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Liula [17]
3 years ago
14

I need help with 9 and 10

Mathematics
1 answer:
adell [148]3 years ago
7 0

#9) 180 - 116 = 64

(3x -13) + ( 2x +4 ) + 64 = 180

5x - 9 + 64 = 180

5x = 180 + 9 - 64

5x = 125

x = 25

#10) 3x - 13

= (3× 25) - 13 = 62

m< A = 62°

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Answer:

X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}

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The question we have at hand is, in other words,

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix} - where we have to solve for the value of X

If we have to isolate X here, then we would have to take the inverse of the following matrix ...

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Therefore, we can conclude that the equation as to solve for " X " will be the following,

X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} - First find the 2 x 2 matrix inverse of the first portion,

\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1} = \frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}= \frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix} = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} ,

X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix} - Simplifying this, we should get ...

\begin{bmatrix}1&3\\ 2&4\end{bmatrix} ... which is our solution.

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