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Tema [17]
4 years ago
15

Help!!!!!!!!! i need this asap!

Mathematics
1 answer:
Nutka1998 [239]4 years ago
4 0

A=\left[\begin{array}{ccc}3&0\\2&-1\end{array}\right]\\\\B=\left[\begin{array}{ccc}2&8\\0.6&3\end{array}\right]\\\\A\cdot B=\left[\begin{array}{ccc}3&0\\2&-1\end{array}\right]\cdot\left[\begin{array}{ccc}2&8\\0.6&3\end{array}\right]=\left[\begin{array}{ccc}(3)(2)+(0)(0.6)&(3)(8)+(0)(3)\\(2)(2)+(-1)(0.6)&(2)(8)+(-1)(3)\end{array}\right]\\\\=\left[\begin{array}{ccc}6&24\\3.4&13\end{array}\right] \to\boxed{A.}

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Answer:

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Answer:

Second choice:

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Fifth choice:

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Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

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t=\frac{x}{2}.

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Choice 3:

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y=(x+3)^2+2(x+3)

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y=(x^2)+(6x+2x)+(9+6)

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Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

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