glasskerfu
2021-03-07
Answered

Find the value of x or y so that the line passing through the given points has the given slope. (9, 3), (-6, 7y), $m=3$

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irwchh

Answered 2021-03-08
Author has **102** answers

Formula for the slope of a line, given two points on that line slope = $m=\frac{(y2-y1)}{(x2-x1)}$

Substituting the known values for this problem and simplifying$3=\frac{(7y-3)}{(-6-9)}$

$3=\frac{(7y-3)}{-15}$

Cross-multiplying and solving

$-45=7y-3$

$-42=7y$

$y=-6$

Substituting the known values for this problem and simplifying

Cross-multiplying and solving

asked 2020-11-05

To graph: The sketch of the solution set of system of nonlinear inequality

asked 2022-04-10

I am trying to interpolate a function defined over a three-dimensional real space:

$f:{R}^{3}\to R\phantom{\rule{0ex}{0ex}}(x,y,z)\to f(x,y,z)$

Let assume I have ${N}_{1}{N}_{2}{N}_{3}$ points in the space which form my grid for this interpolation, and the multivariate series

$F(x,y,z)=\sum _{a=0}^{{N}_{1}-1}\sum _{b=0}^{{N}_{2}-1}\sum _{c=0}^{{N}_{3}-1}{C}_{abc}{x}^{a}{y}^{b}{z}^{c}$

is the chosen interpolator. In order to find the coefficients I should form the following sets of equations:

$\sum _{a=0}^{{N}_{1}-1}\sum _{b=0}^{{N}_{2}-1}\sum _{c=0}^{{N}_{3}-1}{C}_{abc}{x}_{i}^{a}{y}_{i}^{b}{z}_{i}^{c}={f}_{i}=f({x}_{i},{y}_{i},{z}_{i})\phantom{\rule{thinmathspace}{0ex}}{\textstyle \text{, where,}}i=1,\dots ,{N}_{1}{N}_{2}{N}_{3}\phantom{\rule{thinmathspace}{0ex}},$

and then solve for the coefficients. However, for this I first need to write the above set of linear equations in the standard form

$A\overrightarrow{x}=\overrightarrow{b}$

wherein, A is the matrix of coefficients, $\overrightarrow{x}$ is the vector of unknowns, and $b=\{{f}_{i}{\}}_{1}^{{N}_{1}{N}_{2}{N}_{3}}$ is the known vector. For this to be done I would require to expand the multivariate power series in the form of a single variable series, that is,

$\sum _{a=0}^{{N}_{1}-1}\sum _{b=0}^{{N}_{2}-1}\sum _{c=0}^{{N}_{3}-1}{C}_{abc}{x}_{i}^{a}{y}_{i}^{b}{z}_{i}^{c}=\sum _{m=0}^{({N}_{1}-1)({N}_{2}-1)({N}_{3}-1)}{C}_{m}{\eta}_{i}^{m}$

wherein ${\eta}_{i}={\eta}_{i}({x}_{i},{y}_{i},{z}_{i})$ and probably $m=abc$. OF course the expansion needs to be nontrivial and useful.

Is it possible at all? Any suggestion to find the coefficients more practically?

Regards, owari

UPDATE.

Maybe it appears that the most natural way for solving this problem is comprised of the following steps:

1. first solve for the coefficients of

$\sum _{a=0}^{{N}_{1}-1}{E}_{a}(y,z){x}_{i}^{a}=f({x}_{i},y,z)\phantom{\rule{thinmathspace}{0ex}}{\textstyle \text{, where,}}i=1,\dots ,{N}_{1}\phantom{\rule{thinmathspace}{0ex}},$

2. then solve for the coefficients of

$\sum _{b=0}^{{N}_{2}-1}{D}_{ab}(z){y}_{i}^{b}={E}_{a}({y}_{i},z)\phantom{\rule{thinmathspace}{0ex}}{\textstyle \text{, where,}}i=1,\dots ,{N}_{2}\phantom{\rule{thinmathspace}{0ex}}{\textstyle \text{, and,}}a=0,\dots ,{N}_{1}-1$

3. and finally solve for the coefficients of

$\sum _{c=0}^{{N}_{2}-1}{C}_{abc}{z}_{i}^{c}={D}_{ab}({z}_{i})\phantom{\rule{thinmathspace}{0ex}}{\textstyle \text{, where,}}i=1,\dots ,{N}_{3}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}a=0,\dots ,{N}_{1}-1{\textstyle \text{, and,}}b=0,\dots ,{N}_{2}-1$

which gives the coefficients we were looking for. However, this way, the first two steps will be working with function-type coefficients instead of numerical coefficients and that will prevent efficient usage of the available codes in numerical analysis. Solving for each set of coefficients at each step for different grid points would also inevitably increase the number of equations drastically, so any better suggestion to solve for this problem?

$f:{R}^{3}\to R\phantom{\rule{0ex}{0ex}}(x,y,z)\to f(x,y,z)$

Let assume I have ${N}_{1}{N}_{2}{N}_{3}$ points in the space which form my grid for this interpolation, and the multivariate series

$F(x,y,z)=\sum _{a=0}^{{N}_{1}-1}\sum _{b=0}^{{N}_{2}-1}\sum _{c=0}^{{N}_{3}-1}{C}_{abc}{x}^{a}{y}^{b}{z}^{c}$

is the chosen interpolator. In order to find the coefficients I should form the following sets of equations:

$\sum _{a=0}^{{N}_{1}-1}\sum _{b=0}^{{N}_{2}-1}\sum _{c=0}^{{N}_{3}-1}{C}_{abc}{x}_{i}^{a}{y}_{i}^{b}{z}_{i}^{c}={f}_{i}=f({x}_{i},{y}_{i},{z}_{i})\phantom{\rule{thinmathspace}{0ex}}{\textstyle \text{, where,}}i=1,\dots ,{N}_{1}{N}_{2}{N}_{3}\phantom{\rule{thinmathspace}{0ex}},$

and then solve for the coefficients. However, for this I first need to write the above set of linear equations in the standard form

$A\overrightarrow{x}=\overrightarrow{b}$

wherein, A is the matrix of coefficients, $\overrightarrow{x}$ is the vector of unknowns, and $b=\{{f}_{i}{\}}_{1}^{{N}_{1}{N}_{2}{N}_{3}}$ is the known vector. For this to be done I would require to expand the multivariate power series in the form of a single variable series, that is,

$\sum _{a=0}^{{N}_{1}-1}\sum _{b=0}^{{N}_{2}-1}\sum _{c=0}^{{N}_{3}-1}{C}_{abc}{x}_{i}^{a}{y}_{i}^{b}{z}_{i}^{c}=\sum _{m=0}^{({N}_{1}-1)({N}_{2}-1)({N}_{3}-1)}{C}_{m}{\eta}_{i}^{m}$

wherein ${\eta}_{i}={\eta}_{i}({x}_{i},{y}_{i},{z}_{i})$ and probably $m=abc$. OF course the expansion needs to be nontrivial and useful.

Is it possible at all? Any suggestion to find the coefficients more practically?

Regards, owari

UPDATE.

Maybe it appears that the most natural way for solving this problem is comprised of the following steps:

1. first solve for the coefficients of

$\sum _{a=0}^{{N}_{1}-1}{E}_{a}(y,z){x}_{i}^{a}=f({x}_{i},y,z)\phantom{\rule{thinmathspace}{0ex}}{\textstyle \text{, where,}}i=1,\dots ,{N}_{1}\phantom{\rule{thinmathspace}{0ex}},$

2. then solve for the coefficients of

$\sum _{b=0}^{{N}_{2}-1}{D}_{ab}(z){y}_{i}^{b}={E}_{a}({y}_{i},z)\phantom{\rule{thinmathspace}{0ex}}{\textstyle \text{, where,}}i=1,\dots ,{N}_{2}\phantom{\rule{thinmathspace}{0ex}}{\textstyle \text{, and,}}a=0,\dots ,{N}_{1}-1$

3. and finally solve for the coefficients of

$\sum _{c=0}^{{N}_{2}-1}{C}_{abc}{z}_{i}^{c}={D}_{ab}({z}_{i})\phantom{\rule{thinmathspace}{0ex}}{\textstyle \text{, where,}}i=1,\dots ,{N}_{3}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}a=0,\dots ,{N}_{1}-1{\textstyle \text{, and,}}b=0,\dots ,{N}_{2}-1$

which gives the coefficients we were looking for. However, this way, the first two steps will be working with function-type coefficients instead of numerical coefficients and that will prevent efficient usage of the available codes in numerical analysis. Solving for each set of coefficients at each step for different grid points would also inevitably increase the number of equations drastically, so any better suggestion to solve for this problem?

asked 2021-01-15

Let T denote the group of all nonsingular upper triaungular entries, i.e., the matrices of the form, [a,0,b,c] where $a,b,c\in H$

$H=\{[1,0,x,1]\in T\}$ is a normal subgroup of T.

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Determine whether the following statement is true or false. A system of linear equations Ax=b has the same solutions as the system of linear equations Rx=c, where [R c] is the reduced row echelon form of [A b].

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Determine whether the statement is true or false. If the last row of the reduced row echelon form of the augmented matrix of a system of linear equations has only
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Write the vector form of the general solution of the given system of linear equations.

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The reduced row echelon form of the augmented matrix of a system of linear equations is given. Tell whether the system has one solution, no solution, or infinitely many solutions. Write the solutions or, if there is no solution, say the system is inconsistent.