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katrin2010 [14]
2 years ago
14

PLEASE HELP ASAP!!!!!!!!!!!

Mathematics
1 answer:
irina [24]2 years ago
7 0

Answer:

132 in^2 but DOUBLE CHECK MY WORK!!!!

Step-by-step explanation:

Break it down into parts ok. So I see a 10x10 square, two 2x4 right triangles and a 4x6 rectangle.

10x10 is 100 in

1/2(2*4) = 4 x 2 = 8 in

6x4 = 24 in

add them together to get 132 in^2

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The sum of twice a number and 13 is 75 <br> • <br> but in an algebraic equation<br> DONT SOLVE
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Answer:

I won't solve it, because you asked me to.

Step-by-step explanation:

The first question we should ask ourselves is "what are we looking to solve?"

We are looking for some sort of value that satisfies that equation. We'll call it x

Reading from left to right:

The sum of twice a number and 13 is 75

         +            2              x              13 = 75

2x + 13 = 75

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2 years ago
What is the length of WY if XZ=6
olga_2 [115]

Answer:

WY= squar root 55 + 4

Step-by-step explanation:

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Michelle's class is handing out balloon arrangement consisting of 4 balloons each each balloon has and equally likely chance of
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Answer:

The answer is 1/30 or 0.0333333...

Step-by-step explanation:

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The product of 5 and y added to 3
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The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati
makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\

And the second one would be

y = 2-x \,\,\,\,\,,  1 \leq x \leq 2.

If you rotate the first region around the "y" axis you get that

{\displaystyle A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51 }

And if you rotate the second region around the "y" axis you get that

{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }

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If you revolve just the outer curve you get

If you rotate the first  region around the x axis you get that

{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }

And if you rotate the second region around the x axis you get that

{\displaystyle A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472 }

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3 years ago
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