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natima [27]
2 years ago
7

Tell whether the angles are complementary, supplementary, or neither.

Mathematics
1 answer:
Nat2105 [25]2 years ago
4 0

Answer:they are supplementary

Step-by-step explanation:

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An artist has completed 1/4<br> of a penting in 2 weeks. At what rate is she working?
Mrac [35]

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1/8

Step-by-step explanation:

1/4 ÷ 2 = 0.125

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5 0
2 years ago
Please help me with this
blsea [12.9K]

Answer:

9x squared + 12x

Step-by-step explanation:

8 0
3 years ago
Brainliest question please help me now plz I need it
Alenkasestr [34]

Answer:

As shown in picture:

A(-4, 1)

Z(-2, 3)

P(3, -4)

The length of AZ is calculated by:  

L = sqrt((-4 - -2)^2 + (1 - 3)^2) = 2.83

The length of AP is calculated by:

L = sqrt((3 - -4)^2 + (-4 - 1)^2) = 8.60

THe length of ZP is calculated by:

L = sqrt((3 - -2)^2 + (-4 - 3)^2) = 8.60

=>Perimeter of triangle AZP is calculated by:

P = AZ + AP + ZP = 2.83 + 8.60 + 8.60 = 20.3

Hope this helps!

:)

7 0
3 years ago
Please solve the problem ​
jek_recluse [69]

Treat the matrices on the right side of each equation like you would a constant.

Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.

Then you can eliminate <em>Y</em> by taking the sum

4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>

==>   <em>X</em> = (4<em>A</em> + <em>B</em>)/11

Similarly, you can eliminate <em>X</em> by using

-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>

==>   <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11

It follows that

X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}

Similarly, you would find

Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}

You can solve the second system in the same fashion. You would end up with

P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}

3 0
3 years ago
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