the other hand the concept is far too fascinating to be discarded . . . .
Fig. 1: Lundahl LL1693 output transformer
Fig. 4: SuperTriode concept.
Fig. 2: estimated voltages at the plate / collector of SuperTriode circuit discussed
super triode, vinyl, audio, analog, single ended, SE, power amplifier, hybrid, tube, KT66, 6SN7, ECC88, Mosfet, Lundahl, MC phono stage, preamplifier, MM, MC, moving coil, moving magnet, LL1693, LL1667, LL9226, LL1933, RIAA, folded cascode, 2CS5200, MAT12, 2N3810, LL1660S, IXFN32N120P, balancing amplifier | 1,302 | 4,981 | {"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-2024-26 | latest | en | 0.852359 |
https://www.geeksforgeeks.org/program-to-print-solid-and-hollow-square-patterns/ | 1,579,951,429,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672440.80/warc/CC-MAIN-20200125101544-20200125130544-00548.warc.gz | 881,146,779 | 28,747 | # Program to print solid and hollow square patterns
For any given number n, print Hollow and solid Squares and Rhombus made with stars(*).
Examples:
```Input : n = 4
Output :
Solid Square:
****
****
****
****
Hollow Square:
****
* *
* *
****```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
1. Solid Square : Solid Square is easiest among all given patterns. To print Solid square with n rows, we should use two loops iterating n times both. Where the outer loop is used for numbers of rows and the inner loop is used for printing all stars in a particular row.
2. Hollow Square : Hollow Square requires a little bit improvisation. Here we should again use two loops, where 1st and last row must have all n-stars and remaining rows have stars only in 1st and last column.
## C++
`// C++ program to print ` `// hollow and solid square patterns ` ` ` `#include ` `using` `namespace` `std; ` ` ` `// Function for hollow square ` `void` `hollowSquare(``int` `rows) ` `{ ` ` ``int` `i, j; ` ` ``for` `(i=1; i<=rows; i++) ` ` ``{ ` ` ``// Print stars for each solid rows ` ` ``if` `(i==1 || i==rows) ` ` ``for` `(j=1; j<=rows; j++) ` ` ``cout << ``"*"``; ` ` ` ` ``// stars for hollow rows ` ` ``else` ` ``for` `(j=1; j<=rows; j++) ` ` ``if` `(j==1 || j==rows) ` ` ``cout << ``"*"``; ` ` ``else` ` ``cout << ``" "``; ` ` ` ` ``// Move to the next line/row ` ` ``cout << ``"\n"``; ` ` ``} ` `} ` ` ` `// Function for Solid square ` `void` `solidSquare(``int` `rows) ` `{ ` ` ``int` `i, j; ` ` ``for` `(i=1; i<=rows; i++) ` ` ``{ ` ` ``// Print stars after spaces ` ` ``for` `(j=1; j<=rows; j++) ` ` ``cout << ``"*"``; ` ` ` ` ``// Move to the next line/row ` ` ``cout << ``"\n"``; ` ` ``} ` ` ` `} ` ` ` `// Utility program to print all patterns ` `void` `printPattern(``int` `rows) ` `{ ` ` ``cout << ``"\nSolid Square:\n"``; ` ` ``solidSquare(rows); ` ` ` ` ``cout << ``"\nHollow Square:\n"``; ` ` ``hollowSquare(rows); ` `} ` ` ` `// Driver program ` `int` `main() ` `{ ` ` ``int` `rows = 5; ` ` ``printPattern (rows); ` ` ``return` `0; ` `} `
## Java
`// Java program to print ` `// hollow and solid square patterns ` ` ` `class` `GFG ` `{ ` ` ``// Function for hollow square ` ` ``static` `void` `hollowSquare(``int` `rows) ` ` ``{ ` ` ``int` `i, j; ` ` ``for` `(i = ``1``; i <= rows; i++) ` ` ``{ ` ` ``// Print stars for each solid rows ` ` ``if` `(i == ``1` `|| i == rows) ` ` ``for` `(j = ``1``; j <= rows; j++) ` ` ``System.out.print(``"*"``); ` ` ` ` ``// stars for hollow rows ` ` ``else` ` ``for` `(j = ``1``; j <= rows; j++) ` ` ``if` `(j == ``1` `|| j == rows) ` ` ``System.out.print(``"*"``); ` ` ``else` ` ``System.out.print(``" "``); ` ` ` ` ``// Move to the next line/row ` ` ``System.out.print(``"\n"``); ` ` ``} ` ` ``} ` ` ` ` ``// Function for Solid square ` ` ``static` `void` `solidSquare(``int` `rows) ` ` ``{ ` ` ``int` `i, j; ` ` ``for` `(i = ``1``; i <= rows; i++) ` ` ``{ ` ` ``// Print stars after spaces ` ` ``for` `(j = ``1``; j <= rows; j++) ` ` ``System.out.print(``"*"``); ` ` ` ` ``// Move to the next line/row ` ` ``System.out.print(``"\n"``); ` ` ``} ` ` ` ` ``} ` ` ` ` ``// Utility program to print all patterns ` ` ``static` `void` `printPattern(``int` `rows) ` ` ``{ ` ` ``System.out.print(``"\nSolid Square:\n"``); ` ` ``solidSquare(rows); ` ` ` ` ``System.out.print(``"\nHollow Square:\n"``); ` ` ``hollowSquare(rows); ` ` ``} ` ` ` ` ``// Driver program ` ` ``public` `static` `void` `main (String[] args) ` ` ``{ ` ` ``int` `rows = ``5``; ` ` ``printPattern (rows); ` ` ``} ` `} ` ` ` `// This code is contributed by Anant Agarwal. `
## Python3
`# Python3 program to print ` `# hollow and solid square patterns ` ` ` `# Function for hollow square ` `def` `hollowSquare(rows): ` ` ` ` ``for` `i ``in` `range``(``1``, rows ``+` `1``): ` ` ` ` ``# Prstars for each solid rows ` ` ``if` `(i ``=``=` `1` `or` `i ``=``=` `rows): ` ` ``for` `j ``in` `range``(``1``, rows ``+` `1``): ` ` ``print``(``"*"``, end ``=` `"") ` ` ` ` ``# stars for hollow rows ` ` ``else``: ` ` ``for` `j ``in` `range``(``1``, rows ``+` `1``): ` ` ``if` `(j ``=``=` `1` `or` `j ``=``=` `rows): ` ` ``print``(``"*"``, end ``=` `"") ` ` ``else``: ` ` ``print``(end ``=` `" "``) ` ` ` ` ``# Move to the next line/row ` ` ``print``() ` ` ` `# Function for Solid square ` `def` `solidSquare(rows): ` ` ` ` ``for` `i ``in` `range``(``1``, rows): ` ` ` ` ``# Print stars after spaces ` ` ``for` `j ``in` `range``(``1``, rows ``+` `1``): ` ` ``print``(``"*"``, end ``=` `"") ` ` ` ` ``# Move to the next line/row ` ` ``print``() ` ` ` `# Utility program to print all patterns ` `def` `printPattern(rows): ` ` ` ` ``print``(``"Solid Square:"``) ` ` ``solidSquare(rows) ` ` ` ` ``print``(``"\nHollow Square:"``) ` ` ``hollowSquare(rows) ` ` ` `# Driver Code ` `rows ``=` `5` `printPattern (rows) ` ` ` `# This code is contributed by ` `# Mohit kumar 29 `
## C#
`// C# program to print ` `// hollow and solid square patterns ` `using System; ` ` ` `class GFG ` `{ ` ` ``// Function for hollow square ` ` ``static void hollowSquare(int rows) ` ` ``{ ` ` ``int i, j; ` ` ``for (i = 1; i <= rows; i++) ` ` ``{ ` ` ``// Print stars for each solid rows ` ` ``if (i == 1 || i == rows) ` ` ``for (j = 1; j <= rows; j++) ` ` ``Console.Write("*"); ` ` ` ` ``// stars for hollow rows ` ` ``else ` ` ``for (j = 1; j <= rows; j++) ` ` ``if (j == 1 || j == rows) ` ` ``Console.Write("*"); ` ` ``else ` ` ``Console.Write(" "); ` ` ` ` ``// Move to the next line/row ` ` ``Console.WriteLine(); ` ` ``} ` ` ``} ` ` ` ` ``// Function for Solid square ` ` ``static void solidSquare(int rows) ` ` ``{ ` ` ``int i, j; ` ` ``for (i = 1; i <= rows; i++) ` ` ``{ ` ` ``// Print stars after spaces ` ` ``for (j = 1; j <= rows; j++) ` ` ``Console.Write("*"); ` ` ` ` ``// Move to the next line/row ` ` ``Console.WriteLine(); ` ` ``} ` ` ` ` ``} ` ` ` ` ``// Utility program to print all patterns ` ` ``static void printPattern(int rows) ` ` ``{ ` ` ``Console.Write("\nSolid Square:\n"); ` ` ``solidSquare(rows); ` ` ` ` ``Console.Write("\nHollow Square:\n"); ` ` ``hollowSquare(rows); ` ` ``} ` ` ` ` ``// Driver program ` ` ``public static void Main () ` ` ``{ ` ` ``int rows = 5; ` ` ``printPattern (rows); ` ` ``} ` `} ` ` ` `// This code is contributed by vt_m. `
## PHP
` `
Output :
```Solid Square:
*****
*****
*****
*****
*****
Hollow Square:
*****
* *
* *
* *
*****
``` | 2,670 | 7,834 | {"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-2020-05 | longest | en | 0.603024 |
https://forum.allaboutcircuits.com/threads/simple-circuit-to-turn-on-when-above-certain-voltage.182387/ | 1,638,587,396,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00554.warc.gz | 314,952,216 | 26,751 | # Simple circuit to turn on when above certain voltage?
#### MikeA
Joined Jan 20, 2013
272
This is for in-vehicle purposes. Would like the circuit to sense when on-board voltage exceeds say 13V, which means the engine is running and the alternator is spinning, and then provide power to another low power (<1A) circuit.
When the engine is not running, the sensing circuit would cut the power to the secondary circuit, and not use much power for itself while the vehicle is parked.
Any ideas for a circuit that does not use any purpose built ICs?
#### dl324
Joined Mar 30, 2015
13,527
Use a comparator like LM393 to switch power using a MOSFET (high side or low side).
#### MrSalts
Joined Apr 2, 2020
723
Just use two diodes if your device can handle a bit lower voltage.
the 12 volt is used until the switch is closed. Then no current flows out of the 12VDC - only 13.5v.
#### MikeA
Joined Jan 20, 2013
272
Just use two diodes if your device can handle a bit lower voltage.
It can, but I'm not following how this works. This sensing circuit would have only 1 input, that alternates between 12v and 13.8v, depending on if the engine is running or not.
#### MrSalts
Joined Apr 2, 2020
723
so, you're asking for a circuit like used for the cigarette lighter on most import vehicles. The lighter (accessory power) only works when the key is turned to run position (I.e. engine running, typically).
#### MikeA
Joined Jan 20, 2013
272
so, you're asking for a circuit like used for the cigarette lighter on most import vehicles. The lighter (accessory power) only works when the key is turned to run position (I.e. engine running, typically).
In a way, yes. This sensing circuit would be completely autonomous. As @dl324 mentioned, a comparator was something that came to my mind first, but I was wondering if there is something more elegant, and something that would drain less when idle. Something in the 0.1ma range would be ideal, if I decide to have more than a couple around the vehicle.
#### dl324
Joined Mar 30, 2015
13,527
I was wondering if there is something more elegant
Just connect the device to a switched 12V circuit. There should be several fuse circuits that you can use.
#### crutschow
Joined Mar 14, 2008
28,201
If the vehicle has an alternator light, you could use the voltage from that to determine if the engine is running.
#### Juhahoo
Joined Jun 3, 2019
250
Motor oil pressure sensor (=oil pressure light) or as mentioned alternator light. Both turn OFF when motor is running.
#### MikeA
Joined Jan 20, 2013
272
If the vehicle has an alternator light, you could use the voltage from that to determine if the engine is running.
The reason for this circuit is I want to have devices around the car that only turn on by themselves, or can be turned on manually, when the engine is running. The purpose is to not drain the battery.
Tapping into a circuit that is only on when the engine is running, or sensing some other variable is not feasible as not tearing up the interior and running wires is the goal.
Say I want to put a fairly bright LED into the dome, but want it to work only when the vehicle is running. Running a wire up there would be tricky. But a little PCB the size of a penny would fit just fine.
#### ElectricSpidey
Joined Dec 2, 2017
1,970
Be aware the battery voltage will not drop to 12 volts as soon as you shut off the engine.
Maybe you could use a circuit to sense the "noise" on the line when the engine is running.
#### Tonyr1084
Joined Sep 24, 2015
6,459
Maybe I'm the only one who's lost here; you want a circuit to become active when the voltage exceeds 13V (engine running) OR a circuit that can be manually turned ON? Then you speak of a dome light that is on or can be turned on manually. That means it's always on when engine is running or on when you turn it on. Maybe I need to eat a snickers bar. What's the problem with using an accessory switched source? You don't want to tear up the dash and run wires? Then how are you going to run wires if you don't want to run them? Oh, wait - I think I see something in that question: You want something that is on an existing circuit that comes on when engine runs. OK. The mud and fog is clearing a little. But now I want to know "Why?" If you have a light that can be turned on manually via the wires that already exist OR turn it on automatically when voltage rises above 13V. If it can be turned on and off manually then if it's "OFF" how is it going to see a voltage? Even if the dome light is switched on the ground side, how are you going to sense the voltage without ground reference?
Sorry, I have more questions than answers for you. Perhaps some clarity will come with greater information that is precise and accurate. Right now I just don't fully understand the need or the desired solution.
#### BobaMosfet
Joined Jul 1, 2009
1,898
In a way, yes. This sensing circuit would be completely autonomous. As @dl324 mentioned, a comparator was something that came to my mind first, but I was wondering if there is something more elegant, and something that would drain less when idle. Something in the 0.1ma range would be ideal, if I decide to have more than a couple around the vehicle.
@MikeA - "drain less when idle" you say-- it's not voltage that is drained, but current. Few things will draw less current than an OpAmp or comparator (choose carefully)...
#### crutschow
Joined Mar 14, 2008
28,201
Below is the LTspice simulation of a comparator circuit using a low-cost, programmable TL431 voltage reference as a comparator to turn a MOSFET on when the battery is above 13.5V and off when it drops below 13V.
Potentiometer U2 allows adjustment of the trip point.
R4 adds some hysteresis to make the turn-on voltage (green trace) about 1/2V higher than the turn-off voltage, which prevents oscillation at the trip point.
The circuit takes about a half mA (blue trace) when the load is turned off. That is much less than the normal self-discharge current of a vehicle battery and would take years to drain it.
The M1 P-MOSFET can be just about any that has at least a 50v voltage rating, and a low enough on-resistance that it dissipates less than 1W with your load current.
It can be constructed on a small PCB or perfboard.
Last edited:
#### dcbingaman
Joined Jun 30, 2021
498
Here is another possibility:
This can handle up to a 2 Amp load with no issues. I would recommend the caps for filtering as the noise on the 12V of a car is significant. Recommend breadboarding it first, you may have to fine tune the parts some. Q2 is a 2N3904, forgot to change that.
#### Attachments
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#### LowQCab
Joined Nov 6, 2012
1,340
Here's my 2-cents ........
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.
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#### dcbingaman
Joined Jun 30, 2021
498
Here is another version. The quiescent current less than 1mA. You may use other part numbers for M1. This one can handle up to 5 Amp loads.
#### Attachments
• 2.8 KB Views: 0
#### crutschow
Joined Mar 14, 2008
28,201
Be aware the battery voltage will not drop to 12 volts as soon as you shut off the engine.
True.
But it shouldn't take long to drop below 13V for even a small drain on the battery, leaving it still essentially fully charged.
#### MikeA
Joined Jan 20, 2013
272
You want something that is on an existing circuit that comes on when engine runs. OK. The mud and fog is clearing a little. But now I want to know "Why?" If you have a light that can be turned on manually via the wires that already exist OR turn it on automatically when voltage rises above 13V. If it can be turned on and off manually then if it's "OFF" how is it going to see a voltage? Even if the dome light is switched on the ground side, how are you going to sense the voltage without ground reference?
So the issue I have is a vehicle which is not driven enough and the battery is always somewhere below 60%. The engine just doesn't run enough, and there's lots of parasitic load from factory gadgets when the vehicle is parked. That's in addition to self discharge of a lead acid battery. I also have a bunch of aftermarket equipment, that taps into constant on circuits around the vehicle. For example the vehicle has 360 degree parking camera system, but in the dark I can't see anything on the sides of the vehicle. So I added lights under the doors to light up the area when it's dark.
The dome light(s) has its own timer circuit, so when the vehicle turns off, it will keep the dome on for several minutes, and probably even indefinitely as long as doors keep opening and closing and the car thinks someone is loading or unloading things. So, whenever anyone uses the car, I'd like to have all those accessories, all powered differently, to only be operable when the engine is running (well, it's a hybrid, so when the vehicle is not turned off).
So in the case of the dome, it has an off, auto, and on. But no matter what, I don't want it to be on when the vehicle is off. The circuit will sit before any accessory and will prevent it draining the lead acid battery to any degree. | 2,216 | 9,055 | {"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-2021-49 | latest | en | 0.932776 |
https://brainly.in/question/81454 | 1,485,095,712,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281426.63/warc/CC-MAIN-20170116095121-00180-ip-10-171-10-70.ec2.internal.warc.gz | 796,164,861 | 9,583 | # ABCD is a llgm and T is the midpoint of AD. If ar of quadrilateral ABCT is 45 cm sq. Then find ar of triangle ACD.
1
by niya
## Answers
2015-02-17T17:16:30+05:30
In quad. abcd
t is mid-pt. of ad
if there is mid-pt. s on cb
ar. of abst =ar. of stdc
therefore ar. of cst=45/3=15cm sq.
therefore ar. sdc=15cm sq.
sc is median in triangle adc
ar. of adc=15*2=30 cm sq. | 149 | 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} | 3.90625 | 4 | CC-MAIN-2017-04 | latest | en | 0.762566 |
https://assignmentoverflow.com/python-set-intersection-with-codes/ | 1,675,299,685,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499954.21/warc/CC-MAIN-20230202003408-20230202033408-00299.warc.gz | 140,125,504 | 34,023 | Python Set Intersection with Codes
As people are well aware that python is a booming programming language for today’s generation. Due to easy to learn syntax and versatility it is being used on various platforms and organisations. In this article we will focus on one of the functions of python set Intersection().
Here is a brief overview on what python is. Рythоn is а dynamic, interрreted lаnguаge. There аre nо tyрe deсlаrаtiоns оf variables, раrаmeters, funсtiоns, оr methods in sоurсe соde. This mаkes the соde short and flexible, and yоu lоse the соmрile-time tyрe сheсking оf the sоurсe соde. It can be used to build websites, softwares, automate tasks and also conduct data analysis.
What is Set Intersection?
The intersection оf two sets саn be defined аs the fоrmаtiоn оf the new set соntаining the соmmоn elements оf both the given sets. Set intersection is nоt а constraint between two sets, and therefore, yоu саn have аs many sets аs yоu want tо find the соmmоn elements between them.
Python Set Intersection
There are many ways to implement set in Python. But before seeing how to create a set in python.
Creating Set in Python
А set is аn unordered collection оf items. Every set element is unique (nо duрliсаtes) and must be immutаble (саnnоt be сhаnged).
It can be сreаted by рlасing all the items (elements) inside curly brасes {}, seраrаted by соmmа, оr by using the built-in set() funсtiоn.
Example:
``````# Different tyрes оf sets in Рythоn
# set оf integers
my_set = {1, 2, 3}
рrint(my_set)
# set оf mixed dаtаtyрes
my_set = {1.0, “Hellо”, (1, 2, 3)}
рrint(my_set)``````
Output:
{1, 2, 3}
{1.0, (1, 2, 3), ‘Hello’}
Now you can easily create the set in python and implement python set intersection. There are many approaches to implement python set intersection(), three of them are listed below.
• Using intersection() function
• Using intersection operator(&)
• Using empty set for set intersection
Using intersection() function
Рythоn рrоvides а large collection оf inbuilt funсtiоns, one оf which is the intersection() methоd. It is one оf the соmmоn methods by which yоu саn perform the python set intersection and is highly used among рrоgrаmmers. The intersection methоd takes а раrаmeter аs the set variable fоr which the соmmоn element is fоund. It returns аs the new set containing the соmmоn elements оf all the sets(set B, set С, set D…).
Syntax:
set A.intersection(set B, set C, set D, …)
Example:
``````X = {1, 3, 5, 7}
Y = {2, 3, 5, 8}
print(X.intersection(Y))``````
Using interseсtiоn орerаtоr(&)
Apart frоm the рythоn inbuilt funсtiоn, yоu саn аlsо perform set intersection using the ‘&’ орerаtоr. Here, the орerаtоr is рlасed between both sets, and аs а result, the completely new set is fоrmed аs the output соntаining the соmmоn elements оf sets, аs shоwn in the belоw exаmрle.
Example:
``````X = {1, 3, 5, 7, 8}
Y = {2, 3, 5, 8}
Z = {0, 2, 4, 8}
рrint(X & Y)
рrint(Y & Z)
рrint(X & Y & Z``````
Using emрty set fоr set interseсtiоn
Set intersection returns the соmmоn elements оf any two sets, but what if the sets аre emрty and dо nоt contain any elements in them. The output after рerfоrming intersection will аlsо be аn emрty set аs there аre nо соmmоn elements in the given sets.
Example:
``````X = {}
Y = {}
рrint(set(X).interseсtiоn(set(Y)))``````
You can also check : Python Switch Statement, and Python Programming Help.
Scroll to Top | 1,101 | 3,432 | {"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} | 2.609375 | 3 | CC-MAIN-2023-06 | longest | en | 0.767597 |
https://math.stackexchange.com/questions/1544014/langle-ab-1-a-b-text-prime-rangle-is-not-a-finitely-generated-subsemig | 1,627,487,417,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00607.warc.gz | 388,583,901 | 38,624 | # $\langle ab + 1 : a,b \text{ prime}\rangle$ is not a finitely generated subsemigroup of $\Bbb{Z}^{\times}$.
Let $T \equiv PP + 1 \equiv \{ ab + 1 : a,b \text{ are prime }\} \subset \Bbb{Z}^{\times}$. Consider the subsemigroup generated by $T$. How can I show that it is not finitely generated, by that I mean there doesn't exist a finite set of integers $\{z_i\}$ such that each element of $\langle T \rangle$ can be written $t = z_1^{e_1}\cdots z_n^{e_n}$ for some $e_j \geq 0$?
I suppose I could do it by showing that there are infinitely many primes $p = 1 + ab$, but how do I do that?
• What's the source of this problem, please? – Gerry Myerson Nov 24 '15 at 6:59
• @GerryMyerson prime numbers and all the unsolved problems therein. – ExercisingMathematician Nov 24 '15 at 7:02
• $ab+1$ is even and hence composite, unless $a$ or $b$ is $2$. In the latter case, you have Sophie Germain primes. It is conjectured that there are infinitely many of those. – vadim123 Nov 24 '15 at 7:06
• In other words, I'm out of luck... :'| – ExercisingMathematician Nov 24 '15 at 7:10
Hint: you only need to show that there are infinitely many primes dividing numbers in $T$. To start, let $S$ be a finite set of primes and compare $\pi_2(x)$ with how many numbers less than $x$ are products of primes in $S$.
One can show that for each prime $p$ there is a number $ab+1\in T$ with $p\mid ab+1$. As this implies that a generating set must contain a multiple of $p$, $T$ cannot be finitely generated. So, given $p$ pick aprime $b\ne p$ and see what simple condition for $a$ you get ...
• I like this one. So you're saying pick any prime $p$ and some multiple $m$ such that $pm - 1 = ab$. Not sure how to do that though. – ExercisingMathematician Nov 24 '15 at 7:19
• Pick a prime $p$ and consider all solutions to $ab = -1 \pmod p$? – ExercisingMathematician Nov 24 '15 at 7:24 | 595 | 1,871 | {"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} | 3.59375 | 4 | CC-MAIN-2021-31 | latest | en | 0.906741 |
https://es.mathworks.com/matlabcentral/cody/problems/165-woodall-number/solutions/2126916 | 1,600,845,679,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209999.57/warc/CC-MAIN-20200923050545-20200923080545-00124.warc.gz | 372,004,175 | 16,741 | Cody
# Problem 165. Woodall number
Solution 2126916
Submitted on 13 Feb 2020 by Ryo Ishido
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### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; tf_correct = true; assert(isequal(woodall(x),tf_correct))
2 Pass
x = 2; tf_correct = false; assert(isequal(woodall(x),tf_correct))
3 Pass
x = 3; tf_correct = false; assert(isequal(woodall(x),tf_correct))
4 Pass
x = 7; tf_correct = true; assert(isequal(woodall(x),tf_correct))
5 Pass
x = 23; tf_correct = true; assert(isequal(woodall(x),tf_correct))
6 Pass
x = 63; tf_correct = true; assert(isequal(woodall(x),tf_correct))
7 Pass
x = 159; tf_correct = true; assert(isequal(woodall(x),tf_correct))
8 Pass
x = 383; tf_correct = true; assert(isequal(woodall(x),tf_correct))
9 Pass
x = 895; tf_correct = true; assert(isequal(woodall(x),tf_correct))
10 Pass
x = 1000; tf_correct = false; assert(isequal(woodall(x),tf_correct))
11 Pass
x = 2000; tf_correct = false; assert(isequal(woodall(x),tf_correct))
12 Pass
x = 2047; tf_correct = true; assert(isequal(woodall(x),tf_correct))
13 Pass
x = 3000; tf_correct = false; assert(isequal(woodall(x),tf_correct))
14 Pass
x = 3001; tf_correct = false; assert(isequal(woodall(x),tf_correct))
15 Pass
x = 4607; tf_correct = true; assert(isequal(woodall(x),tf_correct))
16 Pass
x = 10239; tf_correct = true; assert(isequal(woodall(x),tf_correct))
17 Pass
x = 22527; tf_correct = true; assert(isequal(woodall(x),tf_correct))
18 Pass
x = 7516192767; tf_correct = true; assert(isequal(woodall(x),tf_correct))
19 Pass
x = 7516192766; tf_correct = false; assert(isequal(woodall(x),tf_correct)) | 575 | 1,745 | {"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-2020-40 | latest | en | 0.37497 |
http://docplayer.net/22546595-Step-coordination-algorithm-of-traffic-control-based-on-multi-agent-system.html | 1,521,751,128,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648000.93/warc/CC-MAIN-20180322190333-20180322210333-00006.warc.gz | 86,652,605 | 29,850 | # Step-coordination Algorithm of Traffic Control Based on Multi-agent System
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1 International Journal of Automation and Computing 6(3), August 2009, DOI: /s z Step-coordination Algorithm of Traffic Control Based on Multi-agent System Hai-Tao Zhang 1 Fang Yu 1, Wen Li 2 1 Electronic Information Engineering College, Henan University of Science and Technology, Luoyang , PRC 2 Materials Science and Engineering College, Zhengzhou University, Zhengzhou , PRC Abstract: Aiming at the deficiency of conventional traffic control method, this paper proposes a new method based on multi-agent technology for traffic control. Different from many existing methods, this paper distinguishes traffic control on the basis of the agent technology from conventional traffic control method. The composition and structure of a multi-agent system (MAS) is first discussed. Then, the step-coordination strategies of intersection-agent, segment-agent, and area-agent are put forward. The advantages of the algorithm are demonstrated by a simulation study. Keywords: Traffic control, coordination algorithm, multi-agent system (MAS), traffic control system, agent. 1 Introduction The conventional control method is mainly suitable for linear and static systems. It is difficult to build a precise mathematical model for a non-linear, dynamic, and nondefinitude system. The multi-agent system (MAS) is considered one of the most significant achievements in artificial intelligence. Its aim is to divide the large system into several small systems that can communicate and coordinate with each other and can be managed easily. Agent technology has been applied to control field for a few years, and its characteristics are autonomy, mobility, reactivity, and intelligence. This paper applies MAS technique to the traffic control system. The MAS organizes and controls intersection-agents, segment-agent, area-agent, and centeragent so as to keep the traffic flow in a rational state. There are mainly three coordination ways at present: 1) Set a special coordination-agent to coordinate the acts between different agents. 2) Different coordination tasks are planted into different functional modules. The system completes the coordination automatically. 3) Combine the central method with the distributed method. Thus, the system can complete some coordination and accept the rules made by the higher-level agent at the same time. The traffic control based on MAS was first introduced into traffic flow allocation, and an iterative algorithm of allocation was put forward in [1]. Wang [2] put forward a bottom-up holistic artificial transportation system to solve the problems of complex transportation systems effectively. Du and Wu [3] presented a model of the multi-agent collaboration based on game theory and the basic structure of the traffic agents. Ma et al. [4] combined explicit coordination Manuscript received June 20, 2008; revised October 10, 2008 *Corresponding author. address: based on game theory with implicit coordination based on social rules/knowledge. Liu et al. [5] put forward a coordination algorithm based on the game theory. This paper puts forward a step-coordination algorithm, which coordinates the traffic step by step according to the order of segment-agent, intersection-agent, area-agent, and center-agent. 2 Architecture of the traffic control system based on multi-agent system The MAS is an important branch of artificial intelligence. It decomposes a complex system to several small systems that can communicate and coordinate with each other and can be managed easily. In MAS, each agent is an autonomous system and can interact with other agents to complete their common goals. MAS coordination means that an autonomous agent rationally arranges its objectives, resources, and mental states, and adjusts its decision-making and behavior to accomplish the goals. MAS is distributed, intelligent, flexible, and interactive. MAS can be used to organize and control traffic flow, and the coordination of different agents can keep the city traffic flow in a rational state. On the basis of the multi-agent technique, there are five types of agents in the system: centeragent (CA in Fig. 1), area-agent (AA1 and AA2 in Fig. 1), segment-agent (SA1 and SA2 in Fig. 1), intersection-agent (IA1 IA8 in Fig. 1), and vehicle-agent (VA1 and VA2 in Fig. 1). All types of agents have their own structures, information, goals, and tasks. They control and manage the traffic by communication, coordination, and cooperation. Agents on the same layer can conduct real-time data communication with adjacent agents so as to adjust the control strategy. The basic function of various types of agent is described as follows: 1) The vehicle-agent once it enters a certain road, it must
2 H. T. Zhang et al. / Step-coordination Algorithm of Traffic Control Based on Multi-agent System 309 register in the segment-agent of the road to make the agent know its arrival, so that the segment-agent can know the traffic flow at any moment. When a vehicle enters a segment, the downstream detector detects the vehicles arrival and sets COUNT = COUNT + 1; when a vehicle drives out of the segment, the upstream detector sets the COUNT = COUNT 1. 2) The intersection-agent is the bottom layer of the coordination system. It is responsible for monitoring the signal of the intersection, conducting real-time communication with adjacent intersections or regional control center, and completing the control tasks. It is the major agent to carry out traffic control. When the intersection-agent cannot solve the traffic problem by adjusting its own decisionmaking, it coordinates with the adjacent intersection-agent. Once the coordination fails, it will turn to the area-agent for help. 3) The segment-agent is responsible for coordinating with the adjacent intersection agents. It is on the same layer as the intersection-agent and is controlled by the area-agent too. 4) The area-agent is the middle layer of the control system, responsible for monitoring the intersection-agents and segment-agents. When the area-agent receives a request, it adjusts the behaviors of the intersection-agents according to the principle of acquiring the largest overall interests. 5) The center-agent is the highest-level of the control system, responsible for managing the whole system, monitoring different area-agents, and assessing the running state of the city road so as to realize the optimal control of the entire city traffic control system. 3 Coordination algorithm First, we introduce a few related hypotheses. Suppose the number of vehicles of intersection i at the moment t is Q i(t) = {Q i,e(t), Q i,s(t), Q i,w(t), Q i,n(t)} (1) where Q i,e(t), Q i,s(t), Q i,w(t), and Q i,n(t) are the numbers of waiting vehicles on the east, south, west, and north directions of intersection i, respectively. Q i,j(t)(j = e, s, w, n) is a vector Q i,j(t) = {q j,a(t), q j,r(t), q j,l (t)} (2) where the variable q j,a(t) means the number of vehicles going ahead at the moment t, q j,r(t) means the number of vehicles turning right at the moment t, and q j,l (t) means the number of vehicles turning left at the moment t. Then, assume the threshold value vector of the numbers of waiting vehicles of intersection i is Q i = {Q i,e, Q i,s, Q i,w, Q i,n} (3) where Q i,e, Q i,s, Q i,w, and Q i,n are the maximum numbers of the vehicles waiting at the east, south, west, and north directions of intersection i, respectively. It is a static vector, the value will not change as time goes by, but it can be modified at the initial time. 3.1 Determination of priority Suppose an intersection-agent is a quaternion A, Act, R, Φ, in which A is the set of agents; Act(t) is the set of optional actions at the moment t; act i(t) Act(t) (i A) is the action of agent i: Act(t) = (act i(t)) (4) R(t) is the resource of the intersection-agent at moment t; and Φ is the profit function. Definition 1. The profit function of agent i under the action act i(t) is Φ : A T Act A(T ) R [0, 1]. (5) The value of Φ(i, t, act i(t)) is (prof, est), where prof means the profit of agent i to execute act i(t) at moment t, and est is the probability. However, Definition 1 does Fig. 1 Structure of the traffic control system based on agent
3 310 International Journal of Automation and Computing 6(3), August 2009 not take the other agents into account, so it cannot predict the profit accurately. It needs another function to predict the actions of other agents. other(i, t, act i(t)) is a set of the other agents behaviors when agent i implements act i(t) and act A i = (act j(t)). (6) Definition 2. The profit function of Agent i, which is under the action act i(t), considers other agents action as Φ (i, t, act i(t)) = {Φ(i, t, act i(t) act A i act A i other(i, t, act i(t))). (7) When determining the priority, we first calculate Φ. Then, set Fig. 2 Coordination of segment-agents Φ(i, t, act A(t)) = max act A (t) Act A (t) {Φ (i, t, act A(t))}. (8) The joint actions which make the establishment of equation (8) acquire the highest priority. 3.2 Coordination of the intersectionagents We set detectors on the upstream and downstream of each segment, such as D1, D2, and D3 in Fig. 2. The detectors are responsible for collecting the information of the road segment and sending the information to the segmentagent [6]. Each intersection-agent has a variable COUNT to record the traffic flow of each direction where COUNT = {Count e, Count s, Count w, Count n}. If the values of COUNT are less than the thresholds Q i,e, Q i,s, Q i,w, and Q i,n, respectively, then the different phases are coordinated by the time slices circular scheduling. The time is divided into a series of equal-length time slices, which are transmitted in the entire loop. The entire loop can be regarded as a shift register, all phases waiting for the time slice. The phase can let the vehicle go only after it has received a time slice. When the traffic flow is very busy and the traffic is not balanceable at different phases, coordinations are done by sending a token, and each phase can pass it only after having a token. When the vehicles of one direction of agent i are over the threshold Q i,j, a request is sent to agent i. If agent i receives the request, then it produces a token and sends it to the relevant phase. If more than one phase ask for the green time, then all of their priorities are calculated, and the phase of the highest priority acquires the token. Slice and token are first transmitted to segment-agent and then the segment-agent allocates it to the relevant phases. The intersection-agent is allowed to produce more than one token at the same time and allocates them to nonconflicting phases. This not only will guarantees no conflict but also will not cause waste of resources. The transmission and distribution processes are as shown in Fig. 3. If the intersection-agents cannot resolve the conflicts by coordination, then resort to segment-agents to coordinate with the adjacent intersection-agents. Fig. 3 Transmission of time slice and token of intersection-agent 3.3 Coordination of the segment-agents When the intersection-agents cannot solve the problem by themselves, they must coordinate with the adjacent intersection-agents. Liu and Wang [7] put forward the consulting strategy of intersection-agents and segment-agents. We suppose that intersection-agent IA 1 needs to coordinate with IA 3. Considering the traffic flow between IA 1 and IA 3, the strategies made by IA 1 include two factors as follows: 1) The request IA1 to IA 3; 2) The self-strategy of IA 1. The strategies made by IA 3 include two factors as follows: 1) The request IA 3 to IA 1; 2) The self-strategy of IA 3. The main role of SA 1 is to compare the above strategies, choose the optimal strategy and send the final strategy to IA 1 and IA 3. There are four possible strategies: 1) IA 1 finishes ST 31; IA 3 finishes ST 13. 2) IA 1 finishes ST 31; IA 3 finishes self-strategy. 3) IA 1 finishes self-strategy; IA 3 finishes ST 13. 4) IA 1 finishes self-strategy; IA 3 finishes self-strategy. ST ij means that IA i asks IA j to do the strategy ST ij. For each intersection-agent, we set IA i={e,s,w,n}, where
6 H. T. Zhang et al. / Step-coordination Algorithm of Traffic Control Based on Multi-agent System 313 brought in. If traffic flow is not great and the flows of different phases are in balance, the time slice is used to coordinate. If the traffic flows of the different phases are not in balance, then the profit function is calculated and priorities are set, and the highest priority phase gains the token. When the traffic flow is too heavy to resolve by themselves, they coordinate with the adjacent intersection-agents or send requests to area-agents for coordination. Simulation shows that the specific algorithm of coordination is better than the traditional control methods. References [1] R. E. Allsop. Some Possibilities for Using Traffic Control to Influence Trip Distribution and Route Choice. In Proceedings of the 6th International Symposium Transportation and Traffic Theory, Elsevier, New York, USA, pp , [2] F. Y. Wang. Agent-based Control for Networked Traffic Management Systems. IEEE Intelligent Systems, vol. 20, no. 5, pp , [3] R. H. Du, Q. Y. Wu. Research on Multi-agent-game in City Area Traffic Coordination Control. Computer Engineering & Science, vol. 29, no. 4, pp , (in Chinese) [4] S. F. Ma, Y. Li, B. Liu. Agent-based Traffic Coordination Control Method for Two Adjacent Intersections. Journal of Systems Engineering, vol. 18, no. 3, pp , (in Chinese). [5] H. X. Liu, W. Wei, C. Peng. City Traffic Control and Route Guidance Based on Multi-agent. Highways & Automotive Applications, vol. 5, pp , (in Chinese) [6] F. Y. Wang. Agent-based Control for Fuzzy Behavior Programming in Robotic Excavation. IEEE Transactions on Fuzzy Systems, vol. 12, no. 4, pp , [7] X. M. Liu, F. Y. Wang. Study of City Area Traffic Coordination Control on the Basis of Agent. In Proceedings of IEEE International Conference on Intelligent Transportation Systems, IEEE Press, Singapore, pp , [8] H. J. Gao, G. J. Yu, Z. L. Li. Agent-based Urban Traffic Signal Control. Control and Decision, vol. 19, no. 7, pp , (in Chinese) Hai-Tao Zhang graduated from Henan University of Science and Technology, PRC in He received the M. Sc. degree from Henan University of Science and Technology in 1997, and the Ph. D. degree from Institute of Automation, Chinese Academy of Sciences, PRC in He is currently an associate professor at Henan University of Science and Technology. His research interests include intelligent control and computer application technology. Fang Yu graduated from Huzhou Teachers College, PRC in She is currently a master student in Henan University of Science and Technology, PRC. Her research interests include intelligent control and traffic control. Wen Li graduated from Zhengzhou University, PRC in He is currently a master student in Zhengzhou University of Science and Technology, PRC. His research interests include intelligent control and materials processing.
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A Multi-Agent-based Approach to Improve Intrusion Detection Systems False Alarm Ratio by Using Honeypot Babak Khosravifar Dept. Comp. Eng. Concordia University, Montreal, Canada b khosr@encs.concordia.ca | 10,314 | 43,450 | {"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-2018-13 | latest | en | 0.89738 |
http://stackoverflow.com/questions/10320632/sort-and-number-within-levels-of-a-factor-in-r?answertab=active | 1,444,202,641,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736682773.27/warc/CC-MAIN-20151001215802-00069-ip-10-137-6-227.ec2.internal.warc.gz | 292,483,052 | 18,974 | # sort and number within levels of a factor in r
if i have the following data frame G:
``````z type x
1 a 4
2 a 5
3 a 6
4 b 1
5 b 0.9
6 c 4
``````
I am trying to get:
``````z type x y
3 a 6 3
2 a 5 2
1 a 4 1
4 b 1 2
5 b 0.9 1
6 c 4 1
``````
I.e. i want to sort the whole data frame within the levels of factor `type` based on vector `x`. Get the length of of each level `a = 3 b=2 c=1` and then number in a decreasing fashion in a new vector `y`.
My starting place is currently with `sort()`
``````tapply(y, x, sort)
``````
Would it be best to first try and use sapply to split everything first?
-
There are many ways to skin this cat. Here is one solution using base R and vectorized code in two steps (without any `apply`):
1. Sort the data using `order` and `xtfrm`
2. Use `rle` and `sequence` to genereate the sequence.
``````dat <- read.table(text="
z type x
1 a 4
2 a 5
3 a 6
4 b 1
5 b 0.9
6 c 4
``````
Two lines of code:
``````r <- dat[order(dat\$type, -xtfrm(dat\$x)), ]
r\$y <- sequence(rle(r\$type)\$lengths)
``````
Results in:
``````r
z type x y
3 3 a 6.0 1
2 2 a 5.0 2
1 1 a 4.0 3
4 4 b 1.0 1
5 5 b 0.9 2
6 6 c 4.0 1
``````
The call to `order` is slightly complicated. Since you are sorting one column in ascending order and a second in descending order, use the helper function `xtfrm`. See `?xtfrm` for details, but it is also described in `?order`.
-
hey both great solutions. I had not seen xtfrm before. very useful. i wanted the highest number in `x` to have they highest `y` so deleted the `-` from `xtfrm` and its a perfect result thank you – user1322296 Apr 25 '12 at 17:54
@Andrie I hadn't seen `xtfrm` either but don't really get what it does.What is it doing here that the negative won't do? [the help file isn't that terrific on this function] – Tyler Rinker Apr 25 '12 at 18:04
@user1322296 OK, in that case you don't need the `xtfrm` at all - it will just slow things down. – Andrie Apr 25 '12 at 18:25
@TylerRinker The `xtfrm` idiom will also work for strings, but in this case you are correct, one doesn't need it. – Andrie Apr 25 '12 at 18:25
I like Andrie's better:
``````dat <- read.table(text="z type x
1 a 4
2 a 5
3 a 6
4 b 1
5 b 0.9
``````
Three lines of code:
``````dat <- dat[order(dat\$type), ]
x <- by(dat, dat\$type, nrow)
dat\$y <- unlist(sapply(x, function(z) z:1))
``````
I Edited my response to adapt for the comments Andrie mentioned. This works but if you went this route instead of Andrie's you're crazy.
-
You need to replace that last `rep(x,x)` with `seq_len(x)` and then it should work. Nice solution. – Andrie Apr 25 '12 at 17:47
@Andrie, I think it works as it is. The 2nd line gives me the lengths of each factor (what your use of `rle` does) which is `c(3, 2, 1)`. Then I use `rep` to repeat each one of the lengths that number of times. Try running it and see. – Tyler Rinker Apr 25 '12 at 17:50
Nevermind I misunderstood what the poster wanted. I see it now. – Tyler Rinker Apr 25 '12 at 17:55
+1 for self-deprecating comments! – Andrie Apr 25 '12 at 18:26 | 1,118 | 3,281 | {"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} | 3.109375 | 3 | CC-MAIN-2015-40 | latest | en | 0.88828 |
http://www.homebrewtalk.com/f39/weak-ginger-beer-368951-print/ | 1,409,399,278,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500834883.60/warc/CC-MAIN-20140820021354-00064-ip-10-180-136-8.ec2.internal.warc.gz | 424,344,046 | 2,821 | Home Brew Forums (http://www.homebrewtalk.com/forum.php)
- Beginners Beer Brewing Forum (http://www.homebrewtalk.com/f39/)
- - Weak ginger beer.. (http://www.homebrewtalk.com/f39/weak-ginger-beer-368951/)
nvs-brew 11-20-2012 09:38 AM
Weak ginger beer..
My brew has been on over 100hrs and it's only sitting at 1.4%
I took the mid strength recipe and bulked up a 1/3 to take it up to a estimated 4-5%
(3.5% was ment to be 1kg of sugars, I put 1.5kgs)
Why and how could it only be sitting at 1.5%?
Is there anything that can be done?
nvs-brew 11-20-2012 09:51 AM
I did a hydro test 2 days ago and it hasn't changed..
I
RM-MN 11-20-2012 10:06 AM
You're reading your hydrometer wrong. That scale is meant for wine and it shows the potential alcohol. You need to read the beer scale and then it takes some math to get the amount of alcohol.
What was the original gravity? It should be something like 1.054, not 17% alcohol. With that and the final gravity you can calculate the amount of alcohol in your beer.
nvs-brew 11-20-2012 10:10 AM
Og = 1040
2nd and 3rd reading = 1030
(it's a wine/beer hydrometer)
nvs-brew 11-20-2012 10:12 AM
1 Attachment(s)
But advise me if I'm doing it wrong
Please find pic of hydrometer attached
RM-MN 11-20-2012 03:55 PM
My mistake. Why would your ferment stop so soon? Got a recipe so we can see what you have for fermentables. Include the yeast type too.
nvs-brew 11-20-2012 10:22 PM
Recipe as follows...(sorry in Australia so measurements are in kgs/ltrs)
Coopers ginger beer lme
Supplied yeast
500g raw
500g castor
500g dex
To make a 20l batch
nvs-brew 11-20-2012 11:51 PM
Anyone got any ideas why?
The different sugars?
??!
mahe 11-20-2012 11:59 PM
Maybe its been fermenting too cold? and the yeast has gone to sleep?
Johnnyhitch1 11-21-2012 12:06 AM
Rouse the yeast, warm it up, pitch more yeast. If all else fails pitch amylase enzyme
All times are GMT. The time now is 11:47 AM. | 620 | 1,957 | {"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-2014-35 | latest | en | 0.91561 |
https://metanumbers.com/183063 | 1,627,619,245,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153931.11/warc/CC-MAIN-20210730025356-20210730055356-00616.warc.gz | 409,346,279 | 10,945 | ## 183063
183,063 (one hundred eighty-three thousand sixty-three) is an odd six-digits composite number following 183062 and preceding 183064. In scientific notation, it is written as 1.83063 × 105. The sum of its digits is 21. It has a total of 3 prime factors and 8 positive divisors. There are 120,888 positive integers (up to 183063) that are relatively prime to 183063.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 21
• Digital Root 3
## Name
Short name 183 thousand 63 one hundred eighty-three thousand sixty-three
## Notation
Scientific notation 1.83063 × 105 183.063 × 103
## Prime Factorization of 183063
Prime Factorization 3 × 139 × 439
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 183063 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 183,063 is 3 × 139 × 439. Since it has a total of 3 prime factors, 183,063 is a composite number.
## Divisors of 183063
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 246400 Sum of all the positive divisors of n s(n) 63337 Sum of the proper positive divisors of n A(n) 30800 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 427.859 Returns the nth root of the product of n divisors H(n) 5.9436 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 183,063 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 183,063) is 246,400, the average is 30,800.
## Other Arithmetic Functions (n = 183063)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 120888 Total number of positive integers not greater than n that are coprime to n λ(n) 10074 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16549 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 120,888 positive integers (less than 183,063) that are coprime with 183,063. And there are approximately 16,549 prime numbers less than or equal to 183,063.
## Divisibility of 183063
m n mod m 2 3 4 5 6 7 8 9 1 0 3 3 3 6 7 3
The number 183,063 is divisible by 3.
## Classification of 183063
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (183063)
Base System Value
2 Binary 101100101100010111
3 Ternary 100022010010
4 Quaternary 230230113
5 Quinary 21324223
6 Senary 3531303
8 Octal 545427
10 Decimal 183063
12 Duodecimal 89b33
20 Vigesimal 12hd3
36 Base36 3x93
## Basic calculations (n = 183063)
### Multiplication
n×i
n×2 366126 549189 732252 915315
### Division
ni
n⁄2 91531.5 61021 45765.8 36612.6
### Exponentiation
ni
n2 33512061969 6134818600231047 1123058297414096156961 205590421099516684781751543
### Nth Root
i√n
2√n 427.859 56.7806 20.6847 11.2855
## 183063 as geometric shapes
### Circle
Diameter 366126 1.15022e+06 1.05281e+11
### Sphere
Volume 2.56975e+16 4.21125e+11 1.15022e+06
### Square
Length = n
Perimeter 732252 3.35121e+10 258890
### Cube
Length = n
Surface area 2.01072e+11 6.13482e+15 317074
### Equilateral Triangle
Length = n
Perimeter 549189 1.45111e+10 158537
### Triangular Pyramid
Length = n
Surface area 5.80446e+10 7.22995e+14 149470
## Cryptographic Hash Functions
md5 8c871a3cca7d6f62792a225ef886fca5 5482d4e61313e1a86e1995899ba48560d54d4ff2 fc543898186cfdb060b08a669f3a22c72fe9e9877c1b9e0aecf6b829cbc9b503 552b980912588887db7f5220cbe8cb465f35ca45c16ebcf3f1808fabc8e16e62f843ef5d6767dda7be69912fc7e451e2abdc4e53d93fe840a7a43d897527d2f5 18e1d2755a359e279823cca5b6beb986029065fe | 1,462 | 4,239 | {"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} | 3.40625 | 3 | CC-MAIN-2021-31 | latest | en | 0.814847 |
http://www.mathpuzzle.ca/m/Puzzle/The-Nine-Counters.html | 1,556,065,166,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578616424.69/warc/CC-MAIN-20190423234808-20190424020808-00481.warc.gz | 277,144,272 | 3,823 | THE NINE COUNTERS.
[Illustration:
(1)(5)(8) (7)(9)
(2)(3) (4)(6)
]
I have nine counters, each bearing one of the nine digits, 1, 2, 3, 4,
5, 6, 7, 8 and 9. I arranged them on the table in two groups, as shown
in the illustration, so as to form two multiplication sums, and found
that both sums gave the same product. You will find that 158 multiplied
by 23 is 3,634, and that 79 multiplied by 46 is also 3,634. Now, the
puzzle I propose is to rearrange the counters so as to get as large a
product as possible. What is the best way of placing them? Remember both
groups must multiply to the same amount, and there must be three
counters multiplied by two in one case, and two multiplied by two
counters in the other, just as at present.
THE NINE ALMONDS. THE NINE SCHOOLBOYS.
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### PRODUCT DESCRIPTION
This is a generic graphic organizer to get our kids to use several strategies to determine proportionality.
I will be bringing in advertisements from my Sunday newspaper. I will ask students to select one (for example, the Dollar General ad shows a snack pack of 6 bags of M&Ms for a dollar). We will then compute the unit rate and complete the table based upon unit rate.
Next, students will choose a strategy to check for equivalent ratios (proportionality).
Finally, students will graph the relationship. If the line is straight and passes through the origin the graph represents a proportional relationship.
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# X atoms are present at corners in a cube and Y atoms are at face centres of a cube. Then the number of the X and Y atoms present in the cube is:(A) $8,8$(B) $6,8$(C) $8,6$(D) $6,6$
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Hint:Each atom at the corner is shared by eight adjacent unit cells and each atom at the centre of faces is shared between two adjacent unit cells. There are eight corner atoms and six face-centred atoms in a simple cubic unit cell.
Now in the question, it is given that X atoms are present in the corners and Y atoms at the face-centred.
We know that among the categories of unit cells, face-centred unit cells contain one particle present at the centre of each face besides the atoms that are at its corners.
So from the question, we can understand that the packing is fcc where X atoms are at corners and Y atoms at face centres of the cube.
Therefore a face-centred unit cell contains atoms at all the corners and the centre of all faces of the cube.
In the question, it is given that the X atoms are present at the corners in a cube. We know that there are eight corners in a cube and atoms will be present there. So eight atoms at the corners.
Since X atoms are present at corners of the cube, the number of X atoms is eight.
We also know that there are six faces for a cube and the Y atoms will be present at each face of the cube.
Since six faces are there, the number of Y atoms present in the faces of the cube is six
Thus the answer is an option (C) $8,6$ where the number of X atoms is eight and the number of Y atoms is six.
Note:
We can calculate the number of X atoms in corners and the number of Y atoms at face centres simply by knowing how many corners and how many faces are there in a cube. There are eight corners and six faces for a cube. Thus the number of X atoms present at the corners is eight and the number of Y atoms present at faces is six. | 494 | 2,025 | {"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-2024-38 | latest | en | 0.954243 |
http://www.chegg.com/homework-help/questions-and-answers/1-battery-75-v-output-voltage-24-ohm-internal-resistance-battery-used-boot-small-electric--q3149400 | 1,472,393,827,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982939917.96/warc/CC-MAIN-20160823200859-00060-ip-10-153-172-175.ec2.internal.warc.gz | 364,538,748 | 13,855 | 1) The battery has 7.5 V output voltage and 24 Ohm internal resistance. The battery is used to boot a small electric motor with 105 Ohm resistance. Find the power delivered to the motor. 2) Two batteries with 2.3 voltage and 8.5 Ohm internal resistance each are connected in series. This battery pack is used to boot a small electric motor with 160 Ohm resistance. Find the power delivered to the motor. 3) The power dissipated in the resistor R0 is 56 W when connected to a 10V source. Find the total dissipated power if 3 resistors identical to R0 are connected in series to a 10V source. (UNIT:W) 4) The power dissipated in the resistor R0 is 37 W when connected to a 10V source. Find the total dissipated power if 3 resistors identical to R0 are connected in parallel to a 10V source. (UNIT:W) 5)A 90 V DC power supply is connected to a 100 Ohm load resistor. The tested current flowing through the resistor is 0.89 A. find the internal resistance of the DC power suppy. 6) The power dissipated in the resistor is 80 W when connected to a 10V source. Find the total dissipated power when the same resistor is connected to a 5V source.(UNIT:W) 7) The battery is made by connecting 7 unit cells in series and then by connecting 9 of these series cells in parallel. Each unit cell has the voltage of 1.25 V and the internal resistance if 2 Ohm. What is the maximum power that can be delivered to the load by this battery? (Units: W) 8) Some kind of AA battery has 1.3 V power supply voltage and 0.29 Ohm internal resistance. Theoratically, how much current can be drawn out of a battery of such kind at most? How much will the maximum current be if there're two batteries connected in series serving as one supply? How much will the maximum current be if the two batteries are connected in paralell? | 458 | 1,800 | {"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} | 3.265625 | 3 | CC-MAIN-2016-36 | latest | en | 0.927067 |
https://www.calendar-uk.co.uk/frequently-asked-questions/is-undefined-a-horizontal-line | 1,713,046,579,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00763.warc.gz | 641,486,255 | 9,032 | # Is undefined a horizontal line?
The slope of a line can be positive, negative, zero, or undefined. A horizontal line has slope zero since it does not rise vertically (i.e. y1 − y2 = 0), while a vertical line has undefined slope since it does not run horizontally (i.e. x1 − x2 = 0). because division by zero is an undefined operation.
## Is a vertical line undefined?
Remember the Slope Formula
Vertical lines have an undefined slope because the horizontal change is 0 — you cannot divide a number by 0.
## Is undefined a straight line?
A line has a slope that is considered undefined when it does not have any horizontal run. It will be a straight vertical line.
## What direction is an undefined line?
The slope of a line is undefined if the line is vertical. If you think of slope as rise over run, then the line rises an infinite amount, or goes straight up, but does not run at all.
## What is an example of undefined?
In mathematics, undefined means a term that is mathematically inexpressible, or without meaning. Anything divided by zero is considered undefined by the rules of mathematics. Take a fraction such as: 82=x 8 2 = x .
## Vertical and Horizontal Lines, Slope of Zero and Undefined Slopes
45 related questions found
### What is a horizontal line?
In coordinate geometry, horizontal lines are defined as lines parallel to the x-axis. If two points on a line have the same y-coordinate points then the line is said to be horizontal.
### How do you tell if a line is horizontal or vertical?
Horizontal lines go side to side and have a slope of 0. Vertical lines go up and down and have a slope that is undefined. Graphs of horizontal lines are parallel to the x-axis.
### Are perpendicular lines undefined?
Vertical lines and horizontal lines are perpendicular to each other. The slope of the perpendicular line in this case would be the slope of a vertical line which would be undefined. The slope of the parallel line is 0 and the slope of the perpendicular line is undefined.
### Is a straight up and down line undefined?
The slope of a vertical line does not exist! We can't divide by zero, which is of course why this slope value is "undefined". This relationship is always true: a vertical line will have no slope, and "the slope is undefined" or "the line has no slope" means that the line is vertical.
### What does undefined mean in math?
Broadly speaking, undefined means there is no possible value (or there are infinite possible values), while indeterminate means there is no value given the current information.
### Are undefined lines parallel?
Answer and Explanation: Yes, lines with undefined slopes can be parallel. In fact, all lines with undefined slope are parallel.
### Is a horizontal line 0?
Horizontal lines have no steepness at all. Because any calculation of the slope of horizontal lines means dividing zero by another number, the slope of a horizontal line is always zero.
### What does a horizontal line look like?
A horizontal line is a straight line that goes from left to right or right to left. In coordinate geometry, a line is said to be horizontal if two points on the line have the same Y- coordinate points. It comes from the term “horizon”. It means that the horizontal lines are always parallel to the horizon or the x-axis.
### What is horizontal vs vertical?
The terms vertical and horizontal often describe directions: a vertical line goes up and down, and a horizontal line goes across. You can remember which direction is vertical by the letter, "v," which points down.
### Is undefined and 0 perpendicular?
Because horizontal and vertical lines are always perpendicular, then lines having a zero slope and an undefined slope are perpendicular.
### Is perpendicular a straight line?
A perpendicular line is a straight line through a point. It makes an angle of 90 degrees with a particular point through which the line passes. Coordinates and line equation is the prerequisite to finding out the perpendicular line.
### Are all horizontal lines perpendicular?
Horizontal and vertical lines are always perpendicular: therefore, two lines, one of which has a zero slope and the other an undefined slope are perpendicular.
### What is a line that is not horizontal or vertical?
Lines are slanting if they don't go straight across or straight up. They look like a slope and go both up and down, and across too. Example.
### How do you prove a horizontal line?
We can apply the horizontal line test by drawing a horizontal line across our graph. If it touches two or more points then we say it fails the horizontal line test and its inverse is not a function.
### Is a horizontal line undefined or 0?
The slope of a line can be positive, negative, zero, or undefined. A horizontal line has slope zero since it does not rise vertically (i.e. y1 − y2 = 0), while a vertical line has undefined slope since it does not run horizontally (i.e. x1 − x2 = 0).
### What are the five horizontal lines?
staff, also spelled stave, in the notation of Western music, five parallel horizontal lines that, with a clef, indicate the pitch of musical notes.
### What is a straight horizontal line called?
A horizontal line, also called a sleeping line, is a line in which all points have the same y - coordinate. It is parallel to the x-axis of the plane, and the slope of the horizontal line is zero. It is referred to as the sleeping line because it rules left to right without ever crossing the X-axis. | 1,155 | 5,491 | {"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} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.927054 |
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e102_mid_term_solution - UNIVERSITY OF ~ ALIFORNIA DAVIS...
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Unformatted text preview: UNIVERSITY OF ~- ALIFORNIA, DAVIS Dept. of Mechanical a-»=s;.t Aerospace Engineering E-102 Mid-Term Exam (Solutiorfi 1 Hour and 50 Minutes Closed Book and Notes I.) The block shown is observed to have a velocity V] =20 fi/sec as it passes point A and a velocity v; = 10 ft/sec as it passes point B on the incline. 5% a.) Neatly sketch a free body diagram of the system 20% b.) Calculate the coefficient of kinetic friction pk between the block and the incline if x = 30 ft and 9 =15 deg. , i la) 25;? M4! ",‘r‘flkwr'lfw'VfiLn /<5° = M4, (D , ,, , 3 eqjlfi) gunhnownlfi’ ,ZF’y—1M9y , ax}! )IJ 4Macmls°+u= 0 ® . I L V :30 E , 7 ,1, ,, , egg-m1: PM! Q) i V O i SOIQTHWLM 7 3° . ,_—.,’100 _ sm|5°+ M co to“) d4 ’—00 a - We ”2% 4 J, O Swat lat: anae 2.) The crank AB has a constant clockwise angular velocity of 200 revolutions per minute. Determine: 15% a.) the angular velocity of bar BP at this instant 10% b.) the velocity of point P on the piston at this instant 2\ “use, 7 0\L‘>\ 7 Us; obeqloch, nae-Lion 7L: figmse +L\$9 Coce’ =WP= ~Lm5 Smog- ngmtil? é = l' 02?") Wallace ~' 790-9" voJ/£ee QUE-".1. ' ogcocg’ = .y. (65—20039 0 ’19 I , , Lee i r \/=‘,‘Lk5 C039 bu-L ,Lms 3‘; 2"”; 23a lit: -J.F7 5994+ :‘ C05\$=l or 97:46“ C06 :.- or eat? ‘12 f .597 2M Mal/sec fl 3.) The device below consists of two 3.2 kg point mass. 5: {particles) on a light, rigid bar. The device is swinging through the vertical position with a clockwise angular velocity (1) = 6 rad/sec, when a 0.05 kg bullet traveling at v = 300 m/sec in the direction shown strikes the lower mass and becomes imbedded in it. 5% a.) Neatly sketch all free body diagrams necessary to solve the problem 20% b.) Determine the angular velocity (0' which the pendulum has immediately after impact 77 Give 4,54!— .H33 _ HI: ' ’ m3 :; +9. « (seem-am?) 51-“ czalzaiwfir-‘DVW (423 “(imam Ml W'svfiém 4,! I i 1 41", {WNW (flu ’l 4.) A jet aircraft with a mass of 1‘ 104 kg is flying horizontally at a constant velocity of 1000 km/hr. The jet engine is developing a constant thrust T, and the aerodynamic drag is R. At some instant, the pilot starts two rocket-assisted units attached to the aircraft, each of which develops an additional thrust T0 = 8 kN for 9 sec. At the end of 9 see, the aircraft’s velocity is 1050 km/hr. 5% a.) Neatly sketch :1 free body diagram of the system 20% b.) Determine the time-average increase in drag AR over the 9 sec period o 4] rhea, MISéC. _ jézwgfi + (2.1m? =, (/0601: 2713\«19 -(zoc¢)(.m'/t",, o 9 , <7 7 i , 7’ , , 4} mu - fame =’flak/'60:[email protected]? ©fl ...
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Ask a homework question - tutors are online | 1,129 | 3,308 | {"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} | 3.359375 | 3 | CC-MAIN-2018-13 | latest | en | 0.660156 |
https://calculatoruniverse.com/long-addition-calculator/ | 1,721,125,879,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514742.26/warc/CC-MAIN-20240716080920-20240716110920-00892.warc.gz | 130,606,939 | 36,540 | Result:
Detailed Calculation:
Calculation History:
Long addition, also known as column addition or vertical addition, is a method of adding two or more numbers by aligning them vertically and performing addition digit by digit, starting from the rightmost (units) digit and moving towards the left, carrying over any excess values to the next column as needed. It is one of the basic arithmetic operations and is taught in elementary mathematics education.
## All Formulae Related to Long Addition Calculator
Long addition, also known as column addition or vertical addition, doesn’t involve specific formulas like more complex mathematical operations. Instead, it’s a straightforward method for adding numbers. Here’s a step-by-step guide on how to perform long addition:
Step 1: Align the numbers vertically, with their rightmost digits (units place) in the same column.
Step 2: Add the digits in the units place (rightmost column) together. If the sum is 10 or greater, write down the ones digit and carry over the tens digit to the next column to the left.
Step 3: Move to the next column to the left and add the digits in that column, along with any carried-over tens digits from the previous step. Write down the ones digit of the sum and carry over the tens digit if necessary.
Step 4: Repeat Step 3 for each subsequent column to the left until you have added all the digits.
Step 5: If there is a carry-over digit after adding the leftmost column, write it down as part of the result.
Here’s an example of long addition:
`456 + 789 ______`
1. Add the units place: 6 + 9 = 15. Write down 5 and carry over 1.
`456 + 789 ______ 5`
1. Move to the tens place: 1 (carry-over) + 5 + 8 = 14. Write down 4 and carry over 1.
`456 + 789 ______ 45`
1. Move to the hundreds place: 1 (carry-over) + 4 + 7 = 12. Write down 2.
`456 + 789 ______ 1245`
This is the result of the long addition.
## Practical Uses of Long Addition
Long addition, while a fundamental arithmetic operation, has numerous practical uses in everyday life and various fields. Here are some practical applications of long addition:
1. Financial Transactions: Long addition is used in banking and personal finance for tasks such as balancing checkbooks, reconciling account statements, and calculating monthly expenses.
2. Shopping and Budgeting: Consumers use long addition to track their expenses, calculate discounts, and create budgets to manage their finances effectively.
3. Receipts and Invoices: Businesses and individuals use long addition to calculate totals on invoices, bills, and receipts for goods and services.
4. Taxes: Long addition plays a role in computing income tax returns, as taxpayers add up their sources of income and deductions.
5. Inventory Management: In retail and manufacturing, long addition helps calculate the total value of inventory and reconcile it with sales records.
6. Timekeeping: Long addition is used in time calculations, such as adding hours and minutes to determine the total time elapsed or to calculate payroll hours.
7. Cooking and Recipes: When following recipes that require adjusting ingredient quantities, long addition is useful for calculating the total amounts needed.
8. Measurement and Conversions: In fields like construction and engineering, long addition is used to sum measurements or quantities of materials required for a project. It can also be used to convert units of measurement.
9. Travel Planning: Travelers use long addition to calculate the total cost of a trip, including expenses such as transportation, accommodation, and meals.
10. Grading and Averaging: Educators use long addition to calculate and record students’ scores, as well as to compute grade point averages (GPAs).
11. Medical Calculations: Healthcare professionals use long addition for tasks like calculating medication dosages, totaling patients’ bills, or summing up laboratory results.
## Applications of Long Addition Calculator in Various Fields
Long addition calculators, which are basic calculators capable of performing addition with long numbers, can be useful tools in various fields for performing arithmetic operations efficiently and accurately. Here are some applications of long addition calculators in different domains:
1. Finance and Accounting:
• Calculate the total expenses, revenues, or profits for a financial period.
• Add up individual transactions or entries in accounting ledgers.
• Compute the total costs, revenues, or investments in business planning and analysis.
• Summarize data in economic reports or research.
• Calculate students’ final grades by adding their scores for different assignments and exams.
• Determine class averages and grade point averages (GPAs).
4. Construction and Engineering:
• Add up measurements, dimensions, or quantities of materials required for construction projects.
• Calculate project costs and budgets.
5. Inventory and Stock Control:
• Update and reconcile inventory levels by adding or subtracting items in stock.
• Compute the total value of inventory.
6. Retail and Sales:
• Calculate the total amount of sales or revenue generated in a retail store.
• Add up prices to determine the cost of a shopping cart.
7. Manufacturing and Production:
• Sum up production costs, including raw materials, labor, and overhead expenses.
• Calculate the total output of manufactured goods.
8. Medicine and Healthcare:
• Add up quantities of medications or medical supplies needed for patient care.
• Compute the total cost of medical procedures or treatments.
9. Scientific Research:
• Sum data points from experiments or observations in various scientific fields.
• Calculate cumulative values or totals in research studies.
10. Logistics and Supply Chain Management:
• Determine the total weight, volume, or quantity of goods in logistics operations.
• Calculate transportation costs and shipping expenses.
## Benefits of Using the Long Addition Calculator
Using a long addition calculator offers several benefits across various domains and scenarios. Here are some of the key advantages of using a long addition calculator:
1. Accuracy: Long addition calculators provide precise and error-free results, reducing the risk of calculation mistakes that can occur with manual addition, especially when dealing with long numbers or large datasets.
2. Efficiency: Calculators can perform long addition quickly, saving time for users. This is particularly valuable when adding multiple numbers or performing repetitive addition tasks.
3. Reduced Mental Load: Long addition calculators relieve users from mentally tracking and carrying over digits, allowing them to focus on understanding the problem and its context.
4. Complex Calculations: Calculators can handle complex addition tasks, including adding long numbers with multiple decimal places or performing addition with many terms, simplifying otherwise time-consuming calculations.
5. Consistency: Long addition calculators provide consistent results across multiple calculations, ensuring accuracy and reliability in various applications.
6. Reduced Fatigue: When dealing with a large amount of numerical data, calculators can help reduce mental fatigue and maintain concentration, especially during extended periods of calculations.
7. Streamlined Data Entry: In fields like accounting, finance, and data analysis, calculators enable users to input numbers efficiently, improving the overall workflow and data entry process.
8. Error Detection: Calculators can help users identify errors in their addition process by immediately displaying incorrect inputs or solutions, allowing for quick correction.
9. Versatility: Long addition calculators can handle a wide range of addition tasks, from basic arithmetic to more complex financial or scientific calculations.
10. Educational Tool: Long addition calculators can be valuable aids in teaching and learning mathematics, helping students practice addition skills and verify their work.
Last Updated : 27 February, 2024
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# NCERT Solutions for Class 10 Maths Chapter 2: Polynomials - Exercise 2.2
Last updated date: 07th Sep 2024
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## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
NCERT Solution for Class 10 Maths PDF includes the solved Exercise 2.2 from Chapter 2 Polynomials. These solutions were developed by math professionals who have updated based on the current academic year 2024-25. The Answers for Class 10 Math Ex 2.2, Chapter 2 are provided here in an accessible and step-by-step manner. Also, This article has covered every crucial topics, such as Zeroes and Coefficients of Polynomials. After going over these NCERT Solutions for Class 10 Maths, students would undoubtedly be able to tackle these problems with ease.
Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
2. Glance on Class 10 Maths Exercise 2.2 Polynomials Chapter 2
3. Topics Covered in Class 10 Maths Chapter 2 Exercise 2.2
4. Access Class 10 Maths Chapter 2 Exercise 2.2 Solutions – Polynomials
4.1Exercise 2.2
5. NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2
5.1Definition
5.2Degree of a Polynomial
5.3Types of Polynomial
5.4Value of a Polynomial
5.5Zero of a Polynomial
5.6Graph of a Polynomial
5.7Meaning of the Zeroes of a Quadratic Polynomial
5.8The Following Observations Can Be Made On Class 10 Maths ex 2.2
5.9Note in Exercise 2.2 Class 10 Math:
6. NCERT Solutions Class 10 Maths Chapter 2 Exercises
7. Other Related Links for Chapter 2 Polynomials Of Class 10
8. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs
## Glance on Class 10 Maths Exercise 2.2 Polynomials Chapter 2
• Class 10 Maths Exercise 2.2 dives into polynomials, which are algebraic expressions formed with variables and their non-negative integer powers.
• This exercise concentrates on quadratic polynomials, which are of degree 2.
• The focus is on finding the zeroes of these polynomials. The zeroes are the values of the variable (x in this case) that make the polynomial equal to zero.
• The exercise also explores the relationship between the zeroes and the coefficients of a polynomial.
• There's a surprising connection – the sum of the zeroes is equal to the negative of the coefficient of the first-degree term (linear term), and the product of the zeroes is equal to the constant term.
• Exercise 2.2 helps you with identifying polynomials
• Understanding the concept of degree of a polynomial
• Finding zeroes of quadratic polynomials
• Discovering the relationship between zeroes and coefficients
## Topics Covered in Class 10 Maths Chapter 2 Exercise 2.2
1. Understanding Polynomials
2. Types of Polynomials
3. Degree of a Polynomial
Competitive Exams after 12th Science
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## Access Class 10 Maths Chapter 2 Exercise 2.2 Solutions – Polynomials
### Exercise 2.2
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$
Ans:
Given: ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$.
Now factorize the given polynomial to get the roots.
$\Rightarrow \text{(x}-\text{4)(x+2)}$
The value of ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$ is zero.
when $\text{x}-\text{4=0}$ or $\text{x+2=0}$. i.e., $\text{x = 4}$ or $\text{x = }-\text{2}$
Therefore, the zeroes of ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$ are $\text{4}$ and $-2$.
Now, Sum of zeroes$\text{=4}-\text{2= 2 =}-\dfrac{\text{2}}{\text{1}}\text{=}-\dfrac{\text{Coefficient of x}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$
$\therefore Sum\,of\,zeroes=-\dfrac{Coefficient\,of\,x}{Coefficient\,of\,{{x}^{2}}}$
Product of zeroes $\text{=4 }\!\!\times\!\!\text{ (}-\text{2)=}-\text{8=}\dfrac{\text{(}-\text{8)}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$
$\therefore \Pr oduct\,of\,zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$ .
(ii) $4{{s}^{2}}-4s+1$
Ans:
Given: $\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}$
Now factorize the given polynomial to get the roots.
$\Rightarrow {{\text{(2s}-\text{1)}}^{\text{2}}}$
The value of $\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}$ is zero.
when $\text{2s}-\text{1=0}$, $\text{2s}-\text{1=0}$. i.e., $\text{s =}\dfrac{\text{1}}{\text{2}}$ and $\text{s =}\dfrac{\text{1}}{\text{2}}$
Therefore, the zeroes of $\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}$ are $\dfrac{\text{1}}{\text{2}}$ and $\dfrac{\text{1}}{\text{2}}$.
Now, Sum of zeroes$\text{=}\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{2}}\text{=1=}\dfrac{\text{(}-\text{4)}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$
$\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$
Product of zeroes$=\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}=\dfrac{\text{1}}{\text{4}}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$
$\therefore Product of zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$ .
(iii) $6{{x}^{2}}-3-7x$
Ans:
Given: $\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }$
$\Rightarrow \text{6}{{\text{x}}^{\text{2}}}-\text{7x}-\text{3}$
Now factorize the given polynomial to get the roots.
$\Rightarrow \text{(3x+1)(2x}-\text{3)}$
The value of $\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }$ is zero.
when $\text{3x+1=0}$ or $\text{2x}-\text{3=0}$. i.e., $\text{x =}\dfrac{-\text{1}}{\text{3}}$ or $\text{x =}\dfrac{\text{3}}{\text{2}}$.
Therefore, the zeroes of $\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }$ are $\dfrac{\text{-1}}{\text{3}}$ and $\dfrac{\text{3}}{\text{2}}$.
Now, Sum of zeroes$\text{=}\dfrac{-\text{1}}{\text{3}}\text{+}\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{\text{7}}{\text{6}}\text{=}\dfrac{-\text{(}-\text{7)}}{\text{6}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$
$\therefore Sum of zeroes\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$
Product of zeroes$\text{=}\dfrac{-\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{-\text{3}}{\text{6}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$
$\therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$
(iv) $4{{u}^{2}}+8u$
Ans:
Given: $\text{4}{{\text{u}}^{\text{2}}}\text{+8u}$
$\Rightarrow \text{4}{{\text{u}}^{\text{2}}}\text{+8u+0}$
$\Rightarrow \text{4u(u+2)}$
The value of $\text{4}{{\text{u}}^{\text{2}}}\text{+8u}$ is zero.
when $\text{4u=0}$ or $\text{u+2=0}$. i.e., $\text{u = 0}$ or $\text{u =}-\text{2}$
Therefore, the zeroes of $\text{4}{{\text{u}}^{\text{2}}}\text{+8u}$ are $\text{0}$ and $\text{-2}$.
Now, Sum of zeroes$\text{=0+(}-\text{2)=}-\text{2=}\dfrac{-\text{8}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$
$\therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$
Product of zeroes$\text{=0 }\!\!\times\!\!\text{ (}-\text{2)= 0 =}\dfrac{\text{0}}{\text{4}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$
$\therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$
(v) ${{t}^{2}}-15$
Ans:
Given: ${{\text{t}}^{\text{2}}}-\text{15}$
$\Rightarrow {{\text{t}}^{\text{2}}}-\text{0t}-\text{15}$
Now factorize the given polynomial to get the roots.
$\Rightarrow \text{(t}-\sqrt{\text{15}}\text{)(t +}\sqrt{\text{15}}\text{)}$
The value of ${{\text{t}}^{\text{2}}}-\text{15}$ is zero.
when $\text{t}-\sqrt{\text{15}}\text{=0}$ or $\text{t+}\sqrt{\text{15}}\text{=0}$, i.e., $\text{t=}\sqrt{\text{15}}$ or $\text{t=}-\sqrt{\text{15}}$
Therefore, the zeroes of ${{\text{t}}^{\text{2}}}-\text{15}$ are $\sqrt{\text{15}}$ and $-\sqrt{\text{15}}$.
Now, Sum of zeroes$\text{=}\sqrt{\text{15}}\text{+(}-\sqrt{\text{15}}\text{)=0=}\dfrac{-\text{0}}{\text{1}}\text{=}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$
$\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$
Product of zeroes$\text{=}\left( \sqrt{\text{15}} \right)\times \left( -\sqrt{\text{15}} \right)\text{=}-\text{15=}\dfrac{-\text{15}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$
$\therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$ .
(vi) $3{{x}^{2}}-x-4$
Ans
Given: $\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$
Now factorize the given polynomial to get the roots.
$\Rightarrow \left( 3x-4 \right)(x+1)$
The value of $\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$ is zero.
when $\text{3x}-\text{4=0}$ or $\text{x+1=0}$, i.e., $\text{x=}\dfrac{\text{4}}{\text{3}}$ or $\text{x=}-\text{1}$
Therefore, the zeroes of $\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$ are $\dfrac{\text{4}}{\text{3}}$ and $\text{-1}$.
Now, Sum of zeroes$\text{=}\dfrac{\text{4}}{\text{3}}\text{+(}-\text{1)=}\dfrac{\text{1}}{\text{3}}\text{=}\dfrac{-\text{(}-\text{1)}}{\text{3}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$
$\therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$
Product of zeroes$\text{=}\dfrac{\text{4}}{\text{3}}\times \text{(}-\text{1)=}\dfrac{-\text{4}}{\text{3}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$
$\therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$ .
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $\dfrac{1}{4},-1$
Ans:
Given: $\dfrac{\text{1}}{\text{4}}\text{,-1}$
Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.
Then,
$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =}\dfrac{\text{1}}{\text{4}}$
$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =}-\text{1}$
Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.
$\Rightarrow {{x}^{2}}-\dfrac{1}{4}x-1$
$\Rightarrow 4{{x}^{2}}-x-4$
Therefore, the quadratic polynomial is $\text{4}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$.
(ii) $\sqrt{2},\dfrac{1}{3}$
Ans:
Given: $\sqrt{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}$
Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.
Then,
$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =}\sqrt{\text{2}}$
$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =}\dfrac{\text{1}}{\text{3}}$
Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.
$\Rightarrow {{\text{x}}^{\text{2}}}-\sqrt{2}\text{x+}\dfrac{1}{3}$
$\Rightarrow 3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}$
Therefore, the quadratic polynomial is $3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}$.
(iii) $0,\sqrt{5}$
Ans:
Given: $\text{0,}\sqrt{\text{5}}$
Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.
Then,
$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ = 0}$
$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =}\sqrt{\text{5}}$
Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.
$\Rightarrow {{\text{x}}^{\text{2}}}-0\text{x+}\sqrt{5}$
$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{5}$
Therefore, the quadratic polynomial is ${{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{5}}$.
(iv) $1,1$
Ans:
Given: $\text{1,1}$
Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.
Then,
$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =1}$
$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =1}$
Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.
$\Rightarrow {{\text{x}}^{\text{2}}}-\text{1x+1}$
Therefore, the quadratic polynomial is ${{\text{x}}^{\text{2}}}-\text{x+1}$.
(v) $-\dfrac{1}{4},\dfrac{1}{4}$
Ans:
Given: $-\dfrac{\text{1}}{\text{4}}\text{,}\dfrac{\text{1}}{\text{4}}$
Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.
Then,
$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =}-\dfrac{\text{1}}{\text{4}}$
$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =}\dfrac{\text{1}}{\text{4}}$
Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.
$\Rightarrow {{\text{x}}^{\text{2}}}-\left( -\dfrac{1}{4} \right)\text{x+}\dfrac{1}{4}$
$\Rightarrow \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1}$
Therefore, the quadratic polynomial is $\text{4}{{\text{x}}^{\text{2}}}\text{+ x +1}$ .
(vi) $4,1$
Ans
Given: $\text{4,1}$
Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.
Then,
$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ = 4}$
$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =1}$
Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.
$\Rightarrow {{\text{x}}^{\text{2}}}-\text{4x+1}$
Therefore, the quadratic polynomial is ${{\text{x}}^{\text{2}}}-\text{4x+1}$.
## NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2
### Definition
If x be a variable and x be a positive integer with a1, a2, a3, ……., an be constants then, f(x) = anxn + an-1x-1 + ….. + a1x + a0 is called a polynomial in x.
In this polynomial anxn + an-1x-1, ………, a1x and a0 are called the terms of the polynomial, and an, an-1,…., a1, a0 are called their coefficients.
Example: p(x) = 4x -1 is a polynomial in variable x.
q(y) = 2y2 – 3y + 5 is a polynomial in variable y.
Note: Expressions like 3x2 - 2√x + 7, 1/(3x-5), etc. are not polynomials as some powers of x are not integers.
### Degree of a Polynomial
The exponent of the highest degree term in a polynomial is called its degree.
Example:
1. f(x) = 4x -1 is a polynomial in x of degree 1.
2. q(y) = 2y2 - 3y + 5 is a polynomial in y of degree 2.
### Types of Polynomial
Let us see the different types of polynomials.
1. Constant Polynomial: It is a polynomial of degree zero.
Ex: f(x) = 5, g(x) = 2, etc.
The constant polynomial f(x) = 0 is called the zero polynomial. The degree of the zero polynomial is not defined as f(x) = 0, g(x) =0x, h(x) = 0x2, p(x) = 0x3, etc. are all equal to the zero polynomial.
2. Linear Polynomial: It is a polynomial of degrees.
Ex: f(x) = 4x -1.
A polynomial q(y) = 3y2 + 4 is not linear as it has degree 2.
In general, any linear polynomial in x with real coefficients is of the form f(x) = ax + b where a & b are real numbers and a 0.
Note: A linear polynomial may be a monomial (having only one term. Ex: q(x) = 5x) or a binomial (having 2 terms. Ex: f(x) = 4x-1)
3. Quadratic Polynomial: It is a polynomial of degree 2.
Ex: f(x) = 5x2 - 3x + 4.
In general, a quadratic polynomial is of the form f(x) = ax2 + bx + c, where a, b & c are real numbers and a 0.
4. Cubic Polynomial: A polynomial of degree 3 is called a cubic polynomial.
Ex: f(x) = 2x3 – 5x2 + 7x – 9.
The general form of a cubic polynomial is f(x) = ax3 + bx2 + cx + d, where a, b, c, d are real numbers and a 0.
### Value of a Polynomial
If f(x) is a polynomial and ‘a’ is a real number then the real number obtained by replacing x by ‘a’ , is called the value f(x) at x = a and is denoted by f(a). For example: if f(x) = 2x2 - 3x -2, then its value
1. at x = 1 is given by f(1) = 2(1)2 - 3 1 -2 = 2-3-2 = -3
2. at x = -2, is given by f(-2) = 2(-2)2 -3(-2)-2 = 8 + 6 -2 = 12
### Zero of a Polynomial
Consider the cubic polynomial f(x) = x3 – 6x2 + 11x - 6. The value of this polynomial at x =1, 2, 3 are respectively
f(1) =13 – 6 * 12 + 11 * 1 – 6 = 1-6 + 11 -6 =0
f(2) = 23 – 6 * 22 + 11 * 2 – 6 = 8 – 24 + 22 – 6 = 0
f(3) = 33 – 6* 32 + 11 * 3 -6 = 27 -54 + 33 -6 = 0
1, 2, 3 are called the zeroes of the cubic polynomial f(x) = x3 -6x2 + 11x - 6.
A real number is 0 of a polynomial f(x), if f() = 0.
To find this we solve the equation f(x) = 0.
For Example: if the polynomial is f(x) = x -3 then putting x -3 = 0, we get x = 3
Then the real number 3 is the zero of the polynomial f(x) = x -3.
There can be more than one zero of a polynomial as can be seen in the example f(x) = x3-6x2 + 11x - 6.
### Graph of a Polynomial
The graph of a polynomial f(x) is the set of all points (x, y), where y = f(x). This graph is a smooth free hand curve, passing through the points (x1, y1), (x2, y2), (x3,y3), ……, etc. y1, y2, y3, …… are the estimated values of the polynomial f(x) at x1, x2, x3,..... respectively.
Geometrical meaning of the zeroes of a linear polynomial
We take an example f(x) = 4x - 2
Let y 4x - 2
We make a table with values of y corresponding to different values of x.
A B C x 0 2 ½ y -2 6 0
The points A(0, 1) and B (2, 6) are plotted on the graph paper on a suitable scale. A line drawn runs through these points to derive the graph of the given polynomial. It can be seen that the graph intersects the x-axis at C (½, 0), where y = f(x) = 0 so that the zero of the given polynomial is ½.
### Meaning of the Zeroes of a Quadratic Polynomial
Graph of a quadratic polynomial f(x) = x2 - 2x - 8
Let y = x2 - 2x - 8
We construct a table containing the values of y corresponding to various values of x:
x -4 -3 -2 -1 0 1 2 3 4 5 6 y = x2 - 2x - 8 16 7 0 -5 -8 -9 -8 -5 0 7 16
We now plot the points (-4, 16), ( -3, 7), (-2, 0) , (-1, -5), (0 -8), (1, 9), (2, 8), (3, -5), (4, 0) , (5, 7) and (6, 16) on a graph paper and draw a smooth free hand curve passing through these points. Hence, the curve derived demonstrates the graph of the polynomial f(x) = x2 -2x -8. Such a curve is called a parabola. The lowest point of the graph (1, 9) is called the vertex of the parabola.
### The Following Observations Can Be Made On Class 10 Maths ex 2.2
1. The graph is symmetrical about the line parallel to the y-axis through the vertex (1, -9). This line is called the axis of the parabola.
2. The concavity of the parabola is towards the positive direction of the y-axis (this happens when the coefficient of x2 is positive.
3. As x2 -2x -8 = (x-4) (x + 2), the polynomial f(x) has two factors (x-4) & (x+2), so that the parabola cuts the x-axis at two distinct points (4,0) & (-2, 0). The x-coordinates of these two points: 4 & -2 are the two zeroes of f(x).
### Note in Exercise 2.2 Class 10 Math:
1. The number of zeroes of a polynomial depends on its degree. For a polynomial of degree n, the number of zeroes can be utmost n or less. For example, for a quadratic, polynomial the no. of zeroes can be 2 or less.
2. For a quadratic polynomial, its graph (ex: a parabola) may cut the x-axis at two points in fig (i), or at one point in fig (ii) or at no point at all in fig (iii).
## Conclusion
Your one-stop resource for learning how to solve these expressions might be Vedanta's Ex 2.2 Class 10 NCERT Solutions. These solutions simplify difficult subjects like factorization and finding zeroes into simple steps with illustrations. Pay particular attention to comprehending how a polynomial's zeroes and coefficients relate to one another.You'll come across a number of polynomial problems throughout the chapter, and this is the key to solving them. By using these solutions for practice, you'll get more confident and ace the test!
## NCERT Solutions Class 10 Maths Chapter 2 Exercises
Chapter 2 - Polynomials All Exercises in PDF Format Exercise 2.1 1 Question & Solutions
## Chapter-Specific NCERT Solutions for Class 10 Maths
Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
## FAQs on NCERT Solutions for Class 10 Maths Chapter 2: Polynomials - Exercise 2.2
1. State the condition in which the graph of quadratic polynomial p(x) = ax2 + bx + c is an upward parabola or downward parabola.
For a > 0, the parabola is upward and for a < 0, the parabola is downward.
2. Where will I get the best reference for NCERT Solutions for Class 10 Chapter 2 -- Polynomial Exercise 2.2?
You will get the best reference for NCERT Solutions for Class 10 Chapter 2 - Polynomial Exercise 2.2 on the official website of Vedantu, the leading education portal in India. The solutions and the reference notes for NCERT Solutions Chapter 2 - Polynomials are created by the in-house subject experts stepwise as per the latest guidelines of NCERT (CBSE). These notes and solutions will give you a better understanding of the topic and will help you with revision for your exams.
3. Can I download the NCERT Solutions for Class 10 Chapter 2 -- Polynomials Exercise 2.2 from the Vedantu app?
You can definitely download the free pdf of NCERT for Class 10 Chapter 2 -- Polynomials Exercise 2.2 from the Vedantu app. You can also register for Maths online tuitions on www. Vedantu.com.
4. What are the main topics in Exercise 2.2 of Chapter 2 of Class 10 Maths to learn?
Exercise 2.2 of Chapter 2 of Class 10 Maths textbook focuses on polynomials and their types. Values, degrees, and how to plot them on the graph are also covered. Polynomial equations are also solved and answers are explained. The solutions PDF has answers to all the NCERT questions and also extra questions from practice developed by experts. The notes and answers in the solutions PDF will help you understand the concepts better and easily.
5. How many questions are there in Exercise 2.2 of Chapter 2 of Class 10 Maths?
Exercise 2.2 from the Chapter 2 of Class 10 Maths book has a few questions. The answers to all these questions and more for practice can be found in the solution PDF by Vedantu. Before the exercise, the examples are also explained in depth in the PDF. This helps the students understand the concept better. The PDF explains the concepts required to answer each question. All the questions and answers are developed by experts.
6. How many examples are there in the NCERT textbook before Exercise 2.2 in Chapter 2 of Class 10 Maths?
There are five examples before Exercise 2.2 of Chapter 2 of Class 10 Maths in the NCERT textbook. These examples discuss the relationship between coefficients and zeros. This is the topic on which Exercise 2.2 of Chapter 2 is based. All examples are explained step by step and give an overview of Exercise 2.2 of Chapter 2 through this. The NCERT Solutions PDF also has extra questions in addition to the exercise questions to help students understand the concept.
7. What is the sum and the product of the zeros of the polynomial, according to Exercise 2.2 of Chapter 2 of Class 10 Maths?
The zeroes of a polynomial are the values of the equation. They are the factors of p(x). In the concept covered before Exercise 2.2 of Chapter 2 of Class 10 Maths, we understand that the sum of the zeros of an equation is equal to -(coefficient of x) divided by the coefficient of x square. The product of the zero is the constant term divided by the coefficient of x square. The concept is explained with the help of examples in the textbook before the Exercise and the NCERT Solution PDF.
8. How can I practice more questions like in Exercise 2.2 of Chapter 2 of Class 10 Maths NCERT textbook?
Students can refer to the NCERT Solutions PDF for Exercise 2.2 of Chapter 2 of Class 10 Maths free of cost on the Vedantu website and the Vedantu app, for NCERT Solutions and extra questions. The solutions PDF provides solutions for all the questions asked in the exercise. It has enough questions for the students to practice. The PDF also explains concepts in an easy-to-understand language by experts for the students to grasp them better. For extra practice, students should refer to the pdf and practice as many questions as possible. | 8,420 | 25,090 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.875559 |
https://cs.stackexchange.com/questions/66246/calculating-the-process-time-for-a-synchronous-network-system | 1,627,714,785,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154053.17/warc/CC-MAIN-20210731043043-20210731073043-00306.warc.gz | 199,180,493 | 37,052 | # Calculating the process time for a synchronous network system
Okay I am trying to understand why asynchronous systems are better that synchronous systems in terms of throughput.
Relating to the calculation done in this answer. Can someone tell if my reasoning is correct ?
Lets say I have a system that has 1000 live connections and each connection takes 5 packets to complete and a packet is received every millisecond. Processing time is 0.001 ms(1 micro second). The system can only process 1 packet at a given time, if there are multiple packets then remaining packets must wait until the processing is complete for the present packet.
My reasoning and calculation-:
So in the 1st millisecond the system receives 1000 packets, all 1st packets of the 5 packets from each of the 1000 connections. Am I right ?
If I am correct then the process time should be calculated in the following fashion-:
1st packet of 1st connection
Waiting Time for first packet-: 0ms
Process Time-: 0.001ms
Time taken to completion of this packet-: 0.001 (wait time + process time)
1st packet of 2nd connection
Waiting Time : time taken for completion of previous packet = 0.001ms
Process time = constant = 0.001ms
Time taken to completion of this packet = (wait time + process time) = 0.001+0.001 ms = 2*0.001 = 0.002ms
So in the same way processing all 1st packets of the 1000 connections would take = 1000*0.001 = 1ms
So in a total I will be receiving 5*1000 packets = 5000 packets, so processing them will take 5000*0.001 = 5ms
Am I correct in my above reasoning and calculation ? If not please correct me where I am going wrong.
• Typical packet size is 1500 bytes. To receive one packet per microsecond you need 12 Gb / sec assuming zero overhead. I'd enquire what is meant by "processing time". – gnasher729 Nov 20 '16 at 12:44
• @gnasher729 Given the information is my arithmetic correct. Because the answer that I have linked to is coming up with a different answer. – Kramer786 Nov 20 '16 at 17:01
• We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. – D.W. Nov 20 '16 at 18:25 | 629 | 2,530 | {"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-2021-31 | latest | en | 0.919256 |
https://coolprofessors.com/according-to-the-bureau-of-labor-statistics-it-takes-an-average-of-22-weeks-for-someone-over-55-to-find-a-new-job-compared-with-16-weeks-for/ | 1,685,528,414,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646457.49/warc/CC-MAIN-20230531090221-20230531120221-00429.warc.gz | 184,796,171 | 18,706 | # According to the Bureau of Labor Statistics, it takes an average of 22 weeks for someone over 55 to find a new job, compared with 16 weeks for
Journal,
September 2, 2008). Assume that the probability distributions are
normal and that the standard deviation is 2 weeks for both distributions.
(a) What is the probability that it takes a worker over the age of 55 more
findajob. 19 weeks to find a than
(b) What is the probability that it takes a younger worker more than 19
weeks to find a job?
(c) What is the probability that it takes a worker over the age of 55 between
23 and 25 weeks to find a job?
(d) What is the probability that it takes a younger worker between 23 and
25 weeks to find a job?
(e) Let’s say that the Journal defined the top 5% fastest of younger workers
who found jobs as “Rising Stars”. If you are a younger worker, how fast
do you need to find a job to be considered as a “Rising Star”?
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https://www.pnginsight.com/MathsExamResources/grade-9-maths-distance-and-open-learning-modular-units/ | 1,713,732,477,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00019.warc.gz | 833,282,624 | 16,668 | Posted in: FODE Maths, Open and Distance Learning
# Grade 9 Distance and Open Learning Modules
The Grade 9 distance and open learning maths modules (also called modular strands) provided in this article are for students who are studying at a FODE in the provincial centres. We simplify the 6 mathematics modules into topics to make it easier for students to follow. It is an important outline to use as a study guide or when revising for the Grade 10 mathematics examination.
## Grade 9 distance and open learning modules/strands
The Grade 9 distance learning mathematics units continue from the Grades 7 and 8 units. That is why it is important to understand the lower primary strands before moving on to Grade 9 and 10 maths.
Take a look at the checklist for Grade 8 modular units (strands and sub-strands).
In Grade 9 Mathematics, there are 6 material aids or PDF books for students. Each modular unit corresponds to the core mathematics strands students study via the distance learning mode. And further broken down into sub-strands, FODE info page.
The general guide to Grade 9 Distance Learning maths content areas are listed below.
## Grade maths distance learning topics
Grade 9 Module 1: Maths in Community
• numbers and operations
• money and percentages
• ratio and rates
• measurements
Grade 9 Module 2: Patterns and Change
• directed numbers
• indices
• algebraic expressions
• equations
Grade 9 Module 3: Working with Data
• organising of data
• presentation of data on graphs
• measures of central tendency
Grade 9 Module 4: Statistics, Graphs and Probability
• straight-line graphs
• probabilities
• statistical estimation
• points, lines, planes and angles
• polygons
• area
• surface area and volume
• transformation and symmetry
• similarity and congruence
• circles
• constructions
## Grade 9 maths modular units and topics
There is no major difference between the mathematics topics inscribed in the syllabus as far as the mainstream schools and distance learning modes are concerned.
The main difference is that the main schools have teachers there 2-4-7 – all the time to teach and guide the students. Whereas, distance learning such as the FODE concept is self-paced where students learn at their pace.
The main syllabus topic for Grades nine and ten in mainstream schools can be found here.
## Know the distance and open learning units
In the FODE course booklet, each strand is divided into 4 sub-strands and each sub-strand consists of 5 to 7 lessons. That is 6 strands and 24 sub-strands altogether. In the coursebook, the students will find the:
• Exercises,
• Practice Questions, and
• a Summary of Lessons.
Note that the answers to the Practice Questions are usually given at the end of the main strand.
Each strand has one assignment. For the 6 main modular strands in the Grade 9 FODE course, students will do 6 assignments. Here is FODE’s expectation:
You will study Strand 1 and do Assignment 1 at the same time. Then you will study Strand 2 and do Assignment 2 at the same time, and so on up to Strand 6 and Assignment 6.
When you have completed your assignment for each strand, you must submit it immediately for marking.
FODE is flexible about when to submit your completed assignment. But as a self-paced student, you should aim to complete your work and submit it promptly to your study centre for marking.
Students who are studying at a registered Study Centre must give their completed assignments to their Supervisor for marking. Other students who study at home or in the provinces must send their completed assignments to their Provincial Study Centres for marking.
The study centres are also where students follow up on their assessment marks.
Students who wish to get practise maths questions and answers can download them here. Get the question booklet: Grade 10 past exam papers, PDF. Though the papers are from the mainstream schools, they are also the best questions to practise for your mathematics examination at the end of Grade 10.
The latest Grade 10 mathematics exam papers are password-protected. We will release them password before the Grade 10 mathematics examination. Check them out.
## FODE application and enrollment info
We compiled some information about application and enrollment for distance education at FODE. The article contains specific details about how to apply, and provide further information on Grades 7 – 10 course programs. Click here.
The information is for those students who are starting as new students. So, if you have enrolled, you can use the resources we provide here for study and revisions purposes.
We hope you find this resource useful as you strive to get the education you deserve. PNG Insight Mathematics is an independent website. We do not represent the Department of Education or FODE. We provide the resource here as learning guides only. | 1,015 | 4,877 | {"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} | 3.640625 | 4 | CC-MAIN-2024-18 | latest | en | 0.870366 |
https://undergroundmathematics.org/polynomials/r7108/solution | 1,660,150,818,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00630.warc.gz | 538,747,443 | 5,297 | Review question
# When is $x^n + 1$ divisible by $x + 1$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R7108
## Solution
Show that $x^n + 1$ is divisible by $x + 1$ when, and only when, $n$ is an odd integer.
By the factor theorem, given a polynomial $p(x)$ with real coefficients and some real number $a$, we know that $x-a$ is a factor of $p(x)$ if and only if $p(a) = 0$.
If we let $p(x) = x^n + 1$ and $a = -1$, by the factor theorem we have that \begin{align*} \text{x^n + 1 is divisible by x + 1} &\iff (-1)^n + 1 = 0 \\ &\iff (-1)^n = -1 \\ &\iff \text{n is an odd integer}. \end{align*} | 227 | 690 | {"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": 3, "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-2022-33 | latest | en | 0.829403 |
https://www.thestudentroom.co.uk/showthread.php?t=4043361 | 1,513,207,418,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948531226.26/warc/CC-MAIN-20171213221219-20171214001219-00309.warc.gz | 806,495,742 | 42,313 | You are Here: Home >< Maths
# S3 Watch
1. https://771a1ec81340d97ae9ed29694f73...XZCVmc/CH1.pdf
On Page 10, for part d the summation part, I have that it is 30 by doing
3 * (Var(Y)+0.25Var(X)
So I get 3(9+2) = 33 which does not match solution
Is it not this way, they use Y1+y2 etc.
2. (Original post by L'Evil Wolf)
https://771a1ec81340d97ae9ed29694f73...XZCVmc/CH1.pdf
On Page 10, for part d the summation part, I have that it is 30 by doing
3 * (Var(Y)+0.25Var(X)
So I get 3(9+2) = 33 which does not match solution
Is it not this way, they use Y1+y2 etc.
Puzzled as to why you're multiplying the Var(X) by 3.
Hence
3. (Original post by ghostwalker)
Puzzled as to why you're multiplying the Var(X) by 3.
Hence
How come it is not (Y1-0.5X)+(Y2-0.5X)+(Y3-0.5X)?
4. (Original post by L'Evil Wolf)
How come it is not (Y1-0.5X)+(Y2-0.5X)+(Y3-0.5X)?
I added an extra line to my previous post.
It's not what you've just posted because of the way S is defined. There's only one lot of X involved.
5. (Original post by L'Evil Wolf)
How come it is not (Y1-0.5X)+(Y2-0.5X)+(Y3-0.5X)?
The summation sign only applies to what's immediately following.
6. (Original post by ghostwalker)
I added an extra line to my previous post.
It's not what you've just posted because of the way S is defined. There's only one lot of X involved.
ah because X is not changing right?
7. (Original post by L'Evil Wolf)
ah because X is not changing right?
It's not because the X isn't changing.
It's really an extension of BIDMAS, or BODMAS or whatever you use. The summation is effectively a function, like log or cosine, etc.
log 6 + 7 means (log 6) + 7, not log (6+7)
8. (Original post by ghostwalker)
The summation sign only applies to what's immediately following.
Thank you very much I now understand.
9. (Original post by L'Evil Wolf)
Thank you very much I now understand.
10. (Original post by ghostwalker)
The summation sign only applies to what's immediately following.
I know this is a while after lol but:
S = Yi - 0.5X + Y2-0.5X + Y3 - 0.5X
= 3[Yi - 0.5X]
= 3Var(Yi)+0.25Var(X)
which still yields 28 in that regard!
So it is a conincindence that both ways that you stated result in the same answer?
11. (Original post by L'Evil Wolf)
I know this is a while after lol but:
S = Yi - 0.5X + Y2-0.5X + Y3 - 0.5X
= 3[Yi - 0.5X]
= 3Var(Yi)+0.25Var(X)
which still yields 28 in that regard!
So it is a conincindence that both ways that you stated result in the same answer?
How did you go from the first bolded line to the second bolded line?
12. (Original post by Zacken)
How did you go from the first bolded line to the second bolded line?
Took a factor of 3 out and then applied the variance of it. I assume you consider it to be wrong?
The idea stemmed from integration, how we take out constants and then integrate if that makes sense?
13. (Original post by L'Evil Wolf)
Took a factor of 3 out and then applied the variance of it. I assume you consider it to be wrong?
The idea stemmed from integration, how we take out constants and then integrate if that makes sense?
By that logic, why don't you "take out the 0.5 and then applied the variance of X"?
14. (Original post by Zacken)
By that logic, why don't you "take out the 0.5 and then applied the variance of X"?
Do you mean, have:
3[Yi - 0.5X]
3/2[2Yi - X]
3/2[ 4Var(Yi) + Var(X)]
3/2 [4(9)+4]
3/2[40}
=60
which would be incorrect.
15. (Original post by L'Evil Wolf)
Do you mean, have:
3[Yi - 0.5X]
3/2[2Yi - X]
3/2[ 4Var(Yi) + Var(X)]
3/2 [4(9)+4]
3/2[40}
=60
which would be incorrect.
No, I'm saying, why are you doing:
Var(3(Yi - 0.5X)) = 3Var(Yi - 0.5X).
16. (Original post by Zacken)
No, I'm saying, why are you doing:
Var(3(Yi - 0.5X)) = 3Var(Yi - 0.5X).
I was doing from concepts of integration, as in the factoring out a constant, but this wouldn't apply in this different mathematical operation.
17. (Original post by L'Evil Wolf)
I was doing from concepts of integration, as in the factoring out a constant, but this wouldn't apply in this different mathematical operation.
I'm afraid I don't understand your point, but Ghostwalker's way is the correct version and yours is simply an (incorrect) fluke.
18. (Original post by Zacken)
I'm afraid I don't understand your point, but Ghostwalker's way is the correct version and yours is simply an (incorrect) fluke.
Okay, thank you/
19. (Original post by L'Evil Wolf)
I know this is a while after lol but:
S = Yi - 0.5X + Y2-0.5X + Y3 - 0.5X
OK, you're using the second definition of S (not the one in the original question)
= 3[Yi - 0.5X]
This is meaningless; with the Yi, rather than Y1+Y2+Y3
If you're going to follow through with the second definition, you'd have:
S=Y1+Y2+y3 - 1.5X
assuming we're dealing with just one observation of X, and not three independent ones.
Working out the variance you'd have:
Var(S) = Var(Y1)+Var(Y2)+Var(Y3)+2.25Var( X)
= 3 Var(Y) + 2.25Var(X)
which won't yield 28.
IF you had three independent observations of X instead; i.e. X1,X2,X3, you'd get
S=Y1+Y2+y3 - 0.5X1-0.5X2-0.5X3
Then Var(S) = Var(Y1)+Var(Y2)+Var(Y3)+0.25Var( X1)+0.25Var(X2)+0.25Var(X3)
=3Var(Y)+0.75Var(X)
Again, no 28.
20. (Original post by ghostwalker)
OK, you're using the second definition of S (not the one in the original question)
This is meaningless; with the Yi, rather than Y1+Y2+Y3
If you're going to follow through with the second definition, you'd have:
S=Y1+Y2+y3 - 1.5X
assuming we're dealing with just one observation of X, and not three independent ones.
Working out the variance you'd have:
Var(S) = Var(Y1)+Var(Y2)+Var(Y3)+2.25Var( X)
= 3 Var(Y) + 2.25Var(X)
which won't yield 28.
IF you had three independent observations of X instead; i.e. X1,X2,X3, you'd get
S=Y1+Y2+y3 - 0.5X1-0.5X2-0.5X3
Then Var(S) = Var(Y1)+Var(Y2)+Var(Y3)+0.25Var( X1)+0.25Var(X2)+0.25Var(X3)
=3Var(Y)+0.75Var(X)
Again, no 28.
Oh right, yes. Thank you.
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In my other question I was looking for a relation between the forward speed and the thrust (lift) of an auto-gyro.
I have come to the conclusion that for an auto-gyro in auto-rotation the forward speed will rotate the rotor and with that regulate its rpm. The faster an auto-gyro flies the faster the rotor rotates resulting in a higher thrust.
What I hope to accomplish is forward braking where the auto-gyro flies faster than needed to provide sufficient thrust, resulting in a higher rotation speed which will then result in more thrust and lift the auto-gyro up. This additional thrust isn't needed as the same level is required. By braking the rotor, using a generator, the rotor speed is reduced till the point it is still able to maintain level using auto-rotation. For this theory I need to understand the relation between the forward speed and the rotation speed.
But it is still not completely clear how this forward speed relates to the rotation speed. I hope someone can explain it to me or point me into the right direction.
I might be completely wrong and came to the wrong or inconclusive conclusions if so don't hesitate to tell me. | 258 | 1,262 | {"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-2021-31 | longest | en | 0.949931 |
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# Practice Problems
Lesson 5. Unit 5. Grade 6 Open Up Resources
Open Up ResourcesVaries
This Problem Set is a part of the Lesson 5, Unit 5, Grade 6. Decimal Points in Products Use understanding of place value and fractions to reason about multiplication of decimals and the placement of the decimal point in a product. Let's look at products that are decimals. Learning targets: students can use place value and fractions to reason about multiplication of decimals. | 103 | 481 | {"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} | 2.640625 | 3 | CC-MAIN-2020-45 | latest | en | 0.859026 |
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MOUNTAIN VIEW, Calif. (Sept. 20, 2012) – One of the most common questions that flits across a math student’s mind is, “How will I use this?” Understandably, it may be difficult to see at a glance how finding a solution to the equation 2x2 + 5x – x = 17 might directly correlate to one being able to tie their shoelaces. (By the way, if you really are using this equation for such a purpose, it may be time to invest in less complicated shoelaces.)
Thank goodness for calculus. Admittedly, some of the concepts are tough, but the great thing about it is that you can easily see its practical applications. Rather than simply juggling numbers and variables, you are beginning to apply the mathematical concepts you have picked up along the way to solve real-world problems. Suddenly, math has a purpose! Huzzah!
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Okay, before I start with the question I would like to point out that I am aware that there are many proofs of this question online. However, I am interested in showing this with the definitions of my university script which are somewhat different from most of what I've seen online.
Show that $P \subseteq NP$.
Definition 1: A set of words $A$ is in $NP$ if and only if there is a polynomial $p$ and a set $B\in P$ such that
$$x\in A \iff \exists y(|y| \le p(|x|)\land x\#y \in B)$$
Definition 2: $$P = \bigcup_{polynomial \; p}TIME(p(n))$$
Definition 3: $$TIME(f(n)) = \{A\subseteq \Sigma^*:\text{there exists a multitape Turing machine M which computes} \;\chi_A \text{with}\;time_M(x)\le f(|x|) \}$$
Definition 4: $$time_M:\Sigma^* \to \Bbb N$$ is the function that gives the amount of "computing steps" for a word $x$.
Definition 5: $$\chi_A$$ is the indicator (also called characteristic) function of $A$.
My attempt:
The way I understand definition 1 $$P\subseteq NP :\iff$$ $$x\in P \iff (\exists polynomial \; p)(\exists B \in P)(\exists y \in \Sigma^*)(|y| \le p(|x|)\land x\#y \in B)$$
Proof:
(i) "$\Rightarrow$"
Let $x\in P$ and $|x|=n$ for an $n\in \Bbb N$
Define
$p(t):=n$ for all $t\in \Bbb N$
$y:=x$
$B:=\{x\#x\} = \{x\#y\}$
Then
$|y|\le p(|x|)$ since $|y|=|x|=n$ and $p(|x|)=p(n)=n$
and $\{x|\#y\}\in B$
(ii) "$\Leftarrow$"
Let $p$ be a polynomial, $B\in P$, $y\in \Sigma^*$ such that
$$|y| \le p(|x|)\land x\#y \in B \text{ for an x\in\Sigma^*}$$
Now I have to show that $x\in P$ but I do not know how.
Furthermore, I do not understand is that in (i) I could have let $x$ be an element of any set and this implication would still hold. For example $x\in S$ for a set $S:NP\subset S$ with the rest of the proof being identical. This would mean that every set is in NP which is not the case. My guess is that my understanding of definition 1 is wrong.
This raises my first question: How come is not every set is a subset of NP using the definitions from above?
My second question: How do I go about solving (ii)?
• We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. – Discrete lizard May 5 '18 at 10:28
• @Discretelizard Thanks for your feedback, I have edited the question to avoid yes/no answers – Travis May 5 '18 at 10:48 | 850 | 2,745 | {"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} | 3.84375 | 4 | CC-MAIN-2019-26 | longest | en | 0.873598 |
https://jp.mathworks.com/matlabcentral/answers/276944-creating-2d-scatter-plot-with-different-dot-size-and-color-accoridng-to-vlaues-in-matrices?s_tid=prof_contriblnk | 1,685,387,607,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00238.warc.gz | 387,233,235 | 27,435 | # Creating 2d scatter plot with different dot size and color accoridng to vlaues in matrices
7 ビュー (過去 30 日間)
SMA 2016 年 4 月 4 日
I have two matrices, one is say 5x5 double with magnitude and the other is 5x5 of direction. The 'y-axis' or values along the column represent a length while the values along the row represent years.
Matrix1 =
1 3 4 5 8
2 7 3 9 1
2 4 5 4 3
7 1 9 8 6
Matrix2 =
-0.1 -0.93 0.4 0.65 0
.25 .71 .83 -.59 -0.7
.23 -.14 -.55 .49 .53
74. .71 -.19 .58 -.76
Row(x-axis) =
2000 2001 2002 2003 2004
Column(y-axis) =
50
40
30
20
10
I want to create a scatter plot with x and y axis as above with matrix1 giving size of the dot (say everything smaller than 5 as one size and larger than 5 as another) and Matrix2 giving the color (say negative sign as red, positive as blue, yellow otherwise).
I am having a hard time implementing this with scatter plot. It is not a 3d plot, basically it says that in 2000, the 50th member had magnitude of 1 and negative direction.
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### 採用された回答
Azzi Abdelmalek 2016 年 4 月 4 日
M1=[1 3 4 5 8
2 7 3 9 1
2 4 5 4 3
7 1 9 8 6]
M2=[-0.1 -0.93 0.4 0.65 0
.25 .71 .83 -.59 -0.7
.23 -.14 -.55 .49 .53
74. .71 -.19 .58 -.76]
R1=[2000 2001 2002 2003 2004]
C1=[50 40 30 20 10 ]
[ii,jj]=meshgrid(R1,C1)
cl=sign(M2)+2
col='byr'
for k=1:numel(M1)
h=scatter(ii(k),jj(k),'MarkerFaceColor',col(cl(k)),'linewidth',M1(k))
hold on
end
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Translated by | 608 | 1,611 | {"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-2023-23 | longest | en | 0.64148 |
https://www.nagwa.com/en/worksheets/823172574929/ | 1,550,780,068,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247508363.74/warc/CC-MAIN-20190221193026-20190221215026-00133.warc.gz | 919,008,199 | 46,408 | # Worksheet: Arcs and Arcs Measures
Q1:
Given that is a diameter in circle and , determine .
Q2:
Given that is a diameter in circle and , determine .
Q3:
Given that is a diameter in circle and , determine .
Q4:
Suppose that . Determine .
• A
• B
• C
• D
Q5:
Find the measure of the arc that represents of the circumference of a circle.
Q6:
Find the measure of the arc that represents of the circumference of a circle.
Q7:
What is ?
Q8:
An arc has a measure of .
What is the measure of the central angle?
What is the measure of the inscribed angle?
What is the measure of the circumscribed angle?
Q9:
Find and .
• A ,
• B ,
• C ,
• D ,
Q10:
Given that is a diameter in circle M and , determine .
Q11:
Given that is a diameter in circle M and , determine .
Q12:
Given that is a diameter in circle M and , determine .
Q13:
Given that is a diameter in circle M and , determine .
Q14:
An arc measures of the circumference of a circle. What angle does the arc subtend at the center?
• A
• B
• C
• D
Q15:
Which of these is equal to the measure of an arc?
• Athe measure of the central angle subtended by that arc
• Bthe inscribed angle subtended by that arc
Q16:
Suppose that points and on a circle center make times its reflex angle. What is the minor arc ?
• A
• B
• C
• D
Q17:
Given that and , find .
Q18:
Given that measure arc the measure of arc , find .
Q19:
Find . | 374 | 1,410 | {"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-2019-09 | longest | en | 0.898294 |
http://www.physicsforums.com/showthread.php?t=688907 | 1,410,754,297,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657104119.19/warc/CC-MAIN-20140914011144-00306-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 730,586,110 | 8,262 | # Why aren't these functions the same?
by Whistlekins
Tags: functions
P: 21 I have f(x) = (x^2+x-2)/(x-1) and g(x) = x+2 Now everyone would agree that f has a domain R\{1} and g has a domain R. Yet I can write (x^2+x-2)/(x-1) = x+2 So why wouldn't g have a domain R\{1} if I rewrite the expression, and vice versa for f? What mathematical principle is behind this?
HW Helper
P: 3,557
Quote by Whistlekins I have f(x) = (x^2+x-2)/(x-1) and g(x) = x+2 Now everyone would agree that f has a domain R\{1} and g has a domain R. Yet I can write (x^2+x-2)/(x-1) = x+2 So why wouldn't g have a domain R\{1} if I rewrite the expression, and vice versa for f? What mathematical principle is behind this?
If you evaluate f(1) then you get an undefined value 0/0. Of course we know that
$$\lim_{x\to 1}f(x) = 3$$
But just because the limit exists doesn't mean that the function is defined at that point.
P: 21 I understand that. But why can't I write g(x) = (x^2+x-2)/(x-1) = x+2 ?
P: 835
Why aren't these functions the same?
Quote by Whistlekins I have f(x) = (x^2+x-2)/(x-1) and g(x) = x+2 Now everyone would agree that f has a domain R\{1} and g has a domain R. Yet I can write (x^2+x-2)/(x-1) = x+2
Only if x is not equal to 1. So you need to write ##\forall x \neq 1, \,\frac{x^2+x-2}{x-1} = x+2##
So why wouldn't g have a domain R\{1} if I rewrite the expression, and vice versa for f? What mathematical principle is behind this?
But you didn't rewrite the expression. If I define
##h : \mathbb{R} \to \mathbb{R}## with ##h(x) = \frac{x^2+x-2}{x-1}## and ##h(1) = 3## then that is indeed equal to g(x), but not equal to f(x).
There is a difference between you can do something, and you did something.
Mentor
P: 18,346
Quote by Whistlekins I have f(x) = (x^2+x-2)/(x-1) and g(x) = x+2 Now everyone would agree that f has a domain R\{1} and g has a domain R.
I wouldn't agree with this. The domain is something chosen by the person who defines it. All we can say is that the maximal possible domain of ##f## is ##\mathbb{R}\setminus \{1\}##. But the domain can possibly be much smaller if we choose it to be.
Yet I can write (x^2+x-2)/(x-1) = x+2 So why wouldn't g have a domain R\{1} if I rewrite the expression, and vice versa for f? What mathematical principle is behind this?
The equation
$$\frac{x^2 + x - 2}{x-1} = x+2$$
is only valid for ##x\in \mathbb{R}## with ##x\neq 1##. For ##x=1##, it is not true. So we have that ##f(x) = x+2## for all ##x\in \mathbb{R}\setminus \{1\}##. The value ##f(1)## still isn't defined.
That ##f(1)=3## somehow, is false. However, this is why limits are invented. So we can say that
$$\lim_{x\rightarrow 1} f(x) = 3$$
So although ##f(1)## doesn't make sense, we can take the limit. The limit denotes the value that ##f(1)## would have been if it were defined in ##1## and if ##f## were to be continuous.
P: 835
Quote by micromass I wouldn't agree with this. The domain is something chosen by the person who defines it. All we can say is that the maximal possible domain of ##f## is ##\mathbb{R}\setminus \{1\}##. But the domain can possibly be much smaller if we choose it to be.
In fact it could also be larger. No body said x couldn't be complex.
Mentor
P: 18,346
Quote by pwsnafu In fact it could also be larger. No body said x couldn't be complex.
Very true!
P: 21 So if I previously define the domain, I can't change that domain unless I write an entirely new function? Would it be true that if h(x) = x+2 , for all x in R\{1}, then f = h?
Mentor
P: 18,346
Quote by Whistlekins So if I previously define the domain, I can't change that domain unless I write an entirely new function? Would it be true that if h(x) = x+2 , for all x in R\{1}, then f = h?
Yes to both.
P: 21 Cool, thanks for helping me clear my confusion. I guess that never really got explained to me by anyone and I never picked up on it.
Related Discussions Introductory Physics Homework 19 General Physics 20 Classical Physics 2 General Discussion 7 General Discussion 43 | 1,282 | 4,002 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2014-41 | latest | en | 0.946358 |
visionpointzone.online | 1,623,785,669,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621519.32/warc/CC-MAIN-20210615180356-20210615210356-00614.warc.gz | 48,091,650 | 5,560 | Skewness In Box Plot | visionpointzone.online
A box and whisker plot sometimes called a boxplot is a graph that presents information from a five-number summary. In a box and whisker plot: the ends of the box are the upper and lower quartiles, so the box spans the interquartile range. the median is marked by a vertical line inside the box. It shows information about the location, spread, skewness as well as the tails of the data. However, when the data are skewed, usually many points exceed the whiskers and are often erroneously declared as outliers. An adjustment of the boxplot is presented that includes a robust measure of skewness in the determination of the whiskers. In a boxplot, the interquartile range is represented by the width of the box Q3 minus Q1. In the chart above, the interquartile range is equal to about 7 minus 3 or about 4. And finally, boxplots often provide information about the shape of a data set.
Box-plot grafico a scatola Come si costruisce un box-plot Si calcola per la variabile statistica X il primo ed il terzo quartile, la mediana, il valore minimo e quello massimo della distribuzione riportando i valori in un piano cartesiano asse delle ascisse o delle ordinate in modo indifferente, nel seguito tutti i valori. Regarding the box plot itself, it is a crude summary of a distribution and not necessarily good enough to allow the skewness to be estimated. Indeed, there can be quite different data sets, with quite different skewness, that will have the same boxplot.
The box of the plot is a rectangle which encloses the middle half of the sample, with an end at each quartile. The length of the box is thus the interquartile range of the sample. The other dimension of the box does not represent anything in particular. A line is drawn across the box at the sample median. Another way to characterize a distribution or a sample is via a box plot aka a box and whiskers plot. Specifically, a box plot provides a pictorial representation of the following statistics: maximum, 75 th percentile, median 50 th percentile, mean, 25 th percentile and minimum. On a box and whisker diagram, outliers should be excluded from the whisker portion of the diagram. Instead, plot them individually, labelling them as outliers. Skewness. If the whisker to the right of the box is longer than the one to the left, there is more extreme values towards the positive end and so the distribution is positively skewed. But in any case the way to plot a distribution is to look up its skewness and kurtosis and plot them. If the skewness and kurtosis are fixed, just plot that point and label it. However, note that some distributions may not have both skewness and kurtosis being finite if kurtosis is finite then skewness must be too, and if skewness is not. seaborn.boxplot x=None, y=None,. Draw a box plot to show distributions with respect to categories. A box plot or box-and-whisker plot shows the distribution of quantitative data in a way that facilitates comparisons between variables or across levels of a categorical variable.
When a box plot is left-skewed, values gather at the upper end, making a short and tight section there. To the left of that crowd, data points spread out, creating a longer tail. Box plots are like the base of distribution curves. Skewness suggests that data may not be normally distributed. Limitations of box. | 759 | 3,379 | {"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-2021-25 | latest | en | 0.840102 |
https://www.doglink.pt/chase-daniel-cwckpzm/pn93tcs.php?908327=3-phase-voltage-calculation-formula | 1,627,952,400,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00485.warc.gz | 747,056,359 | 12,381 | BC=651.85A, AC=651.84A, supply voltage 220 line to line. Formula for Voltage | How to calculate Potential Difference. calculations or measurements and from either phase parameters or line parameters. To protect the windings of a transformer against overcurrent, use the percentages listed in Table 450.3(B) and its applicable notes. If sending end voltage and load PF are known. I'm not sure if this helpful, but I will derive the formulae using the above method so you can see the two tie together. I've always been bad a remembering formulae, so I prefer to remember concepts/methods. In a three phase system, the amount of current going through one of the phases is not a fraction of the amount of current going through all three? I expect it to stay the same. Initially, we explored the idea of three-phase power systems by connecting three voltage sources together in what is commonly known as the “Y” (or “star”) configuration.. is it possible to find the power factor given the reactive power(VAR) and real power(KW) by this equation? Through a circuit, a current of 9A flows through that carries a resistance of 10 $$\Omega$$. This configuration of voltage sources is characterized by a common connection point joining one side of each source. If you are looking at it on a phase by phase you may need to be a little careful. Sir, when a voltage dip occurs current increases, so instead of dimming, a bulb should glow brighter but the reverse thing happens, why? The current is simply the kVA divided by the voltage. Only emails and answers are saved in our archive. If you have some expert knowledge or experience, why not consider sharing this with our community. Voltage Drop Formulas and Calculation. Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! VD = $$\frac{2 LRI}{ 1000 \times 0.866}$$ Here, L = refers to the length of the circuit R = refers to the resistance in Omega ($$\Omega$$) I = refers to the load current in amperes. What normally happens in star is that because the voltage is less across each winding, you end up delivering less power then you would in delta (less power is less current). Wapplied = 3 Uln I cos Φ (2), For pure resistive load and power factor = 1 the real power in a 400/230 voltage (line to line / line to neutral) 20 amps circuit can be calculated as, WBHP = 31/2 U I PF μ / 746 (3). Similarly a transformer (with three windings, each identical) supplying a given kVA will have each winding supplying a third of the total power. Assuming 577 V is Line Voltage and load is balanced. You would have 208 volts across each of the loads you gave and can find the current using Ohm’s law (i.e. I know that it is right but how can i used it to find the value of complex power?? By measuring voltages from Line 1 to Line 2, Line 2 to Line 3, and Line 3 to Line 1, and performing a few simple calculations, a technician can determine if there is a voltage unbalance problem. I try not to think like that as it confuses me as well and there is sometimes more to it than that. That 60A is total rms current for a single phase, not the peak current, correct? To protect the windings of a transformer against overcurrent, use the percentages listed in Table 450.3(B) and its applicable notes. This is done all the time (fridge or toilet extract fan in your house for example). To calculate kVA, you need to enter the known values of voltage and the current into the respective fields. When selecting a cable, the performance of the cable under fault conditions is an important consideration. three phase power is 36 kW, single phase power = 36/3 = 12 kW The... Here’s list of some famous scientists. The first formula is used more often because it is generally more convenient to obtain line values of volt- age and current, which can be measured with a voltmeter and clamp-on ammeter. Because you have an unbalanced system, you will have some current in the neural line. When Wire length is in meters. Also depends on if it is three phase or single phase and the voltage. If the system is perfectly balanced I again have no issues figuring out the amps drawn on each leg. So is it as simple as multiply by 3 to get the total, 3-phase rms current? Single & Three Phase Line kVA calculator is an online tool used in electrical engineering to measure the unknown quantity by two known quantities applied to the below formulas for single phase and three phase connection. From All About Circuits: In balanced “Y” circuits, line voltage is equal to phase voltage times the square root of 3 In balanced “Δ” circuits, line voltage is equal to phase voltage: *ignore Iline and Iphase I'm not sure of where you are getting your 491.5A from. I just don't know how to go about it. For example, take a 400 V (VLL) three phase system with the following loads: phase 1 = 80 A, phase 2 = 70 A, phase 3 = 82 A, the line to neutral (phase) voltage VLN = 400/√3 = 230 V The kW in your formula is the three phase kW, so I think you should be using 746 (not 249). Also, thanks again for posting and all of your help. Also, if this was a Delta configuration, then the line current would be A-N, B-N and C-N). Please read AddThis Privacy for more information. Similarly for phase B (1/2AB + 1/2BC) and phase C (1/2BC + 1/2AC). Power is related to the voltage and current so regardless of star or delta if the box taking the same power, both voltage and current will be the same. To find the kVA you can use sqrt(kW^2 + kWr^2), where kWr is the reactive power. It works the same a for a consumer of power. Cheers, Nathan. Basic Formula to Calculate Apparent Power in Single and Three Phase Circuits EE. Power factor (cosΦ). phase 3 apparent power = 82 x 230 = 18,860 VA = 18.86 kVA Three-phase Rectification. What I need to know is how many amps I'm pulling on each individual leg. kVA = kW / power factor = 12/0.86 = 13.9 kW (13900 W) So it is the second part of your reply that I don't understand. Typical power factor values. So, VD = 1.732 x 21.2 x 100 x 80/83,690 = 3.51 Voltage Drop Example 3 — Find the size of copper wire needed in a single-phase application to carry a load of 40 amperes at 240 volts a distance of 500 feet with a 2% voltage drop. Normalized waveforms of the instantaneous voltages in a three-phase system in one cycle with time increasing to the right. If you interested in an introduction you can view our post: Network Theory – Introduction and Review. kVA per phase = kW per phase / power factor *Line-to line voltage values to be considered. Most AC power today is produced and distributed as three-phase power where three sinusoidal voltages are generated out of phase with each other. In general when connecting in delta you have a higher voltage on each leg, so I would say you consume more power. If you don't have a power factor use 0.9. L = Length of the circuit from power supply to load. I = 249/(1.73*0.72*380) = 0.52 A By writing an electrical note, you will be educating our users and at the same time promoting your expertise within the engineering community. The voltage drop formula for 3 phase systems is the following: where: VD = The voltage drop of the circuit, in volts. tian, There is always more than one way to do things. Voltage drop formulas. This calculation applicable for both the primary and secondary winding of a transformer. L = Length of the circuit from power supply to load. The Average-Value Rectifier (Three-Phase) block models an average-value, full-wave, six-pulse rectifier. Voltage drop E VD = IR cosθ + IX sinθ where abbreviations are same as below “Exact Method”. Please check it .if u divide kw by p.f u will get kVA not KW. Formula : Three Phase Electric Power = V * I * 1.732 * PF Where , V = Voltage I = Current PF = Power Factor (0.8) Three Phase Electric Power calculation is made easier here using this online electrical calculator. Dear jcdelao, Using the formula, [Sqrt3 x V x I x Cos (Phi)= Power (Watts)] , the current comes to be about 711 A. The three phase power calculator calculates the active and reactive power current from the following parameters: Voltage (V): Enter the phase-to-phase ($$V_{LL}$$) voltage for a 3-phase AC supply in volts. I work in the entertainment industry and definitely work with a star connected (wye) system where all legs measured to neutral is 120V and leg to leg is 208V. I just went through and got a few strange results, and realised that earlier you state VLL = √3 x VLN Hence I wondered why you would then multiply VLL again by √3. Current I = kVA per phase / phase voltage I have a small 400 V three phase 4 wire (star) installation with the following resistive loads on each; I need to calculate the line current in each of the three phases can some one please help with a formula? The power factor is the ratio of the real work to the apparent power. You give the worse case as 80.6A. If you want to promote your products or services in the Engineering ToolBox - please use Google Adwords. = kW / (3 x pf) divided by (VLL/√3) This configuration of voltage sources is characterized by a common connection point joining one side of each source. What is the difference between the two. Add standard and customized parametric components - like flange beams, lumbers, piping, stairs and more - to your Sketchup model with the Engineering ToolBox - SketchUp Extension - enabled for use with the amazing, fun and free SketchUp Make and SketchUp Pro .Add the Engineering ToolBox extension to your SketchUp from the SketchUp Pro Sketchup Extension Warehouse! Calculating Currents in a Balanced Three Phase Delta Circuit –General ..... 23 4D.1 Resistive Loads ..... 23 4D.2 Capacitive Loads..... 27 4D.3 Inductive Loads..... 29 4D.4 Two or More Loads ..... 31 4E. Kindly show the computation. Phase B starts at 0 at 120 degrees and Phase C starts at 0 at 240 degrees. In the post, the motor is a three phase load, where as the socket is single phase - in each case case you have the line current which is common (for example, the current flowing in L1). Also be aware that if draw more power, you equipment may not have been designed for it. I do have my 'Motor Starting - Star Delta' post drafted which explains this in more detail. AddThis use cookies for handling links to social media. The voltage drop formula for 3 phase systems is the following: where: VD = The voltage drop of the circuit, in volts. Thanks. I will also put together a proper answer when I get some time. Generally when people talk about power factor, they are concerned with the overall system. The Three-Phase Voltage System Three-phase voltage systems are composed of … tanθ=Q/P, Hi all, i have problem in total current computation from ;three phase and single phase, the system is 380v 3-phase, line to neutral is 220. i computed the current from line to neutral AN=80.6, BN= 65.8, CN=73.2a three phase current is 491.5a. As an example, consider a loa… If you also know the power factor you can convert between kVA and kW as shown earlier. Voltage Drop - Generally we take the percentage of voltage drop and the most common percentage of voltage drop is 3 percent and 5 percent. V Rated is the rated AC voltage that you specify on the block mask.. P fixed is the fixed power loss that you specify on the block mask.. R fixed is the fixed per-phase series resistance in an equivalent wye-connected load.. i p, i n are the currents flowing into the positive and negative terminals of the rectifier. Add up the three phases = 48720 VA (or 48.72 kVA). Phase A starts at 0 at 0 degree phase angle, rises to 1 at 90 degrees, back to 0 at 180, to -1 at 270 degrees, and back to 1 at 360 degrees. The motor would be running faster due to the frequency. Need your excellent advise Mr. Steven. When I try to remember formulae I always forget them soon or become unsure if I am remembering them correctly. Three-phase electric power is a common method of alternating current electric power generation, transmission, and distribution. It is important that calculations be carried... A lot of our members work in countries where PPE (personal protective equipment) is regulated or they work for companies/organizations which take employee... Modelica is an open source (free) software language for modelling complex systems. THREE-PHASE CIRCUIT CALCULATIONS In the following examples, values of line and phase voltage, line and phase current, … so that i can assign suitable cable and breaker? May 28, 2018 Basic Formulas, Apparent power is defined as the product of current time voltage passing through an AC circuit. Hi Steven, Thanks for sharing your knowledge, it really helped me understanding the theory. I would say you cannot use the motor at 50Hz, 380V. 345/380 = 0.90 I Phase Similar method for the other phases. The power taken by a circuit (single or three phase) is measured in watts W (or kW). The simple formula to calculate the rating of three phase Transformers is: KVA = (√3. Steven, let me give you a few more details. To get started and understand our policy, you can read our How to Write an Electrical Note. This post about the 3 phase motor current calculation formula explanation. Line to line voltage: W applied = 3 1/2 U ll I cos Φ = 3 1/2 U ll I PF (1) where. Power & Amp - The amount of voltage running through the lines is related to the power and amp. You can calculated this be summing the A, B and C phases, taking into account the phase difference (easiest to do this using complex notation). I believe that as you lower the voltage across the lamp, the current would reduce and the brightness (think of the lamp as a resistance). This video is about 3 phase motor amps calculation Or 3 phase motor current calculation guide. Hi Steven, how to get the total KVA of the system if Line AB =642.24amp. Network analysis is not the intent of this note. The new posting/commenting system is way better than the current one and I'm now just waiting to use it. Basic Formula to Calculate Apparent Power in Single and Three Phase Circuits EE. With A-B, B-C and A-C you are looking a loads in a delta connected system. The key to doing this is that the sum of power in each phase is equal to the total power of the system. thank you very much. Each phase is a sine wave. ... Voltage drops are phase-to-phase, for three-phase, three-wire or three-phase, four-wire 60 Hz circuits. Enter the power factor of the load. Back to basic, below are the simple Electric Power formulas for Single Phase AC Circuit, Three Phase AC Circuits and DC Circuits. Phase A starts at 0 at 0 degree phase angle, rises to 1 at 90 degrees, back to 0 at 180, to -1 at 270 degrees, and back to 1 at 360 degrees. Similarly given the power in each phase you could easily find the phase currents. It is a type of polyphase system and is the most common method used by electrical grids worldwide to transfer power. I am calculating the current in a 3 phase 380 V balanced Y system with a single motor load of 746 watts with a power factor of 0.72 and am getting different results when using the approaches suggested above. Where I have issues is when we have loads connected across two phases. The relationship between kVA and kW is the power factor (pf):Single phase system - this is the easiest to deal with. Line and Phase currents are related to each other as: I_line=square_root(3)*I_phase Which means that whatever supply current we have, we need a wire cross-section for 1/ square_root(3) times line current only. The calculation of current in a three phase system has been brought up on our site feedback and is a discussion I seem to get involved in every now and again. The product of the voltage and current is the apparent power and measured in VA (or kVA) . IEC 61439 'Low-voltage switchgear and controlgear assemblies', specifies standard arrangements of switchboard (call forms of internal separation). The voltage measured between any line and neutral is called phase voltage. Most AC power today is produced and distributed as three-phase power where three sinusoidal voltages are generated out of phase with each other. CM = Circular-Mils (conductor wire gauge) Notes: • The National … VA is simply the current times the voltage, so knowing this and the voltage can give the current. For Example :- Each phase of 3 phase machine takes a load of 16 amps, So how do I calculate how many KVA generator is required ? If the 491.5A is an existing or additional loads to the 80.6A, you would need to add to get the total current (i.e. These applications will - due to browser restrictions - send data between your browser and our server. KW, KVA KW is real consumed power turned into heat, and is the product of volts x current x power factor. dilan, the kVA would be ( 642.24 + 651.85 + 651.84) * 220 = 428102 VA = 428.1 kVA the kW would be 428.1 x 0.86 = 368.2 kW For the breaker, I would likely be looking at 800 A (posibly something like a MasterPact NT or NW, depending on other parameters) and setting the overload protection in the range of 710 A (this is plus 10%). I line = 0.90 * 1.73 = 1.57 A For each phase the VA is current x voltage (i.e. Here is a 3 phase generator output voltage formula to calculate the three phase generator output voltage. Open 3-Phase Circuit Formulas: Open Delta Watts = 2 / 3 W DELTA Open Wye Watts = 1 / 2 W WYE Open 4-wire Wye Watts = 2 / 3 W WYE. Ignoring efficiency and assuming single phase 230 V, the VA would be 184 (230 V x 0.8 A) and power factor 110/184 = 0.6. Also make sure you use the larger voltage of the system voltages, ex) if P 1ph = P 3, Single phase apparent power S1ph (VA): ie 10mm cable carrying 50A 3 phase over 30m VD=3.8x50x30 / 1000 = 5.7V or VD=3.8x150x30 / 1000 = 17.1V I think it has to be the first but am a bit mixed up, need to brush up on some 3 phase theory I think. For 3 Phase 3 Wire system. Try to look at the following post. And now I would like to know how does the calculation works when a generator has to be used? For the cable size I'll refer you to our calculator as it depends very much on the type of cable and installation: http://myelectrical.com/tools/cable-sizing-calculator After looking at the cable size calculator, if you have any questions on using it (or questions on the breaker selection), you can ask in the questions section of the site: hi,i would like to ask if getting the current in any configuration (delta or star) is it always the power in kva to be used?.i always had some trouble in this.pls help. Your California Privacy Rights Sorry, your blog cannot share posts by email. We previously studied that voltage is the basic property of electrical circuits . If I have a system having Amperes on each phase as 80A, 70A, 82A and VLL is 400V and VLN is 210 V. Kindly share how to calculate its power consumption in KVA. Now the calculation you gave me gives me the total amps on A-B but it doesn't tell me how many amps I'm pulling on A and how many amps I'm pulling on B. I know when you are dealing with loads connected across phases figuring out the current isn't as straight forward as using ohms law. Kindly I want to know in 3 phase distribution ( example a house), Say I have 29 kVA ... equals to 36,250watts (PF=0.8). Again, assuming equal power ratings of the three single-phase AC sources, the total power available to a connected 3-phase AC load is the product of the 3-phase AC line voltage times the 3-phase AC line current multiplied by the √3. Where: V is the voltage (volts) and I is the current (amps). So (I) would be equal to 72.79A ... is this 72.79 divided by the 3 phase, i.e does each phase has a load of 24.26 A ? alternating-current-circuits Have a read of the note and if you are still unclear, you can add a comment to post. I know this probably is easy to understand, but I got myself to a completely confused state. And the value of the impedance and the reactance if required. You can easily find electric power in watts by using the following electric power formulas in electric circuits. To me the easiest way to solve three phase problems is to convert them to a single phase problem. I = Amps. Mathematical calculation for KVA and MVA for transformer. Let’s see two most common methods for calculation of voltage drop – approximate and exact methods: 1. The current is simply the kVA divided by the voltage. Voltage drop formulas. now simply follow the above single phase method. If you do a site search for "International Voltages" we have a Wiki page which lists some of these (may be slightly out of date). Power (3-phase) = kW, therefore per phase = kW/3 To make the situation confusing, the contractor got the confirmation from the manufacturere. Hope this makes some sense. The single- phase voltages are phase to neutral voltages. Three phase simply means you have three windings instead of one; giving you the A, B and C phases (and if star connected a neutral conductor). I= S 1ph V LN = P 3×pf 3 V LL, Simplifying (and with 3 = √3 x √3): You can target the Engineering ToolBox by using AdWords Managed Placements. Working with single phase, three-phase and DC (direct current circuits) and you quickly need to reference formulas for voltage drops and power calculations for a given conductor? Single-phase three-wire: Also known as Edison system, divided or neutral phase with central intake. You could do the calculation in terms of peak current if you wanted (for a sine wave the relationship between the peak and RMS is the square root of two). Enter the rated volts, amps, power, and efficiency to find the current flowing through the motor. The following formula calculates total power in a three-phase system based on KW and KVAr or voltage and current. Motor rated power is normally the output power, so to be fully accurate you need to consider the efficiency. unbalanced phase shift), then it is necessary to revert to more traditional network analysis. To avoid any misunderstanding, the following post may also help you: http://myelectrical.com/notes/entryid/172/three-phase-power-simplified. The line current is then the kVA divided by the voltage, i.e. The 3 phase load current amps formula is explain with a three phase load voltage induction motor nameplate data. At the fault location, because the phases are shorted, they will be at nearly the same voltage (close to zero depending on the impedance). Back to basic, below are the simple Electric Power formulas for Single Phase AC Circuit, Three Phase AC Circuits and DC Circuits. However, if you prefer you can write, manage... GE's latest thinking on product manufacturing is he Shingijutsu philosophy or Lean production system. Our software is the only cloud-based solution and has been built from the ground up to be fully responsive - meaning you can access your cables from anywhere and on any device, desktop, tablet or smartphone. 21,000 / 230 = 91 A. 2 thoughts on “ Cable Sizing & Voltage Drop Calculations Formula ” Brian July 19, 2019. Single phase to 3 phase power calculation input requirement = The square root of 3 (1.732) x 10 amps = 1.732 x 10 amps = 17.32 Amps. Voltage Drop formula for Single Phase and DC Circuits . I have 3 loads on each phase as L1-2=19.74 VA, L2-3=16.91 VA L1-3=13.81 VA Please tell the capacity of the transformer required. The same definition holds for both three phase and single phase systems. Voltages in Three Phase Circuits - General ..... 18 4B. 380/1.73 = 219 V Line-Neutral (Phase Voltage) How to use the formula and which formula to find it out..??? Easy enough. Balanced three phase system with total power P (W), power factor pf and line to line voltage VLL, Convert to single phase problem: The Three-Phase Voltage System Three-phase voltage systems are composed of three sinusoidal voltages of equal magnitude, equal frequency and separated by 120 degrees. | 5,700 | 23,858 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-31 | latest | en | 0.92927 |
https://www.jiskha.com/questions/1008157/im-a-odd-number-actually-both-my-numbers-are-odd-if-you-substract-my-ones-digit-from-my | 1,576,467,878,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541315293.87/warc/CC-MAIN-20191216013805-20191216041805-00508.warc.gz | 767,117,450 | 5,415 | Math
I'm a odd number, actually both my numbers are odd. If you substract my ones digit from my tens digit the diffrence is four. And I am less than sixty. What number am I?
1. 👍 0
2. 👎 0
3. 👁 83
1. x y
x - y = 4
well, five and one will do
51
1. 👍 0
2. 👎 0
posted by Damon
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asked by Stephanie on November 20, 2008
6. math
Consider the following sums of numbers and how they are formed: 1 odd number } 2 odd numbers 678 3 odd numbers 64748 Predict the following sum and complete the generalization: a. 1 + 3 + 5 + 7 + 9 + 11 = ? b. 1 + 3 + 5 + Á + (2n
asked by Ty on January 20, 2010
7. Math
Use the Distributive Property to prove each statement. Hint: you can write an even number as 2n and odd number as 2n + 1, where n represents any whole number. A. The sum of two even numbers is even. B. The sum of two odd numbers
asked by Bailey on October 17, 2017
8. algebra
I have been working this one for awhile and have become more confused. Please assist. Problem: Find two consecutive odd integers such that 5 times the first interger is 12 more than 3 times the second. The key thing is to realize
asked by ken on April 21, 2007
9. A fun one-math
i am a four -digit number with no two digits the same. my ones digit is twice my thousands digit and one less than my tens digit. my hundreds digit is the difference between my tens digit and my thousands digit. my thousands digit
asked by manny on August 20, 2007
10. SUMS?
what is the sum of the first towo odd numbers? what s the sum of the first three odd numbers? what is the sum of the first 4,5 and 25 odd numbers? what is the sum of any number (n) of odd numbers? explain your reasoning. IM SORRY
asked by Extra homework on November 13, 2008
More Similar Questions | 870 | 3,004 | {"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-2019-51 | latest | en | 0.943582 |
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# The Kabbalah of predictive journalism
This time, the wisdom of the Kabbalah is applied to journalism, math, and the Game of Priors
Illustrative: Reporters and photographers during a press conference at the Prime Minister's Office in Jerusalem in 2011. (photo credit: Mark Israel Sellem/Flash90)
Since this article is about predictive journalism, you may have guessed that Nate Silver would be mentioned. But while the statistical likelihood of mentioning Baye’s Theorem in an article that mentions Nate is high — as Nate is known for his creative use of the theorem — the outcome of this article is not as easy to predict (especially as I myself don’t know the outcome yet).
We corresponded our article about conscious journalism to the third intellectual sefirah of da’at (knowledge). But for this article we will be discussing the first intellectual sefirah of chochmah (wisdom) as the sages teach that the wise knows what will be born from their actions (ro’eh et hanoled).
In brief, Baye’s Theorem considers “Priors” (also called “signals”) – events that we have pre-existing knowledge of – and “Posteriors” – events that we want to predict. While the theorem has again been around for over 250 years, Nate adapted the theorem into what he calls the “Bayesian convergence,” or the method of dispelling myths and opposing opinions as evidence of the most likely outcome is uncovered.
This is the second reason why we corresponded Predictive Journalism to the sefirah of chochmah (wisdom). In addition to knowing what will be born from present-day actions–like the chess master who calculates the upcoming moves 1–this sefirah on the right side of the Tree of Life model is relates to the past, whereas binah (understanding) on the left side is relates to the future. 2
As in our previous articles, our subject of focus will be the concept, marketability, or point of attraction behind Baye’s Theorem. Specifically, how Nate approaches it different from others, so much so that he is now assembling a team to carry out his approach to journalism as part of the relaunch of his site, FiveThirtyEight.com.
But in order for his findings, and those under him, to be statistically accurate, even the most unique journalist pieces need to be based on what Baye’s Theorem calls “Priors.” But while the result is to dispel myths, these disruptive results need to be based on prior statistical findings.
First, this is the example that Wikipedia uses to explain Baye’s Theroem (you can read the equation part here:
Suppose a man told you he had had a nice conversation with someone on the train. Not knowing anything about this conversation, the probability that he was speaking to a woman is 50% (assuming the train had an equal number of men and women and the speaker was as likely to strike up a conversation with a man as with a woman). Now suppose he also told you that his conversational partner had long hair. It is now more likely he was speaking to a woman, since women are more likely to have long hair than men. Bayes’ theorem can be used to calculate the probability that the person was a woman.
A Game of Priors
The first finding, as a result of corresponding Nate’s work to the sefirah of chochmah, is that the primary task of a past-minded predictive journalist should be to continuously weigh and reanalyze Priors (pre-existing statistic) not Posteriors (event we want to predict).
For events that already occurred, as in the train example, instead of searching for the most probable Posterior, the approach of chochmah, the yiddishe kop, is to search for the one true Anterior (what actually happened).
As we will explain later, Nate still has come up with a good chap (catch) with how he approaches Baye’s Theorem for events that have not yet occurred. But for the meantime, since most examples used to explain Baye’s describe events that happened, these relate to an exercise in discovering the one true outcome, not the most probable.
This story aptly illustrates this point. Appropriately enough, another train story! (reprinted with permission from Jewlarious.com):
After months of negotiation with the authorities, a Talmudist from Odessa was finally granted permission to visit Moscow.
He boarded the train and found an empty seat. At the next stop, a young man got on and sat next to him. The scholar looked at the young man and he thought: This fellow doesn’t look like a peasant, so if he is no peasant he probably comes from this district. If he comes from this district, then he must be Jewish because this is, after all, a Jewish district.
But on the other hand, since he is a Jew, where could he be going? I’m the only Jew in our district who has permission to travel to Moscow.
Ahh, wait! Just outside Moscow there is a little village called Samvet, and Jews don’t need special permission to go to Samvet But why would he travel to Samvet? He is surely going to visit one of the Jewish families there. But how many Jewish families are there in Samvet? Aha, only two — the Bernsteins and the Steinbergs. But since the Bernsteins are a terrible family, so such a nice looking fellow like him, he must be visiting the Steinbergs.
But why is he going to the Steinbergs in Samvet? The Steinbergs have only daughters, two of them, so maybe he’s their son-in-law. But if he is, then which daughter did he marry? They say that Sarah Steinberg married a nice lawyer from Budapest, and Esther married a businessman from Zhitomer, so it must be Sarah’s husband. Which means that his name is Alexander Cohen, if I’m not mistaken.
But if he came from Budapest, with all the anti-Semitism they have there, he must have changed his name.
What’s the Hungarian equivalent of Cohen? It is Kovacs. But since they allowed him to change his name, he must have special status to change it. What could it be? Must be a doctorate from the University. Nothing less would do.
At this point, therefore, the Talmudic scholar turns to the young man and says, “Excuse me. Do you mind if I open the window, Dr. Kovacs?”
“Not at all,” answered the startled co-passenger. “But how is it that you know my name?”
“Ahhh,” replied the Talmudist, “It was obvious.”
Woman with Long Hair
Taking the initial Wikipedia example. Someone had a conversation on the train who we now know had long hair. Now let’s say this person is an observant Jewish male, so now there are two options. Either he is related to this long-haired woman, or he’s engaging some spiritually-minded male soul in a discussion about Judaism. But we know this man’s family is in another city and he is travelling for business needs, so the latter option is more likely, especially as he is Chabad.3 But then we can start wondering about why this long-haired man is travelling on the same train, a train going to a very unique destination. By the end of it, given a few key “Priors,” the astute will not only be able to figure out the gender of this person, but some personal details as well.
Andrew Mason’s New Company
Now just for fun, let’s put our version of predictive analysis to the test, but this time for an event that has not yet happened. The intent is not that we are trying to predict the outcome (something which can easily enter murky waters according to the Torah) but to deduce based on present facts, the most probable outcome.
A few days ago, former CEO of Groupon, Andrew Mason, was in the news again. This time it was because he sold more than half of his Groupon stock. As a predictive journalist we’d like to write about what Andrew plans to do next as a result of this sale. But in order to predict, we need to go back in time.
In a May 2013 blog post, Andrew begins with the following:
I feel very lucky to be alive at a time when someone like me can have a simple idea like Groupon that ends up impacting millions of people. If there’s a silver lining to leaving Groupon, it’s the opportunity to start something new. I’ve accumulated a backlog of ideas over the last several years, my favorite of which I’ll be turning into a new company this fall.
It is now nearing the spring of 2014, and aside from a motivational business song album entitled “Hardly Workin,” seemingly, we are still waiting for that business announcement.
As explained in “Groupon: Turning Daily Deals into Tipping Points,” Groupon was a commercialized version of Andrew’s prior non-profit venture, ThePoint.com.
Here is some of the CrunchBase.com bio for ThePoint.com:
The Point is a website for organizing group actions. Its goal is to help people congregate around the issues they care about and combine forces to make things happen.
But to travel conceptually back another step, in science the “tipping point” is called a phase transition (e.g., the degree at which water becomes ice). Since Andrew moved from Chicago to San Francisco (what many consider the extended version of Silicon Valley), and as at the time of that announcement was planning on volunteering time at the start-up engine Y-Combinator, the prediction (again from an analysis of the past) is that Andrew is planning a new version of ThePoint.com geared to assist start-ups and other innovators.
But whereas that site relied on the “wisdom of the crowd” to select and support new social initiatives, what would we expect to be new about Andrew’s upcoming company?
Recent tweets indication that Andrew is still interested in implementing music in his new venture. But whereas ThePoint.com gave full access for new initiatives to state their case before the public, seemingly Andrew is looking for some way to tip or change the world, through music. For instance:
Gamifying Productivity
Similar to music, games (also mentioned in this tweet) have also been shown to increase productivity in the workplace. One of the leading proponents for this way of thinking, called the “gamification” of reality, is Jane McGonical.5
Given the above, the probability is that the new project is something related to gamifying tipping points with an emphasis on start-ups and soothing music. For instance, to gamify ideas to see which are more likely to be successful and which not (like the chess master who calculates the upcoming turns in advance).
This platform would also have implications for the workplace, where new ideas are suggested in board meetings on a weekly basis. By gaming new ideas, with creative music and engaging graphics, this could help start-ups and creative minded individuals to work through the possible outcomes of their ideas, before hiring a staff, fundraising, etc…
From Probability to Outcome
Especially since Quantum Mechanics, the world has become less deterministic. What we just presented was the convergence between three concepts called “phase transitions (tipping points),” “music” and “games.” But while each of these three represents a line in our conceptual three-dimensional mapping (i.e., classical music on one end of the music line, neo-modern music on the other), the point of convergence is defined as the intersection of these three lines. While the particular outcome can only be determined if the event already occurred, what can be determined is the conceptual mapping of the topic at hand.
This distinction we are now making is what this article about Nate and Baye’s Theorem calls the ability to navigate “big data” by means of “cloud computing.” But instead of relying on computers to do the work for us, as mathematics also relates to the sefirah of the chochmah, we prefer to speak about the unbounded limits of human potential.
The first step is to begin to conceptually map out Big Data, as explained in “The Question Behind the Quora Brand Concept;” the inner hope and attraction behind quora was that they would become a platform for doing this. Then once we have our cloud of interconnected concepts to travel among, the great connectedness between all “lines” of interest will become readily visible.
Morality, Light and Truth
How does a person choose which way to navigate through the content cloud of Big Data? It depends on which line the traveler most wants to travel on at that moment. While in space the three dimensions are up-down, right-left and front-back, according to Kabbalah they correspond to the lines of morality (moral-immoral), light (light-dark) and truth (true-false).6
Given that Andrew began with the tipping point concept, then went to music, we placed the convergence point or axis as gaming, with the other two lines intersecting it. As truth is always the outcome of the prior two–morality and light–then according to this correspondence, if a game was chosen as the next project, then it was also perceived as the project that would promote the most truth in the world. These three lines also inter-include within one other; which means that are nine primary formations, with an infinite numbers of spatial coordinates in between to choose from.
Bayesian Convergence
From the above, we can begin to appreciate why Nate’s takes on the Baye’s Theorem is so marketable. By using it to determine the probability of future events, in a method Nate calls “Bayesian Convergence,” Nate is alluding to the hidden potential behind the theorem that has laid dormant for over 250 years. The concept behind the theorem is best expressed as a method to navigate through the three-dimensional cloud of future possibilities and probabilities. Specifically how groupings of concepts converge with one another, and how travelers decide which line to travel on and in which direction.
As the sum of the probabilities of all possible outcomes is always one, we would like to end with the blessing that all our journeys lead us to come closer to the One, God.
If you are interested in mathematics and statistics, I encourage you to explore this further. As with previous articles, the above was written as an introduction to begin rethinking the world of predictive journalism.
Inspired by Rabbi Yitzchak Ginsburgh’s recent Monday night class in Efrat. Please check back for a link to the transcript.
1. A game invented by King Solomon, the wisest of all men.
2. This can be understood as relating to the present right-left or conservatism-liberalism differentiation in politics. For an entire class on this subject, watch: http://malchuty.org/2010-02-23-15-41-08/1084–q.html
3. While Chabad men speak to women when doing outreach, instead of physically sitting next to the woman, he would lean over from another seat or talk while standing.
4. The returning nature of chochmah is also returning to a greater awareness of the Divine nothingness or truth to be found in every particle of creation (or every story to be told).
5. My friend Rabbi Asher Crispe has written both about the tipping property of music and the ability to change reality by first gamifying it. The music article is called: Rock and Roll Redemption. Gaming: A World Taken by Games: Jane McGonigal at TED and: Upgrading the (Video) Game of Life
6. This was explained by Rabbi Ginsburgh in a class in Crown Heights a handful of years ago. If someone is interested, I could try and locate the master for you.
Photo Credit: CC-SA (Baye’s Theorem spelt out in blue neon at the offices of Autonomy in Cambridge.” by mattbuck, Creative Commons).
Read Start-Up Israel to keep your finger on the pulse of Israeli high-tech and innovation! | 3,304 | 15,356 | {"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} | 2.5625 | 3 | CC-MAIN-2019-13 | latest | en | 0.94895 |
https://www.scribd.com/doc/52928895/Strength-of-Materials-Formula-Sheet | 1,568,531,377,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514570740.10/warc/CC-MAIN-20190915052433-20190915074433-00275.warc.gz | 1,007,391,983 | 52,341 | You are on page 1of 4
# Clarkson University – ES222, Strength of Materials
## Final Exam – Formula Sheet
P F F
Normal Stress: σ = Splice joint: τ ave = Single shear: τ ave =
A A A
F P
Double shear: τ ave = Bearing stress: σ b =
2A td
P P
σ= cos 2 θ , τ = sin θ cosθ
Ao Ao
Factor of Safety = F.S. =
δ
Normal strain: ε = Normal stress: σ = Eε Shear stress: τ = Gγ
L
PL PL
Elongation: δ = Rods in series: δ = ∑ i i
AE i Ai Ei
Thermal elongation: δ T = α ( ∆T ) L Thermal strain: ε T = α ( ∆T )
lateral strain
Poisson’s ratio: ν = −
axial strain
σ x νσ y νσ z
Generalized Hooke’s Law: εx = − −
E E E
νσ x σ y νσ z
εy = − + −
E E E
νσ x νσ y σz
εz = − − +
E E E
τ xy τ yz τ xz
γ xy = , γ yz = , γ xz =
G G G
Units: k = 103 M = 106 G = 109 Pa = N/m2 psi = lb/in2 ksi = 103 lb/in2
## Coordinates of the Centroid: x =
∑ xA i i i
y=
∑ yA i i i
∑A i i ∑A i i
Parallel Axis Theorem: I x ' = I x + Ad , where d is the distance from the x–axis to the x’–axis
2
y
1 3
Iz = bh
12 h
1 z
I y = hb3
12
b
Torsion:
ρφ cφ
γ= γ max =
L L
Tρ Tc
τ= τ max = τ =γG
J J
TL T
φ= solid rod: J = 12 π c 4
JG
hollow rod: J = 12 π ( co4 − ci4 )
Ti Li
Rods in Series: φ = ∑
i J i Gi
y
Pure Bending: x
My Mc M
σx = − σ max = = M M
I I S
y 1 M
εx = − ε y = ε z = −νε x σ = εE =
ρ ρ EI
y
dy
P M y M z
σx = − z + y
A Iz Iy P
P C
! ! ! ! ! ! z x
Mz = dy × P M y = dz × P
## Shear and Bending Moment Diagrams
dV xd
= − w → VD − VC = − ∫ wdx = − (area under load curve between C and D)
dx xc
dM xd
=V → M D − M C = ∫ Vdx = +(area under shear curve between C and D)
dx xc
## Shear Stress in Beams
VQ VQ
τ ave = q= = shear per unit length Q = Ay
It I
Stress Transformation
σ x +σ y σ −σ y
2
+ (τ xy )
2
Principal stresses: σ max,min = ± x
2 2
2τ xy
Principal planes: tan 2θ p =
σ x −σ y
σ x −σ y
Planes of maximum in-plane shear stress: tan 2θ s = −
2τ xy
σ −σ y
2
+ (τ xy ) = R
2
Maximum in-plane shear stress: τ max = x
2
σ +σ y
Corresponding normal stress: σ ' = σ ave = x
2
## Thin Walled Pressure Vessels
pr pr
Cylindrical: Hoop stress = σ 1 = Longitudinal stress = σ 2 =
t 2t
pr
Maximum shear stress (out of plane) = τ max = σ 2 =
2t
pr
Spherical: Principal stresses = σ 1 = σ 2 =
2t
σ2 pr
Maximum shear stress (out of plane) = τ max = =
2 4t
Deflections of Beams
1 M ( x) d 2 y dy M ( x)
= = 2 slope = θ ( x ) = =∫ dx + C1
ρ EI dx dx EI
deflection = y ( x ) = ∫ θ ( x ) dx + C2 = elastic curve
Columns
π 2 EI
Pcr =
L2e
For x > a,
replace x with
(L-x) and
interchange a
with b. | 1,090 | 2,496 | {"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} | 3.25 | 3 | CC-MAIN-2019-39 | latest | en | 0.64697 |
https://www.worksheetscorner.com/subtracting-mixed-numbers-with-unlike-denominators-3/ | 1,713,561,532,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817455.17/warc/CC-MAIN-20240419203449-20240419233449-00351.warc.gz | 962,711,482 | 16,685 | Date: May 14, 2023
# Subtracting Mixed Numbers with Unlike Denominators Worksheet Test
Instructions: Solve the following problems and write your answers in the spaces provided.
### Problem 1
Subtract the mixed numbers:
Mixed Number 1 Mixed Number 2 Difference
3 1/5 1 3/4
5 2/3 2 1/2
7 3/4 4 1/6
9 5/6 6 2/3
11 1/3 8 3/4
### Problem 2
Subtract the mixed numbers:
Mixed Number 1 Mixed Number 2 Difference
4 7/8 3 1/3
6 2/5 1 3/4
9 3/4 5 1/6
11 5/6 7 2/3
14 1/3 9 3/4
### Problem 3
Subtract the mixed numbers:
Mixed Number 1 Mixed Number 2 Difference
5 1/2 2 3/4
7 5/8 4 1/3
10 3/4 6 1/5
13 2/3 8 3/4
15 1/5 11 2/3
### Problem 4
Subtract the mixed numbers:
Mixed Number 1 Mixed Number 2 Difference
2 3/5 1 1/6
4 1/4 1 2/3
6 1/2 3 3/4
8 5/6 5 1/3
11 2/3 7 5/8
### Problem 5
Subtract the mixed numbers:
Mixed Number 1 Mixed Number 2 Difference
3 1/2 1 3/4
5 5/8 2 1/3
8 1/4 4 1/2
10 2/3 6 1/6
12 5/6 9 3/4
### Problem 6
Subtract the mixed numbers:
Mixed Number 1 Mixed Number 2 Difference
4 3/8 2 1/6
6 1/5 2 2/3
9 1/3 5 1/4
12 5/6 7 2/3
14 1/4 9 5/8
### Problem 7
Subtract the mixed numbers:
Mixed Number 1 Mixed Number 2 Difference
2 5/7 1 2/3
4 2/3 1 1/4
7 1/5 3 1/2
9 2/3 5 5/6
11 3/4 8 1/3
### Problem 8
Subtract the mixed numbers:
Mixed Number 1 Mixed Number 2 Difference
3 2/5 1 3/6
5 3/4 2 1/3
8 1/6 4 1/2
10 5/6 6 1/6
12 1/3 9 3/4
### Problem 9
Subtract the mixed numbers:
Mixed Number 1 Mixed Number 2 Difference
4 1/7 2 2/3
6 1/4 1 1/2
9 1/3 5 1/4
11 5/6 7 2/3
13 1/4 9 5/8
### Problem 10
Subtract the mixed numbers:
Mixed Number 1 Mixed Number 2 Difference
2 3/5 1 1/6
4 1/4 1 2/3
6 1/2 3 3/4
8 5/6 5 1/3
11 2/3 7 5/8 | 848 | 1,657 | {"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} | 3.375 | 3 | CC-MAIN-2024-18 | longest | en | 0.506515 |
https://lengthunits.com/convert/mm/m/675 | 1,726,465,082,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651676.3/warc/CC-MAIN-20240916044225-20240916074225-00542.warc.gz | 310,612,970 | 2,812 | # Convert 675 MM To M
Find 675 millimeter in meter :
675 MM = 0.675 M
### How to convert 675 millimeters to meter
Before you start to calculate 675 millimeters, you should know the conversion rate between millimeters and meter.
The rate is 0.001 and the math equation is 675 X 0.001 = 0.675 .
Answer :675 mm is equal 0.675 m.
### 675 MM Converstion Table
From To How To Calculate ?
675 mm 26.574803149606 in 675 × 25.4
675 mm 2.2145669291339 ft 675 × 304.8
675 mm 0.73818897637795 yd 675 × 914.4
675 mm 0.00041942659723862 mi 675 × 1609340
675 mm 675 mm 675 × 1
675 mm 67.5 cm 675 × 10
675 mm 6.75 dm 675 × 100
675 mm 0.675 m 675 × 1000
675 mm 0.0675 dam 675 × 10000
675 mm 0.00675 hm 675 × 100000
675 mm 0.000675 km 675 × 1000000
## Quick conversion chart of millimeters to meters
1 millimeter to meter = 0.001 meter 5 millimeter to meter = 0.005 meter 10 millimeter to meter = 0.01 meter 20 millimeter to meter = 0.02 meter 30 millimeter to meter = 0.03 meter 40 millimeter to meter = 0.04 meter 50 millimeter to meter = 0.05 meter 75 millimeter to meter = 0.075 meter 100 millimeter to meter = 0.1 meter
675 MM For Other Units :
Do you know how to read 675 MM in english or other languages ?
(-1)(+1)
674 676 | 434 | 1,223 | {"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} | 3.234375 | 3 | CC-MAIN-2024-38 | latest | en | 0.444758 |
https://universitypk.com/statistics-for-economists-ma-economics-sargodha-university-past-papers-2015/ | 1,696,478,020,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00253.warc.gz | 639,018,542 | 41,215 | Admissions Syllabus Past Papers Date Sheets Results Latest Jobs
# Statistics for Economists, MA Economics Sargodha University Past Papers 2015
### Sargodha University MA Economics Paper-IV Statistics for Economists Past Papers 2015
Here you can download Past Papers of Paper-IV Statistics for Economists, MA Economics Part One, 1st & 2nd Annual Examination, 2015 University of Sargodha.
Statistics for Economists UOS Past Papers 2015
M.A. Economics Part – I
Paper-IV(Statistics for Economics) 1st Annual Exam.2015
Time: 3 Hours Marks:100
Note: Objective part is compulsory. Attempt any four questions from subjective Part.
Objective Part
Q.1: Write short answers of the following in two lines each.
1. Statistics
2. Ordinal Scale
3. Weighted Index
4. Standard Error
5. Seasonal Variation
6. Non Probability
7. Confidence Coefficient
8. Multiple Correlation
9. Type II Error
10. Experimental Design
Subjective Part
Q.2:(a)
Age(years) 18-19 20-24 25-29 30-34 35-44 45-49 Operators 9 188 160 123 84 15
Represent the above data by a suitable diagram.
b. Determine the relative variation between the prices in rupees and life in hours of certain similar commodities from the following data.
Price in Rupees 8 13 18 23 30 Life in Hours 130 150 180 250 345
Q3: (a)
Year 1930 1931 1932 1933 1934 1935 1936 Declared value 1044 861 702 675 731 756 484 Value on the basis of 1930 price 1044 1069 939 946 991 1012 1077
Construct average price index and volume index number taking 1930 as base year from the above information.
b. Two cards are drawn at random from a full pack of 52 cards. Find the probability that they are both aces, if the first cards is (i) replaced (ii) not replaced)
Q.4: (a) Find coefficient of determination of the following
X 1 2 3 4 5 Y 2 5 3 8 7
b. Given that r12 = 0.492, r13 = 0.927, r23 = 0.758 find all partial and multiple correlation coefficient.
Q.5: (a) Find the probability distribution of the sum of dots when two fair dice are thrown.
b. Find mean and variance of the following distribution.
X 0 1 2 3 P(X) 1/8 3/8 3/8 1/8
Q.6: (a) The weights of 1000 students of a college are normally distributed with mean 68.5Kg and standard deviation 2.7Kg. If 200 random samples of 25 students each, are obtained from this population, find the expected mean and standard deviation of the sampling distribution of means if sampling distribution are done (i) with replacement (ii) without replacement.
b. In how many samples would you expect to find mean between 67.9Kg and 69.2Kg.
Q.7: (a) The records of a certain hospital showed the birth of 723 males and 617 females in a certain week. Do these figures conform the hypothesis, that sexes are born in equal proportions. Use α = 0.02.
(b) Do the following data present sufficient evidence to indicate a difference in the mean achievement for the four teaching techniques? Use α = 0.025.
Techniques 1 65 87 73 79 81 69 – 2 75 69 83 81 72 79 90 3 59 78 67 62 83 76 – 4 94 89 89 88 – – –
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https://quizizz.com/en-us/building-shapes-worksheets-grade-6 | 1,716,347,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00507.warc.gz | 415,329,047 | 27,893 | Decomposing and Composing Shapes
8 Q
6th
Geometry Vocabulary & Volume Quiz
10 Q
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8 Q
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19 Q
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## Explore printable Composing Shapes worksheets for 6th Grade
Composing Shapes worksheets for Grade 6 are an essential resource for teachers looking to enhance their students' understanding of Math and Geometry concepts. These worksheets provide a variety of engaging activities that help students develop their spatial reasoning skills, recognize patterns, and explore the properties of different shapes. By incorporating these worksheets into their lesson plans, teachers can ensure that their Grade 6 students are exposed to a wide range of geometric problems, allowing them to build a strong foundation in this critical area of mathematics. With the right mix of hands-on activities, visual aids, and challenging problems, Composing Shapes worksheets for Grade 6 are a valuable tool for any teacher looking to make the most of their Geometry lessons.
Quizizz is an innovative platform that offers a wide range of educational resources, including Composing Shapes worksheets for Grade 6, to help teachers create engaging and interactive learning experiences for their students. In addition to worksheets, Quizizz also provides teachers with access to a vast library of quizzes, games, and other activities that can be easily integrated into their existing lesson plans. By using Quizizz, teachers can not only ensure that their Grade 6 students are mastering Math and Geometry concepts but also track their progress and identify areas where additional support may be needed. With its user-friendly interface and extensive collection of resources, Quizizz is an invaluable tool for teachers looking to take their Grade 6 Geometry lessons to the next level. | 600 | 2,473 | {"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} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.938823 |
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# Percentage Difference
A Percentage Difference is a difference shown as a percentage.
## Two Meanings!
"Percentage Difference" can have two meanings, but if you are comparing an old value to a new value people commonly use this method:
## Comparing Old to New
Difference means to subtract one value from another. For example the difference from 5 to 3 is: 5-3 = 2. Percentage Difference means to show that difference as a percent of the old value ... so divide by the old value and make it a percentage: So the percentage difference from 5 to 3 is: 2/5 = 0.4 = 40%
This is also called "Percentage Change"
## How to Calculate
Here are two ways to calculate a percentage difference, use whichever method you prefer:
### Method 1
Step 1: Calculate the difference (subtract one value form the other) Step 2: Divide that Difference by the old value (you will get a decimal number) Step 3: Convert that to a percentage (by multiplying by 100 and adding a "%" sign) Note: if the new value is greater then the old value, it is a percentage increase, otherwise it is a decrease.
### Method 2
Step 1: Divide the New Value by the Old Value (you will get a decimal number) Step 2: Convert that to a percentage (by multiplying by 100 and adding a "%" sign) Step 3: Subtract 100% from that Note: if the result is positive it is a percentage increase, if negative, just remove the minus sign and cal it a decrease.
## Examples
Example: A pair of socks went from \$5 to \$6, what is the percentage difference?
• Step 1: \$5 to \$6 is a \$1 increase
• Step 2: Divide by the old value: \$1/\$5 = 0.2
• Step 3: Convert 0.2 to percentage: 0.2×100 = 20% rise.
• Step 1: Divide new value by old value: \$6/\$5 = 1.2
• Step 2: Convert to percentage: 1.2×100 = 120% (ie \$6 is 120% of \$5)
• Step 3: Subtract 100%: 120% - 100% = 20%, and that means a 20% rise.
Another Example: There were 160 smarties in the box yesterday, but now there are 116, what is the percentage difference?
Answer (Method 1): 160 to 116 is a decrease of 44. Compared to yesterday's value: 44/160 = 0.275 = 27.5% decrease. Answer (Method 2): Compare today's value with yesterday's value: 116/160 = 0.725 = 72.5%, so the new value is 72.5% of the old value. Subtract 100% and you get -27.5%, or a 27.5% decrease.
## The Formula
You could also just put the values into this formula:
|New Value - Old Value| × 100% |Old Value|
(The "|" symbols mean absolute value, so negatives become positive)
Example: There were 200 customers yesterday, and 240 today:
|240 - 200| × 100% = (40/200) × 100% = 20% |200|
A 20% increase.
## How to Reverse a Rise or Fall
Some people think that a percentage increase can be "reversed" by the same percentage decrease. But no!
For example, a 10% increase from 100 is an increase of 10, which equals 110 ... ... but a 10% reduction from 110 is a reduction of 11 (10% of 110 is 11), which equals 99 (not the 100 we started with) Because the percentage rise or fall is in relation to the old value: The 10% increase was applied to 100. But the 10% decrease was applied to 110.
To "reverse" a percentage rise or fall, use the right formula here:
To Reverse: Use this Percent: Example 10%
An "x" percent rise:
x/(1+x/100)
10/(1+10/100) = 10/(1.1) = 9.0909...
An "x" percent fall:
x/(1-x/100)
10/(1-10/100) = 10/(0.9) = 11.111...
Or use this handy-dandy calculator (just type in a value and click in the other box) Percent Rise: <=> Percent Fall: | 1,005 | 3,522 | {"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} | 4.8125 | 5 | CC-MAIN-2018-47 | longest | en | 0.879065 |
http://openstudy.com/updates/4fd8db43e4b091dda54063cd | 1,448,877,829,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398461390.54/warc/CC-MAIN-20151124205421-00043-ip-10-71-132-137.ec2.internal.warc.gz | 184,084,625 | 11,795 | ## Blahh23 3 years ago A spherical fish bowl is half-filled with water. The center of the bowl is C, and the length of segment AB is 24 inches, as shown below. Use for 22/7 pi.
1. Blahh23
no choices are given, it is fill in the blank!
2. Blahh23
TO find the volume of a sphere the formula is 4/3*PI(22/7)*r^3
3. Calcmathlete
What are you trying to find? And is there a diagram?
4. ujjwal
where is the segment AB? Draw the figure.
5. Blahh23
no diagram...
6. Blahh23
|dw:1339612133181:dw|
7. Calcmathlete
Is point C on segment AB?
8. ujjwal
what are you supposed to find out?
9. Blahh23
Point C is the CENTER...
10. Blahh23
were trying to find the VOLUME
11. Calcmathlete
Yes, but is point C ON segment AB? If it isn't we have to approach it completely differently.
12. Blahh23
IT IS ON THE SEGMENT...
13. Calcmathlete
Alright. Last question. And you're trying to find the volume of the water or the bowl?
14. Blahh23
What is volume, in cubic inches, of water inside the fish bowl?
15. Blahh23
thatts what it says.
16. Calcmathlete
Alright. The formula for the volume of a sphere is: $V = \frac43πr^{3}$ Now, the formula for the volume of half the bowl which is the water is half o fhte formula. $V = \frac46πr^{3}$ $V = \frac46π(12)^{3}$ Can you solve from here?
17. Blahh23
3620 (rounded to nearest whole.) | 419 | 1,341 | {"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} | 3.6875 | 4 | CC-MAIN-2015-48 | longest | en | 0.91103 |
https://jacerhea.wordpress.com/2012/01/03/project-euler-problem-9-in-f/?shared=email&msg=fail | 1,585,483,669,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494331.42/warc/CC-MAIN-20200329105248-20200329135248-00444.warc.gz | 543,775,183 | 21,687 | # Project Euler Problem 9 in F#
In the continuing series of solving Project Euler problems with F#, here is problem 9 with my solution:
A Pythagorean triplet is a set of three natural numbers, a b c, for which,
a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
1. Enumerate from 1 to 1000. Create tuples of all sets (m, n) where m > n.
2. Create Pythagorean triplets using Euclid’s formula from the tuples.
3. Find the set where a + b + c = 1000.
4. Find the product of this set.
```let problem9 =
let pythagorean_triplets(m:int, n:int) =
let a = m*m-n*n
let b = 2*m*n
let c = m*m+n*n
[a;b;c]
let tops = 1000
[for m in [1..tops] do
for n in [1..m-1] do yield (m, n)] |> Seq.map (fun t -> pythagorean_triplets((fst t, snd t)))
|> Seq.filter (fun x -> x |> Seq.sum = tops) |> Seq.head |> Seq.fold (fun acc elem -> acc * elem) 1
``` | 317 | 934 | {"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-2020-16 | latest | en | 0.687518 |
http://www.docstoc.com/docs/78665382/String-Searching---PowerPoint | 1,398,399,250,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00433-ip-10-147-4-33.ec2.internal.warc.gz | 554,864,006 | 19,238 | # String Searching - PowerPoint
Document Sample
String Searching
• String searching is a critical algorithm used
– by grep in unix/linux to search a directory for a given text
sequence
– in word processors
– in bioinformatics to locate matching gene sequences
• Because the source file or array might be large, we want
to have an efficient algorithm
– how efficient? is O(m*n) reasonable where m = length of
substring and n = length of string? No, consider a 100Kbyte
file to examine for a 20 character sequence, this would require
100,000 * 20 = 2 million comparisons!
– we would prefer O(m + n)
• We look at four algorithms where we will assume
– we have a master string of length n, and a substring of length m
– we want to find the index of the first occurrence of the
substring in master, or return -1 if the substring is not found
Simple Search Algorithm
• The first algorithm is a straight forward i
search
– index j moves across the substring and the
current segment of the master string starting master
at index i
– if(substring[j] = = master[i+j]) then j
increment j
– if we have a mismatch, reset j to 0 and i++ – sub
• that is, start the search over again at the next
character in master
search(sub, master) { // find sub in master or return -1
m = sub.length; n = master.length; i = 0;
while(i + m <= n) { // is there enough of master to search?
j = 0; // start at sub[0] and master[i]
while(master[i + j] = = sub[j] && j < m)
j++; // while we are still matching, continue
if(j >= m) return i; // did we reach the end of sub? If so,
else i++; // return i otherwise move on to next
} // character in master and try again
return -1;
}
Simple String Search Analysis
• The outer loop will iterate as many as n – m + 1 times depending
on when the substring is found (if found at all)
– we have “ – m” because we don‟t have to search the final characters since
we would run out of master string before we had a complete match
• The inner loop will iterate based strictly on how closely the
substring matches this portion of the master string
– consider matching “12345” against “020304050612345”
• we never look at more than a single character in master for the first 10 characters
– we only start working through the inner while loop of the algorithm once we
reach the actual placement of the substring (O(n))
– in the worst case though, we might have to look at m – 1 items before we
find a mismatch
– consider matching “AAAB” in “AAAAAAAAAAB”, at each position in the
master string, we match 3 „A‟ characters
• O(n*m)
• The algorithm‟s best case is O(n), worst case is O(m * (n – m + 1))
– but since we will probably have a large master string (such as a text file) and
a fairly short substring, m << n and so our worst cast comes out to O(m * n)
• The average case probably won‟t have too many close matches so
the complexity will be closer to O(n)
Rabin-Karp
• In our simple search algorithm, we start checking each segment of
master until we mismatch with the substring
– however, if there is evidence that the substring will not match the master
string, we never have to start that inner-loop search
– in order to accomplish this, we will compute the substring‟s “fingerprint”
and we will compute the fingerprint of each segment of the master string
– if our segment‟s fingerprint does not equal the substring‟s fingerprint, there
is no need to start searching the segment and we move on to the master
string‟s next segment
– this algorithm has a complexity that ranges from O(m*n) in the worst case
to O(m + n) in the average and best case
• What can we use for a fingerprint?
– we need to be able to determine a fingerprint quickly (O(1)) or else this
approach will not improve over the straightforward approach
• If few of the substrings have the same fingerprint, then instead of
O(m*n) we have O(m*n*k + m) where k is the percentage of
substrings that contain the fingerprint and the O(m) at the end is to
determine the substring‟s fingerprint
Using Parity for a Fingerprint
• Assume that we are searching a binary string for a binary substring
– even parity exists if the binary string has an even number of 1 bits
• for instance, 010111 has even parity while 000001, 001011, and 111101 all have
odd parity – note that 000000 has even parity
• Compute the parity of the substring (takes O(m)) and the parity of
each substring of length m in the master string (takes O(n)) and
now use the straight forward string matching algorithm but precede
the inner while loop with an if statement:
– if(fingerprint[i] = = substring_fingerprint) while …
• we only bother to start searching at index i in the master string if the fingerprint
of this portion of the master string matches the fingerprint of the substring
Example: Search for 010111 in 0010110101001010011
Substring fingerprint = 1 (even parity)
Fingerprints of master‟s substrings:
Character: 0 0 1 0 1 1 0 1 0 1 0 0 1 0 1 0 0 1 1
Fingerprint: 0 0 1 0 1 0 1 0 1 0 1 1 0 0 x x x x x
x indicates non-applicable starting points
underlined locations are those where we have to start searching
Using non-Parity Fingerprints
• Parity is either even or odd so we might wind up having as
many locations to search as to not search
– k = 50%, or our average case complexity is O(m * n * .5 + m) which is
still O(m*n) – no improvement
• We want a better fingerprint, one in which we can be more
certain as to whether to search or not
– we use the same idea, but rather than even or odd parity, we will do a
more unique form of computation:
m 1 sj is the jth character of the string
s j 2 m 1 j mod q
j 0
q is some prime number, for instance 7
m is the size of the master string
– If we assume binary numbers, then this is actually computing the
decimal equivalent of the binary number starting at position j, mod-ing it
by q, a prime number
Example
Si 0 0 1 0 1 1 0 1 0 1 0 0 1 0 1 0 0 1 1
S 11 22 45 26 53 42 20 41 18 37 10 20 41 19 x x x x x
f 4 1 3 5 4 0 6 6 4 2 3 6 6 5 x x x x x
• Here, the substring is 001011, since it‟s length is 6, we compute fingerprints in
master of length 6, and we use 7 to mod the summations by
• Examples:
– f[0] (fingerprint starting at location 0) is 001011 % 7 = 0*25 + 0*24 + 1*23 + 0*22 +
1*21 + 1*20 = 11 % 7 = 4
– f[5] = 101010 % 7 = 42 % 7 = 0
– f[14] isn‟t computing because we would run out of string before we could match the
substring
• Our substring has a fingerprint of 001011 % 7 = 11 % 7 = 4
– we only need to search at locations i = 0, 4 and 8
• Computing the fingerprints for the entire master string takes 6*n operations, but
we can reduce this to 1*n if we are clever
– how?
• Whether O(6*n) or O(1*n), it is still O(n)
Rabin-Karp Algorithm
• Assume that fingerprints are evenly distributed across possible substring
sequences, then the inner while loop will only be examined (n – m) / q times
where q is the selected mod value
– by selecting a prime number for q, we will further avoid patterns in the data (consider
if q was 10 or 16, we would find a lot more matches)
• The inner loop only iterates while this portion of the master string and the
substring continue to match
– for instance, we stop iterating if we are searching for “ABCDE” and we have
“ABEDE”
• Thus, we can expect in the average case a complexity of O(m + n)
compute fingerprint f (f[i] is fingerprint starting at index i of master)
compute fingerprint of substring
i = 0; m = substring.length; n = master.length;
while(i + m <= n) {
j = 0;
if (f[i] = = fingerprint)
while(master[i + j] = = substring[j] && j < m) j++;
if(j>=m) return i else i++;
}
return -1;
}
Knuth-Morris Pratt
• Consider string matching on the following example:
• Because of the structure of the substring, when we reach
a mismatch at index 3 of Master, we actually do not have
to restart at index 1, but instead at index 2 – why?
– By taking advantage of this idea, we can reduce the complexity
of the string search algorithm in the worst case to be O(n + m)
– The Knuth-Morris-Pratt algorithm does this by creating an
align array which determines where we need to start our string
searching from once we find a mismatch
– The align array is based on the structure, or repetitiveness, of
the substring
• We therefore solve the problem in two parts, compute
align (O(m)) and string search through master (O(n)).
Computing Align
• The idea of the align array is to tell us where to align
the substring to the master string after a mismatch
– Consider as an example: With a mismatch here, we don‟t want
to restart j at 0 and i at 8, we might be
able to do better than that
• The algorithm for computing align is given below
i=2, j=0;
align[0] = -1; align[1] = 0;
while(i<m) {
if(substring[i] = = substring[j]) {
align[i]=j+1; i++; j++; }
else if(j>0) j=align[j];
else { align[i]=0; i++;}
}
Example
• Compute the align array for “papaya”
• Initialize values
– align[0] = -1
– align[1] = 0
– j = 0, i = 2
• Compute for i=2 to 5
– align[2] = j + 1 = 1 (since character at i = character at j), i = 3, j
=1
– align[3] = j + 1 = 2 (since character at i = character at j), i = 4, j
=2
– align[4]: since character at i != character at j, reset j to align[j]
= align[2] = 1, and still character at i != character at j, reset j to
align[j] = align[1] = 0, so align[4] = 0, i = 5, j = 0
– align[5]: character at i != character at j and j is not > 0 so
align[5] = 0, i = 6
• Done
KMP Algorithm
Compute align array
m = 0, s = 0, done = false
while(s < length of substring &&!done) {
if(master[m] = = substring[s]) {
m++; s++; }
else if(s = = 0) m++;
else s = align[s]+1;
done = (length of substring – s > length of master – m);
}
if(s >= length of substring) return (m – length of substring);
else return -1;
• As shown before, the align algorithm is O(m)
• Here, the while loop iterates once per character in the master string until a
match is found, but may iterate more than once per character if the substring
needs to be realigned
– however, the substring will never be realigned to an earlier point in the master
string than the current location
• The complexity here is no greater than O(n – m + 1) and so we have an overall
KMP Example
Boyer-Moore
• Another string searching algorithm tries to take advantage of what
happens in a mismatch between a substring and master string
when the character in master does not exist at all in the substring
– specifically, the substring to master string match occurs right-to-left rather
than left-to-right
– consider:
• master: abcabdabeacd
• substring: abe
– we start by comparing “abe” to “abc” but rather than scanning these left-to-
right, we scan them right-to-left
– since “c” does not exist at all in the substring, we know that the substring
does not match the master string at positions 0, 1, or 2
– in KMP, we realign the substring to the master string at index 1, but here,
we know that “c” is not in any part of the substring, so we can realign the
substring to start at master string index 3
• Question: how do we know that “c” is not in the substring?
– we need an O(1) means of checking this or else the complexity of Boyer-
Moore will not improve over the simple string search
Creating a Jump Array
• In order to determine if a given substring contains a character, we
prepare a “jump” array
• We scan the entire alphabet of characters that are permissible for
a string in the given programming language
– for each character, we store where to realign our substring to based on
whether the character exists in the substring or not
– for all of the characters in the language that do not appear in the substring,
the jump value will be at m+1
• that is, move m locations in the master string since we no longer have to look
at the m characters that align our substring to the master string
– for any character that does exist in the substring, we can only jump to a
location to where the characters matched
char ch; int k;
for(ch=0; ch<alpha; ch++) // assume all characters are not in substring
jump[ch] = m;
for(k=1;k<=m;k++)
jump[sub.charAt(k)]=m – k; // for those in substring, use m - k
Boyer-Moore Search
• Our search algorithm works as follows:
– Start searching in master at location m-1 and substring at location m-1
– If the two characters are not a match and the character in master does not
exist in the substring, then use the jump array to move to the next starting
location in master, which will allow us to skip over m characters in master!
– If the two characters match, then start working toward the left in both
master and substring until either we have a complete match or a mismatch
• on a mismatch, again use the jump array to shift to the right in the master string
– If the two characters do not match but the character in master exists in the
substring, then we have to shift not to the right but to the left by some
amount as determined by the previous algorithm
• Because we might have to move our substring to the left, we might
ultimately have O(m*n) comparions
• More likely though, we use the jump array to move to the right,
and we might have as few as O(m + n / m) comparisons, which is
a good deal less than our KMP algorithm
– HOWEVER, the Boyer-Moore complexity is misleading, to create the Jump
array, we need to look at |S| characters (|S| is the size of the language‟s
alphabet, for instance 256 for Ascii, 65336 for Unicode!)
– So our complexity is really O(|S| + m + n /m)
Approximate String Matching
• As useful as string matching is, there are many reasons for
wanting to perform approximate string matching (such as in spell
checking and bioinformatics)
• How do we approximate how close a string matches another?
– two forms of mismatch
• if the string1.charAt(j) != string2.charAt(j)
• if string1.charAt(j) does not appear in string2 (or vice versa)
– the first case is known as a revise difference
– the second is a delete or insert difference (a character has been deleted or inserted)
• what about two characters that appear out of position?
– consider approximate vs. apporximate? -- is this 1 error or 2?
– we will compute a matrix d that will store the cost (mismatch) when
comparing two strings (d[i][0] and d[0][i] are initialized to 0 for all i)
• matchCost = d[i-1][j-1] if string1.charAt(i) = = string2.charAt(j)
• reviseCost = d[i-1][j-1] + 1 if string1.charAt(i) != string2.charAt(j)
• insertCost = d[i-1][j] + 1
• deleteCost = d[i][j-1] + 1
• d[i][j] = minimum(matchCost, reviseCost, insertCost, deleteCost)
• The value at d[n-1][m-1] is the cost when comparing a string of
size m to a string of size n
– the complexity of this algorithm is O(m*n)
Example
• We use a 2-D table to compute d[i][j]
– d[0][j] is initialized to j and d[i][0] is initialized to I
– we now fill in the rest of the table up through d[length1][length2] where length1 is
the length of the first string and length2 is the length of the second string
• The “cost” of matching string1 to string2 is d[length1][length2]
– the larger the value of the cost, the less likely that we have a match
– below, as an example, we compare “hello” to “hell”, “hallo” and “hall”
Since “hall” scores 2, it is less of a match than either “hell” or “hallo”, each of which
score 1 (hell is missing 1 letter, hallo is off by 1 letter)
visit http://www-apparitions.ucsd.edu/~rmckinle/string/ if you want to play with the
Algorithm through a Java applet
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https://documen.tv/moi-nguoi-giup-em-giai-nhanh-voi-a-1202834-17/ | 1,713,021,643,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816820.63/warc/CC-MAIN-20240413144933-20240413174933-00788.warc.gz | 189,476,255 | 14,923 | # Mọi người giúp em giải nhanh với ạ
Mọi người giúp em giải nhanh với ạ
### 0 thoughts on “Mọi người giúp em giải nhanh với ạ”
1. Đáp án:
$\begin{array}{l} a)Khi:\sin 2x = 0\\ \Rightarrow 2x = k\pi \\ \Rightarrow x = \dfrac{{k\pi }}{2}\\ Pt:2{\cos ^2}2x = – 4\\ \Rightarrow {\cos ^2}2x = – 4\left( {ktm} \right)\\ + Khi:\sin 2x \ne 0 \Rightarrow x \ne \dfrac{{k\pi }}{2}\\ Pt:2{\cos ^2}2x – 3\sqrt 3 .2\sin 2x.\cos 2x – 4{\sin ^2}2x = – 4\\ \Rightarrow \dfrac{{2{{\cos }^2}2x}}{{{{\sin }^2}2x}} – \dfrac{{3\sqrt 3 \cos 2x}}{{\sin 2x}} – 4 = \dfrac{{ – 4}}{{{{\sin }^2}2x}}\\ \Rightarrow 2{\cot ^2}2x – 3\sqrt 3 \cot 2x – 4 = – 4.\dfrac{1}{{{{\sin }^2}2x}}\\ \Rightarrow 2{\cot ^2}2x – 3\sqrt 3 \cot 2x – 4 + 4.\left( {{{\cot }^2}2x + 1} \right) = 0\\ \Rightarrow 6{\cot ^2}2x – 3\sqrt 3 \cot 2x = 0\\ \Rightarrow 3\cot 2x\left( {2\cot 2x – \sqrt 3 } \right) = 0\\ \Rightarrow \left[ \begin{array}{l} \cot 2x = 0\\ \cot 2x = \dfrac{{\sqrt 3 }}{2} \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} 2x = {90^0} + k{.180^0}\\ 2x = {49^0} + k{.180^0} \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = {45^0} + k{.90^0}\\ x = 24,{5^0} + k{.90^0} \end{array} \right.\left( {tmdk} \right)\\ b){\sin ^2}x + \sin 2x – 2{\cos ^2}x = \dfrac{1}{2}\\ + Khi:\cos x = 0 \Rightarrow x = \dfrac{\pi }{2} + k\pi \\ \Rightarrow {\sin ^2}2x = \dfrac{1}{2}\left( {ktm} \right)\\ + Khi:\cos x \ne 0 \Rightarrow x \ne \dfrac{\pi }{2} + k\pi \\ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{2\sin x.\cos x}}{{{{\cos }^2}x}} – 2 = \dfrac{1}{2}.\dfrac{1}{{{{\cos }^2}x}}\\ \Rightarrow {\tan ^2}x + 2\tan x – 2 = \dfrac{1}{2}\left( {{{\tan }^2}x + 1} \right)\\ \Rightarrow \dfrac{1}{2}{\tan ^2}x + 2\tan x – \dfrac{5}{2} = 0\\ \Rightarrow {\tan ^2}x + 4\tan x – 5 = 0\\ \Rightarrow \left( {\tan x – 1} \right)\left( {\tan x + 5} \right) = 0\\ \Rightarrow \left[ \begin{array}{l} \tan x = 1\\ \tan x = – 5 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\pi \left( {tm} \right)\\ x = \arctan \left( { – 5} \right) + k\pi \left( {tm} \right) \end{array} \right. \end{array}$ | 1,033 | 2,116 | {"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} | 4.375 | 4 | CC-MAIN-2024-18 | latest | en | 0.172931 |
https://juicystudio.com/services/readability.php?url=https://lkstyle5.blogspot.com/ | 1,675,340,533,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00074.warc.gz | 362,538,469 | 4,967 | ## Contents
Gunning Fog, Flesch Reading Ease, and Flesch-Kincaid are reading level algorithms that can be helpful in determining how readable your content is. Reading level algorithms only provide a rough guide, as they tend to reward short sentences made up of short words. Whilst they're rough guides, they can give a useful indication as to whether you've pitched your content at the right level for your intended audience.
## Interpreting the Results
This service analyses the readability of all rendered content. Unfortunately, this will include navigation items, and other short items of content that do not make up the part of the page that is intended to be the subject of the readability test. These items are likely to skew the results. The difference will be minimal in situations where the copy content is much larger than the navigation items, but documents with little content but lots of navigation items will return results that aren't correct.
Philip Chalmers of Benefit from IT provided the following typical Fog Index scores, to help ascertain the readability of documents.
Typical Fog Index Scores
Fog Index Resources
6 TV guides, The Bible, Mark Twain
8 - 10 Most popular novels
10 Time, Newsweek
11 Wall Street Journal
14 The Times, The Guardian
Over 20 Only government sites can get away with this, because you can't ignore them.
Over 30 The government is covering something up
The following table contains the readability results for https://lkstyle5.blogspot.com/.
Summary Value
Total sentences 372
Total words 2576
Average words per Sentence 6.92
Words with 1 Syllable 1350
Words with 2 Syllables 657
Words with 3 Syllables 271
Words with 4 or more Syllables 298
Percentage of word with three or more syllables 22.09%
Average Syllables per Word 1.81
Gunning Fog Index 11.61
## Gunning-Fog Index
The following is the algorithm to determine the Gunning-Fog index.
• Calculate the average number of words you use per sentence.
• Calculate the percentage of difficult words in the sample (words with three or more syllables).
• Add the totals together, and multiply the sum by 0.4.
• Algorithm: (average_words_sentence + number_words_three_syllables_plus) * 0.4
The result is your Gunning-Fog index, which is a rough measure of how many years of schooling it would take someone to understand the content. The lower the number, the more understandable the content will be to your visitors. Results over seventeen are reported as seventeen, where seventeen is considered post-graduate level.
The following is the algorithm to determine the Flesch Reading Ease.
• Calculate the average number of words you use per sentence.
• Calculate the average number of syllables per word.
• Multiply the average number of syllables per word multiplied by 84.6 and subtract it from the average number of words multiplied by 1.015.
• Subtract the result from 206.835.
• Algorithm: 206.835 - (1.015 * average_words_sentence) - (84.6 * average_syllables_word)
The result is an index number that rates the text on a 100-point scale. The higher the score, the easier it is to understand the document. Authors are encouraged to aim for a score of approximately 60 to 70.
The following is the algorithm to determine the Flesch-Kincaid grade level.
• Calculate the average number of words you use per sentence.
• Calculate the average number of syllables per word.
• Multiply the average number of words by 0.39 and add it to the average number of syllables per word multiplied by 11.8.
• Subtract 15.50 from the result.
• Algorithm: (0.39 * average_words_sentence) + (11.8 * average_syllables_word) - 15.9
The result is the Flesch-Kincaid grade level. Like the Gunning-Fog index, it is a rough measure of how many years of schooling it would take someone to understand the content. Negative results are reported as zero, and numbers over twelve are reported as twelve. | 883 | 3,885 | {"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-2023-06 | latest | en | 0.888334 |
https://www.khanacademy.org/math/algebra-home/alg-quadratics/alg-quadratics-square-root/v/solving-quadratic-equations-by-square-roots | 1,723,071,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00189.warc.gz | 673,735,005 | 147,273 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
# Solving quadratics by taking square roots: challenge
Sal solves challenging quadratic equations like (4x+1)²-8=0 by taking the square root of both sides. Created by Sal Khan and CK-12 Foundation.
## Want to join the conversation?
• How would you solve -8n(10n-1)=0 using the zero-product property? I'm not really getting exactly how to use this process. Could someone help me with this?
• The point of the zero-product property is this:
If two or more factors are multiplied together to make 0, then one of the factors must = 0.
Think about it, if you want to make 0 by multiplying, you have to have a 0 as a factor:
0 * 8 = 0
125 * 0 = 0
1/4 * 0 = 0
2 * 3 * 4 * 5 * 6 * 0 = 0
So, if you know that a product is 0, then one of the factors MUST be 0.
In this problem:
-8n(10n-1)=0
The product is 0.
The two factors are: -8n and 10n-1
At least one of the factors must be =0.
So, set each one equal to 0.
-8n = 0
-8n/-8 = 0/-8
`n = 0`
or
10n-1 = 0
10n -1 +1 = 0 + 1
10n = 1
`n = 1/10`
`n = 0 OR n = 1/10`
• How come quadratics have 2 answers that both work only? Why not only one or 3? Is that the structure of quadratic equations to get 2 answers only?
• Yes, you are right. The quadratic equation is structured so that you end up with two roots, or solutions. This is because in the quadratic formula (-b+-√b^2-4ac) / 2a, it includes a radical. When taking the square root of something, you can have a positive square root (the principle square root) or the negative square root. For example: √25 = 5, but the √25 = -5 as well because -5*-5 is 25 also. That's where we get the "plus or minus" in the quadratic formula. So in one solution, you add the radical (which is the solution with the principle square root) and in another, you subtract (hence the negative square root). Hope this helped! :)
• For (4x+1) squared why don't we do the negative square root and do only the positive when we do both for the other side?
• You do not have to put the plus or minus on the variable side because it is redundant. You already have the positives
4x+1 = 2sqrt(2)..... and 4x+1 = -2sqrt(2), right?
Now think about what would happen if you looked at the negatives, you have
-(4x+1) = 2sqrt(2) which, if you multiply both sides by -1 gives you 4x+1 = -2sqrt(2)
and the other
-(4x+1) = -2sqrt(2) again, multiply by -1 and you have
4x+1 = 2sqrt(2).
So, putting the plus and the minus on the left side also, gives you expressions equivalent to the expressions if you put it only on the right side. No point in making yourself extra work for no good reason, right?
• At , what is the step-by-step procedure to solving (x-8)(x-2)=0 when solving for x? How did Sal immediately know that the answer would be x=8 or x=2? The only way I see him working it out is if he divided the expression up into two parts like so; (x-8)=0 and (x-2)=0 and then solving for x. If so, why is this allowed? It doesn't make intuitive sense.
• Basically the expression (x+8)(x+2)=0 means that one of those two factor [you have to look at (x+8) as one factor and at (x+2) as another factor] must be equal to zero to satisfy the equation.
Either (x+8) must be zero or (x+2) must be zero, or both. Therefore if x+8=0 then x = -8 and this is one valid solution, similarly if x+2=0 then x= -2 and this is the second valid solution.
When you tackle problems like this remember that, for a product to be equal to zero there must be a factor which is 0. Hope it helped
• General Question. Why are we so concerned with the so called "solutions" of these functions? There can be many other values of x in f(x)=ax^2+bx+c which give a valid output. Why do we need to find which value/s of x give zero as output?
• Why all this trouble to find where functions are zero?
It tells us something about the behavior of the function, for example, once we know where the zeros are, we can look on either side of the zero values and find out if the function is positive or negative, which in turn gives clues to where maximum and minimum values may be lurking. Later, in calculus, you will learn to take the derivatives of functions and try to find where the derivatives are zero as well, and when we find them, it tells us even more about the behavior of the function such as how fast it changes from negative to positive or vice-versa, and where the maximum/minimum value is or if it even exists.
All this procedure is part of a mathematical process called analysis, which is really what math is all about. You see, we can model much of the world via functions and this type of analysis helps us understand the model, which in turns helps us understand the nature of what we are trying to model, which gives us a better understanding of nature itself.
Right now you are honing your skills at algebra, which will never leave your side; no matter how far up the math ladder you want to climb, algebra is there. With some of these "honing" skills, you may not be able to see the reasons, whys and wherefores at the moment, but that will come in time.
Keep studying!
• I still don't get why it is plus or minus the square root.
• It would be because the Parabola of a quadratic equation crosses the x- axis in both the negative values and the positive values. In other words the x-intercepts are both positive and negative. Note that sometimes the square root is so big that the plus or minus does not have an effect on the side. That is why some parabolas are on the right side and do not have a negative solution. Hope this helped!
• Why do you have to put a + and - when you square root something?
• Let's say we take the square root of 4. Both +2 x +2 and -2 x -2 = +4. Unless it is clear that a negative solution has no meaning, you need to show both solutions. (For example, if a floor is square and has an area of 64 square feet, a side could be either +8 feet of -8 feet. However, in this problem -8 feet doesn't have a meaning.)
• I don't understand what the positive square root and negative square root are (also the +/- signs before the square root symbol) .....so like when someone says you can have the positive square root of this or the negative...what does that mean??
• There are two numbers that you can square to get 25. 5^2 = 25, and (-5)^2 = 25. Thus, to solve the equation x^2 = 25, you have to account for both possibilities.
• does anyone have an idea how the formula -b+-root b*-4ac/2a is derived ?
• how would u solve this: (x+2)^2=10
• There are two possible answers here, namely:
`Answer using the positive + square root:x=-2+√10`
`Answer using the negative - square root:x=-2-√10`
Solution:
(x+2)²=10 → this is our original equation
√(x+2)²=√10 → let's take the square roots of both sides
x+2=±√10 → this is what we'll get, then I'll subtract 2 from both sides
`taking the positive square root.....`
x=√10-2 → this is what I got, and now I'll rearrange the order
x=-2+√10 → I rearranged the order and there we have it!
`or taking the negative square`` root.....`
x=-√10-2 → this is what I got, and now I'll rearrange the order
x=-2-√10 → I rearranged the order and there we have it again! | 1,977 | 7,295 | {"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} | 4.34375 | 4 | CC-MAIN-2024-33 | latest | en | 0.925509 |
https://studylib.net/doc/18789546/reading-part-values | 1,642,750,690,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302740.94/warc/CC-MAIN-20220121071203-20220121101203-00333.warc.gz | 500,412,748 | 11,707 | ```READING PART VALUES
This is a quick guide in order to be used as a reference when assembling
kits. We have used the same parts we send with our kits, so it is easier for
you to identify them and read their value.
-----
RESISTORS
Resistors are widely used in effect pedals. They provide a known resistance
to the flow of electric current. Since they are small, printing their value on
them would be difficult; instead, they are marked with colour band code.
Depending on the resistor type, there are more or less bands. Two main
types are used: metal film resistors and carbon film resistors.
1- CARBON FILM RESISTORS
This resistors are brown coloured, and have 4 colour bands: three of them
set the value, and the last one the tolerance (gold coloured). This is the
equivalence between colours and numbers:
To read the resistor value, place it so the gold coloured band is placed in the
right, and then read the bands from left to right:
- The first two colour bands give you the first two numbers of the
resistor value.
- The third is a multiplier, and gives the number of zeros to add to the
first two numbers to get the final value.
- The fourth is the tolerance.
Example:
-
1st Ring: yellow (4)
2nd Ring: violet (7)
3rd Ring: yellow (4)
4th Ring: gold (5%)
Resistor value: (4 7) · 10^4 = 470000 = 470kOhm, with 5% tolerance
2- METAL FILM RESISTORS
This ones are blue coloured, and have 1% tolerance. Determining their value
is done by the same process than with carbon film, but metal film resistors
have one more band:
- The first three colour bands give you the first three numbers of the
resistor value.
- The fourth is a multiplier, and gives the number of zeros to add to the
first three numbers to get the final value.
- The fifth is the tolerance.
Example:
- 1st Ring: brown (1)
- 2nd Ring: black (0)
- 3rd Ring: black (0)
- 4th Ring: red (2)
- 5th Ring (tolerance): brown (1%)
- Resistor value: (1 0 0) · 10^2 = 10000 = 10kOhm, with 1% tolerance
CAPACITORS
Capacitors are the most used elements in effect pedals among with resistors.
There are three main types that are used, but they share a common marking
system. Capacitors are measured in Farads (F), but as this is a very large unit,
the capacitors normally have values between some picofarads and some
milifarads. Unlike resistors, capacitors are marked with numbers related to
their value, but reading the value is not direct neither. By default, the
capacitors are measured in picoFarads, and the last code number sets the
number of zeros to add to get the real value. We will see this for every
capacitor case.
.
Unit conversion:
The capacitor value can be defined with different units, that relate between
each other as follows:
.
1- CERAMIC CAPACITORS
The ceramic capacitors are lentil shaped, and orange/brown coloured. Their
two possibilities:
Example:
- The capacitor has two numbers engraved: in that case, the two numbers
are directly the capacitor value in picoFarads.
Marking: 22 = 22 pF.
- The capacitor has three numbers engraved: in that case, the two first
numbers are the value, and the last one the number of zeros to add (the
value obtained is in picoFarads and may have to be converted to larger units
to be easily used).
Marking: 104 = 10 · (10^4) = 100000 pF = 100 nF = 0.1 μF
2- POLYESTER CAPACITORS
These ones are better than ceramic and electrolytic, but their value is limited
marking than ceramic capacitors with three numbers: the first two are the
value, and the third one the number of zeros to add (the value obtained is in
picoFarads, and usually has to be converted to larger units to be easily used).
Perhaps the capacitor has some other numbers or letters written on it; the
important ones are the three numbers that are consecutive.
Example:
Marking: 2A104J = 10 · 10^4 = 100000 pF = 100 nF = 0.1μF
3- ELECTROLYTIC CAPACITORS
While their characteristics are not as good as ceramic or polyester capacitors,
they are the only available solution when very large values are required. They
can go up to some farads, but for audio they are usually around some
microFarads. Their reading is direct, as they are larger and the value can be
printed on them.
Example:
Marking: 100 μF
``` | 1,111 | 4,212 | {"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 | 0.921273 |
https://www.onlinemath4all.com/angles-in-standard-position-worksheet.html | 1,702,322,479,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679516047.98/warc/CC-MAIN-20231211174901-20231211204901-00447.warc.gz | 986,350,974 | 7,289 | # ANGLES IN STANDARD POSITION WORKSHEET
Draw the following angles in standard position.
1) -340°
2) 150°
3) -185°
4) 300°
5) 220°
State the related angle for the following principal angles.
6)
7)
8)
Find the measures of the following principal angles.
9)
10)
State the quadrant in which the terminal sides of the following angles lie.
11) 300°
12) 40°
13) -222°
14) -30°
15) 103°
-340°
150°
-185°
300°
220°
The related angle for the principal angle shown above is
= 360° - 340°
= 20°
The related angle for the principal angle shown above is
= 360° - 355°
= 5°
The related angle for the principal angle shown above is
= 360° - 220°
= 140°
The measure of principal angle of the angle shown above is
= (-270°) + (-20°)
= -270° - 20°
= -290°
The measure of principal angle of the angle shown above is
= 90° + 30°
= 120°
300° lies between 270° and 360°
So, the terminal side of 300° lies in the fourth quadrant.
40° lies between 0° and 90°
So, the terminal side of 40° lies in the first quadrant.
Because the given angle is negative, we have to find the related angle for -222°.
The related angle for the angle -222° is
= 360° - 222°
= 138°
138° lies between 90° and 180°
So, 138° lies in the second quadrant.
Because the related angle 138° lies in the first quadrant, its principal angle -222° will also lie in the same first quadrant.
Therefore, the terminal side of -222° lies in the second quadrant.
Because the given angle is negative, we have to find the related angle for -30°.
The related angle for the angle -30° is
= 360° - 30°
= 330°
330° lies between 270° and 360°
So, 330° lies in the fourth quadrant.
Because the related angle 330° lies in the fourth quadrant, its principal angle -30° will also lie in the same fourth quadrant.
Therefore, the terminal side of -30° lies in the fourth quadrant.
103° lies between 90° and 180°.
So, the terminal side of 103° lies in the second quadrant.
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