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2 years ago

What is the area of a triangle with vertices at (−4, 1) , (−7, 5) , and (0, 1) ?

1 answer:

3 0

Plot the points on a graph. Connect the dots into a triangle. See that the height of the triangle is from y=5 down to y=1. So the height is 4 units.
Area of a triangle: A = (1/2)BH

You need to find the length of the base. Which is from point (-4,1) to (0,1). You can use the distance formula r just see from the graph that the base is 4 units.

A = (1/2)(4)(4)
A = 8

** Distance formula fyi
d² = (X-x)² + (Y-y)²
with points (X,Y) and (x,y)

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2 years ago

Find the? inverse, if it? exists, for the given matrix. [4 3] [3 6]

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Therefore, the inverse of given matrix is

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Step-by-step explanation:

The inverse of a square matrix $A$ is $A^{-1}$ such that

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$\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}$

$=\frac{1}{\det \begin{pmatrix}4&3\\ 3&6\end{pmatrix}}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}$

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$=\frac{1}{15}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}$

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

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2 years ago

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2 years ago

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alisha [4.7K]

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2 years ago

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2 years ago

What is the area of a triangle with vertices at (−4, 1) , ​ (−7, 5) ​ , and ​ (0, 1) ​ ?

What is the area of a triangle with vertices at (−4, 1) , (−7, 5) , and (0, 1) ?