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Vinil7 [7]
3 years ago
15

Help me please find x for this triangle!!!!

Mathematics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

x=30

Step-by-step explanation:

Since the measure of an exterior angle is equal to the sum of any two angles in a triangle, the equation would be 4x-18=2x+6+36. Solve this, and we get x=30.

Hope this helped!

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Can someone help me with this
Alborosie

ok

3 0
2 years ago
Pls help me answer my worksheet
bearhunter [10]
Step one know that across from 128 is 128 because of vertical angles
step two a straight line =180 so 128 -180 = 52
step three know you know 3k+4=52 so subtract 4 from 52 and that gives you 48 then divide by 3 to find k and 48/3 is 16
so k +16
because 16x3 is 48 then add 4 which is 52 and 52 plus 128 is 180
8 0
3 years ago
Please help!!!!!!!!!!!!!!!!!
nadya68 [22]

Answer: 5 up 10 over

Step-by-step explanation:

I had this on my quiz

3 0
3 years ago
Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
4 0
3 years ago
Q2:
anygoal [31]

Answer: rectangular prism

Step-by-step explanation:

4 0
3 years ago
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