Describe the transformation that moves ABC to its image A'B'C'.

1 answer:

kap26 [50]3 years ago

8 0

All it did was it counter clock wise rotated to make the shape look like it’s a mirror

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V125BC [204]

Answer:

$3A+9B=\begin{bmatrix} 27 & 63 & -60 \\\\ 102 & -57 & -60\\\\ 63 & -18 & 9 \end{bmatrix}$

Step-by-step explanation

$A=\begin{bmatrix} 6 & 6 & 4\\\\7 & -4 & 4\\\\ -6 & 9 & -3\end{bmatrix}\: and \: B=\begin{bmatrix}1 & 5 & -8\\\\ 9 & -5 & -8\\\\ 9 & -5 & 2\end{bmatrix}$

$\rightarrow 3A=3\begin{bmatrix} 6 & 6 & 4\\\\7 & -4 & 4\\\\ -6 & 9 & -3\end{bmatrix},\: \: 9B =9\begin{bmatrix}1 & 5 & -8\\\\ 9 & -5 & -8\\\\ 9 & -5 & 2\end{bmatrix}$

$\rightarrow 3A=\begin{bmatrix} 3*6 & 3*6 & 3*4\\\\3*7 & 3(-4) & 3*4\\\\ 3(-6) & 3*9 & 3(-3)\end{bmatrix},\:\: 9B =\begin{bmatrix}9*1 &9* 5 &9( -8)\\\\ 9*9 & 9(-5) & 9(-8)\\\\ 9*9 & 9(-5) & 9*2\end{bmatrix}$

$\rightarrow 3A=\begin{bmatrix} 18 & 18 & 12 \\\\ 21 & -12 & 12\\\\ -18 & 27 & -9\end{bmatrix},\:\: 9B =\begin{bmatrix} 9 & 45 & -72 \\\\ 81 & -45 & -72\\\\ 81 & -45 & 18\end{bmatrix}$