The frist three terms of a geometric sequence are as follows 80,40,20

Polygon ABCD has vertices A(1,3), B(1,6), C(4,6), and D(5,2)

Galina-37 [17]

Answer: A (0.5, 1.5) B(0.5, 3) C(2, 3) D(2.5, 1)

Step-by-step explanation:

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1 year ago

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SVETLANKA909090 [29]

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

$\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}$

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}$

In matrix multiplication, the number of columns in the 1st matrix is equal to the number of rows in the 2nd matrix.

$\left(\begin{matrix}2&3\\5&4\end{matrix}\right)\left(\begin{matrix}2&0&3\\-1&1&5\end{matrix}\right)$

Multiply each element of the 1st row of the 1st matrix by the corresponding element of the 1st column of the 2nd matrix. Then add these products to obtain the element in the 1st row, 1st column of the product matrix.

$\left(\begin{matrix}2\times 2+3\left(-1\right)&&\\&&\end{matrix}\right)$