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3 years ago

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Given the triangle below, which of the following equations correctly representsthe relationship between a, b, and c?Triangle not

drawn to scale A cz B az 2abcos (90.)b a2 c2 2abcos (90.)

1 answer:

alukav5142 [94]3 years ago

7 0

A) it is the law of cosine

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PLEASE SOMEONE HELP ME I DONT GET IT!!!

Mkey [24]

Answer: b is the answer

Step-by-step explanation:

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3 years ago

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1/2(3/2y+1/3)=-1/2(y-5/2)

jonny [76]

Answer:

y = - 17/3

Step-by-step explanation:

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3 years ago

What is the equation 8q-6=2q-48

FrozenT [24]

Answer:

The equation is 8q-6=2q-48

The answer is -9

Step-by-step explanation:

8q-6=2q-48 Subtract 2q on both sides

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6q=-48. Divide 6 on both sides

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3 years ago

PLEASE I NEED HELP. I Don't get this at all:(

klemol [59]

Answer: :,)

Step-by-step explanation:

so u see were the b is in the triangle it want u to find how much b is worth

8 0

3 years ago

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

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

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

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

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

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

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

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

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

And the sum would be 1.5708+1.0472 = 2.618

7 0

3 years ago