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3 years ago

• Which expression has a term with a coefficient of 6? бу 4p +6 62 20

Mathematics

1 answer:

Oxana [17]3 years ago

3 0

6y has coefficient 6

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Solve this 8x-6(3-2x)

BaLLatris [955]

It would be −47 cause 8 * -6 = -48 then (3-2x) = 1 then -48 + 1 = -47 so that’s the answer

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3 years ago

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Given that ∠A≅∠B , Evelia conjectured that ∠A and ∠B are acute angles.

eimsori [14]

45 and 45 is the answer

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3 years ago

Please help me thank you

laila [671]

Parallel lines have the same slope. The equation of the parallel line is therefore:

y = x + b

Plug in the values you are given to find b:

2 = -3 + b

b = 5

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3 years ago

Find the value of the power. (2/3)3

Andreyy89

Answer:

The value is 27

Step-by-step explanation:

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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