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amm1812
2 years ago
14

Which expressions are equivalent to 4 to the 5 power choose all that apply

Mathematics
1 answer:
Nuetrik [128]2 years ago
7 0

Answer:

since you didn't put the answer choices, this is what I know...

4 to the 5 power is 1024

it can be written as:

4x4x4x4x4 or 4^5

Step-by-step explanation:

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Pls help I hate math rectangular prism
nekit [7.7K]

Answer:

352.5cm^{2}

Step-by-step explanation:

A=3Rectangles+2Triangles\\=3*7.5*13.5+2*.5*6.5*7.5=352.5cm^{2}

8 0
2 years ago
Consider the figure. What is the area of the figure, in square centimeters? ​
Serhud [2]

Answer: The

Step-by-step explanation: First we look at our right angles of the shape, Next you analyze your formula and use it to form your answer(formula: Base X Height).Then for the final step you would multiply 9x5=45.We wouldn't be doing all of them because that would be doing Total Area.

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2 years ago
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Y= 1/2x -3 on a graph<br> I’m confused on how to graph 1/2x - 3
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Answer:

Put your equation into this!

Step-by-step explanation:

desmos.com/calculator

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3 0
3 years ago
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Express the system as AX = B; then solve using matrix inverses
Wittaler [7]

Answer with explanation:

1. The given equations are

3x -5 y=2

-x+2 y= 0

⇒The matrix in the form of , AX=B, is

A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]

\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B

Adj.A=Transpose of cofactor of Matrix A

Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2

2.

The given equations are

x+y-z=2

x+z=7

2 x +y+z=13

⇒The matrix in the form of , AX=B, is

   A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]

\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4

7 0
3 years ago
Please hurry and answer
Katarina [22]

Answer:

64

Step-by-step explanation:

It’s directly adjacent to the other angle.

4 0
2 years ago
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