trkalo84
2022-09-26
Answered

Find the slope of a line perpendicular to the line whose equation is 5x+3y=8

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ahem37

Answered 2022-09-27
Author has **15** answers

If a line has slope =m

then the slope of a line perpendicular to it is $(-\frac{1}{m})$

Rewrite $5x+3y=8$ in a slope-offset format

$y=-\frac{5}{3}x+\frac{8}{3}$

So the given equation has a slope of $(-\frac{5}{3})$

and

a line perpendicular to it has a slope of $\left(\frac{3}{5}\right)$

then the slope of a line perpendicular to it is $(-\frac{1}{m})$

Rewrite $5x+3y=8$ in a slope-offset format

$y=-\frac{5}{3}x+\frac{8}{3}$

So the given equation has a slope of $(-\frac{5}{3})$

and

a line perpendicular to it has a slope of $\left(\frac{3}{5}\right)$

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The slope of a linear equation of the form

I've read that this is because the coefficient of x (coefficient of y) represent the "speed" at which x (y) increases and thus we can get the slope by dividing the "speed of x" by the "speed of y" - but I'm afraid it's not very intuitive to me.

Why do we use this formula and how is this equivalent to the more general slope formula,

asked 2022-05-15

The following question seems to me interesting. it gives solution space and required the corresponding system of equation. The question is the following:

Consider the vectors in ${R}^{4}$ defined by

${a}_{1}=(-1,0,1,2)$, ${a}_{2}=(3,4,-2,5)$, ${a}_{3}=(1,4,0,9)$

Find a system of homogeneous linear equations for which the space of solutions is exactly subspace of ${R}^{4}$ spanned by the three given vectors.

First i check the linear independence of the given vectors to see form of the generated space. But after determining i only obtained the result that the rank of the coefficient matrix of the corresponding homogeneous system of equations is 2. i obtained this result by rank-nullity theorem. But i can't go further. Please help.

Thanks in advance...

Consider the vectors in ${R}^{4}$ defined by

${a}_{1}=(-1,0,1,2)$, ${a}_{2}=(3,4,-2,5)$, ${a}_{3}=(1,4,0,9)$

Find a system of homogeneous linear equations for which the space of solutions is exactly subspace of ${R}^{4}$ spanned by the three given vectors.

First i check the linear independence of the given vectors to see form of the generated space. But after determining i only obtained the result that the rank of the coefficient matrix of the corresponding homogeneous system of equations is 2. i obtained this result by rank-nullity theorem. But i can't go further. Please help.

Thanks in advance...

asked 2022-06-21

Correct me if I am wrong. Find the value(s) of the constant k such that the system of linear equations

$\{\begin{array}{l}x+2y=1\\ {k}^{2}x-2ky=k+2\end{array}$

has:

1. No solution

2. An infinite number of solutions

3. Exactly one solution

Answer:

so the first step is to get row reduction form, which is:

from $\left[\begin{array}{cc}1& 2\\ {k}^{2}& -2k\end{array}\right]$,

to $\left[\begin{array}{cc}1& 2\\ 0& -2k+2{k}^{2}\end{array}\right]$

$\{\begin{array}{l}x+2y=1\\ {k}^{2}x-2ky=k+2\end{array}$

has:

1. No solution

2. An infinite number of solutions

3. Exactly one solution

Answer:

so the first step is to get row reduction form, which is:

from $\left[\begin{array}{cc}1& 2\\ {k}^{2}& -2k\end{array}\right]$,

to $\left[\begin{array}{cc}1& 2\\ 0& -2k+2{k}^{2}\end{array}\right]$

asked 2022-09-14

What does it mean to interpolate a data?