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Svetllana [295]
2 years ago
6

I forgot to send this in my last question, Sorry!

Mathematics
1 answer:
pshichka [43]2 years ago
8 0

1. \:  =  > 4( - 7) \\  = 4 \times  - 7 \\  =  - 28

3. \:  =  > 9( - 12) \\  = 9 \times  - 12 \\  =  - 108

4. \:  =  >  - 6( - 8.75) \\  = 6 \times 8.75 \\  = 52.5

5. =  > 27( - 3) \\  = 27 \times  - 3 \\  =  - 81

7. \:  =  >  - 55(0) \\  - 55 \times 0 \\  = 0

8. \:  =  > ( - 7)( - 7.25) \\  = 7 \times 7.25 \\  = 50.75

11. \:  =  > ( - 1) {}^{4}  \\  = 1 {}^{4}  \\  = 1

13. =  > 42 \div ( - 7) \\  =  - (42 \div 7) \\  =  - 6

15. =  >  - 9 \div 3 \\  =  - (9 \div 3) \\  =  - 3

<h2>Solved ✔︎</h2>

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D) The graph of the line shifts horizontally left 5 units.

Adding a number, n, inside of the parenthesis with x would move the parent function n units to the left.




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Area of a circle = pi r^2

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10. Given matrices A, B, and C below, perform the indicated operations if possible. If the operation is not possible, explain wh
Artemon [7]

Answer(a):

Order of matrix A = 3x3 {because number of rows=3, number of columns=3}

Order of matrix B = 3x3

Order of matrix C = 1x3


3A will be of order 3x3 because multiply by scalar doesn't change the order. Which is same as the order of matrix B.

So 3A+B is possible.

We simply multiply all elements of A with 3 and add them with corresponding elements of B.

3\begin{bmatrix}2 &  -1& 0\\ 0 &  5& 0.3\\ 1 &  4&10 \end{bmatrix}+\begin{bmatrix}5 &  0& 2\\ 1 &  -3& 9\\ 2 &  0& 4\end{bmatrix}= \begin{bmatrix}11 &  -3& 2\\ 1 &  12& 9.9\\ 5 &  12& 34\end{bmatrix}


Answer(b):

2B will be of order 3x3 because multiply by scalar doesn't change the order. Which is NOT same as the order of matrix C.

So 2B+C is NOT possible.

Answer(c):

CA is possible only if

number of columns of C = number of columns of A

3=3

which is true hence CA is possible.

\begin{bmatrix}1 &3  &5 \end{bmatrix}*\begin{bmatrix}2 &  -1& 0\\ 0 &  5& 0.3\\ 1 &  4&10 \end{bmatrix}=\begin{bmatrix}1*2+3*0+5*1 & 1*-1+3*5+5*4 & 1*0+3*0.3+5*10\end{bmatrix}

= \begin{bmatrix}7 &  34& 50.9\end{bmatrix}

<em><u>If the matrix is not visible properly then look at the attached picture.</u></em>

5 0
3 years ago
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