Find the equation written in point slope form if m=25 and f(0)=250

Hope Hancock
2022-09-01
Answered

Find the equation written in point slope form if m=25 and f(0)=250

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oldgaffer1b

Answered 2022-09-02
Author has **9** answers

Point-Slope Form

$y-{y}_{1}=m(x-{x}_{1})$

Since

$\{\begin{array}{l}\text{Point:}\phantom{\rule{1ex}{0ex}}({x}_{1},{y}_{1})=(0,250)\\ \text{Slope:}\phantom{\rule{1ex}{0ex}}m=25\end{array}$

we have

y−250=25(x−0)

$y-{y}_{1}=m(x-{x}_{1})$

Since

$\{\begin{array}{l}\text{Point:}\phantom{\rule{1ex}{0ex}}({x}_{1},{y}_{1})=(0,250)\\ \text{Slope:}\phantom{\rule{1ex}{0ex}}m=25\end{array}$

we have

y−250=25(x−0)

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I am given the following matrix A and I need to find a nullspace of this matrix.

$A=\left(\begin{array}{cccc}2& 1& 4& -1\\ 1& 1& 1& 1\\ 1& 0& 3& -2\\ -3& -2& -5& 0\end{array}\right)$

I have found a row reduced form of this matrix, which is:

${A}^{\prime}=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$

And then I used the formula ${A}^{\prime}x=0$, which gave me:

${A}^{\prime}x=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)$

Hence I obtained the following system of linear equations:

$\{\begin{array}{l}{x}_{1}+3{x}_{3}-2{x}_{4}=0\\ {x}_{2}-2{x}_{3}+3{x}_{4}=0\end{array}$

So I just said that ${x}_{3}=\alpha $, ${x}_{4}=\beta $ and the nullspace is:

$nullspace(A)=\{2\beta -3\alpha ,2\alpha -3\beta ,\alpha ,\beta )\text{}|\text{}\alpha ,\beta \in \mathbb{R}\}$

Is my thinking correct?

$A=\left(\begin{array}{cccc}2& 1& 4& -1\\ 1& 1& 1& 1\\ 1& 0& 3& -2\\ -3& -2& -5& 0\end{array}\right)$

I have found a row reduced form of this matrix, which is:

${A}^{\prime}=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$

And then I used the formula ${A}^{\prime}x=0$, which gave me:

${A}^{\prime}x=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)$

Hence I obtained the following system of linear equations:

$\{\begin{array}{l}{x}_{1}+3{x}_{3}-2{x}_{4}=0\\ {x}_{2}-2{x}_{3}+3{x}_{4}=0\end{array}$

So I just said that ${x}_{3}=\alpha $, ${x}_{4}=\beta $ and the nullspace is:

$nullspace(A)=\{2\beta -3\alpha ,2\alpha -3\beta ,\alpha ,\beta )\text{}|\text{}\alpha ,\beta \in \mathbb{R}\}$

Is my thinking correct?

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Divided it by$dx$ :

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Transposed the$(y+1)$

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The rest of the problems were easily solved but those 2 are very hard to bring to the standard form.

Divided it by

Transposed the

Divided it by

The farthest I can go is:

The rest of the problems were easily solved but those 2 are very hard to bring to the standard form.

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