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olasank [31]
3 years ago
12

9 and 10 PLEASE ITS SO HARD!!!!

Mathematics
2 answers:
RUDIKE [14]3 years ago
4 0
9) Here, Two angles are unknown without any relation not even a variable.
So, we could calculate the sum of that two unknown angles only,not their individual magnitude.

It would be: 50 + 152 + missing angles = 360
m.a = 360 - 202
m.a = 158

In short, Sum of two Unknown angles would be 158

10) Here, ABCD is a Quadrilateral, so their adjacent side would make a linear pair equation, means their sum would be equal to 180

So, ∠1 + ∠2 = 180

Hope this helps!
astraxan [27]3 years ago
3 0

For 9: All the angle should add up  to 360 degrees.

So 152+50=202

360-202=158 degrees (Missing angle measure)

Method being used: All angle should add up to 360 degrees.


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Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:

<em>y</em> = -1 - <em>x</em>  ==>  <em>x</em> = -1 - <em>y</em>

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\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy

The integral with respect to <em>x</em> is trivial:

\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy

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\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0
3 years ago
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34x=0
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X=0/34
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andre [41]

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