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3 years ago

Find the slope of the line passing through the points (-5,5) and (-5,-8)

Mathematics

2 answers:

AysviL [449]3 years ago

6 0

Answer:

Step-by-step explanation:

5-8 -3

----------=

-5--8 -3

-3

-----= 1

-3

Hope this helps!

VARVARA [1.3K]3 years ago

4 0

-8 - 5 = -13
_____ ___

-5 + 5 = 0

So there is no slope because it’s a vertical line. It stays on the x-axis since there was no change in x-values.

The equation would simply be x=-5, and there is no slope.

The equation for slope formula is
(y2-y1)/(x2-x1)

I hope this helped its a little confusing yes

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Shape: Square

Length: Rational

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2 years ago

What eigen value for this matix <br> (1 -2)<br> (-2 0)

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You find the eigenvalues of a matrix A by following these steps:

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So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$

The determinant of this matrix is

$\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4$

Finally, we have

$\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}$

So, the two eigenvalues are

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