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Anastaziya [24]
3 years ago
7

Write using exponent (-2) (-2) (-2) (-2)

Mathematics
2 answers:
Sergio [31]3 years ago
7 0
Your answer is to be 16
poizon [28]3 years ago
3 0
It would be
( - 2) {}^{4}
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Advocard [28]
I hope this helps you

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Tell me what each angle is in the following triangle: angle 1 is 8x, angle 2 is 2x+3, angle 3: 30 degrees. give me the two answe
mr Goodwill [35]
8 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
6 0
4 years ago
Find the distance between the points: (–6, 7) and (–1, –5). In your final answer, include the formula and calculations that you
Arturiano [62]

To find distance between points <span>A(<span>xA</span>,<span>yA</span>)</span> and <span>B(<span>xB</span>,<span>yB</span>)</span>, 

formula:

<span>d(A,B)=</span>√ [(xB−xA)^2+(yB−yA)^2]

so distance between the points: (–6, 7) and (–1, –5)<span>
</span>= √ [<span><span><span><span>(<span>−1−<span>(<span>−6</span>)</span></span>)^</span>2 </span>+ <span><span>(<span>−5−7</span>)^</span>2]
</span></span>= </span>√ (<span><span>25+144)
</span>= </span>√ 169
<span><span><span>= </span></span><span>13

answer
13</span></span>
7 0
3 years ago
Read 2 more answers
If A = 1 2 1 1 and B= 0 -1 1 2 then show that (AB)^-1 = B^-1 A^-1<br><br><br> help meeeee plessss ​
Trava [24]

A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}

where det(<em>A</em>) = 1×1 - 2×1 = -1.

B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}

where det(<em>B</em>) = 0×2 - (-1)×1 = 1. Then

B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

On the other side, we have

AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}

and det(<em>AB</em>) = det(<em>A</em>) det(<em>B</em>) = (-1)×1 = -1. So

(AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

and both matrices are clearly the same.

More generally, we have by definition of inverse,

(AB)(AB)^{-1} = I

where I is the identity matrix. Multiply on the left by <em>A </em>⁻¹ to get

A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}

Multiplication of matrices is associative, so we can regroup terms as

(A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}

Now multiply again on the left by <em>B</em> ⁻¹ and do the same thing:

B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}

7 0
3 years ago
A man that is 6 feet tall is standing so that the tip of his shadow is 20 feet from a light pole. His shadow is 8 feet long. Wha
pychu [463]

Answer:

The answer to your question is: the height of the light pole is 15 ft.

Step-by-step explanation:

See the picture below

Now, we can do proportions to finds the height of the light pole

                   \frac{20}{8} =  \frac{x}{6}

Solve for x

                      x = 6(20) / 8

                     x = 120 / 8

                     x = 15 ft

                   

4 0
3 years ago
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