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Gekata [30.6K]
3 years ago
15

Carol bought two chairs with triangular backs. For what value of x can you use

Mathematics
1 answer:
Step2247 [10]3 years ago
5 0

Answer:

x = 4

Step-by-step explanation:

Two sides, GH and GI, in ∆GHI are of the same side length as corresponding two sides, TU and TV, in ∆TUV.

To prove that both triangles are congruent using the SSS Congruence Theorem, therefore, the their third sides, HI and UV must be equal.

Therefore,

9x - 21 = 15

Add 21 to both sides

9x - 21 + 21 = 15 + 21

9x = 36

Divide both sides by 9

x = 4

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Please solve the problem ​
jek_recluse [69]

Treat the matrices on the right side of each equation like you would a constant.

Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.

Then you can eliminate <em>Y</em> by taking the sum

4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>

==>   <em>X</em> = (4<em>A</em> + <em>B</em>)/11

Similarly, you can eliminate <em>X</em> by using

-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>

==>   <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11

It follows that

X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}

Similarly, you would find

Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}

You can solve the second system in the same fashion. You would end up with

P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}

3 0
3 years ago
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