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Dominik [7]
3 years ago
7

Help me plz I will give brainliest

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
8 0

Answer:

b

Step-by-step explanation:

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Find the definite integral from 0 to 1/16 of arcsin(8x)/sqrt(1-64x^2)dx
kolezko [41]
\displaystyle\int_0^{1/16}\frac{\arcsin8x}{\sqrt{1-64x^2}}\,\mathrm dx

First let y=8x, so that \mathrm dx=\dfrac{\mathrm dy}8 to write the integral as

\displaystyle\frac18\int_0^{1/2}\frac{\arcsin y}{\sqrt{1-y^2}}\,\mathrm dy

Now recall that (\arcsin y)'=\dfrac1{\sqrt{1-y^2}}, so substituting z=\arcsin y should do the trick. The integral then becomes

\displaystyle\frac18\int_0^{\pi/6}z\,\mathrm dz=\frac1{16}z^2\bigg|_{z=0}^{z=\pi/6}=\frac{\pi^2}{576}
8 0
3 years ago
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anyanavicka [17]

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Step-by-step explanation:

7 0
3 years ago
The product of a number and 4 increased by 16 is -2
Thepotemich [5.8K]
This is some simple Algebra
So it's 4x+16=-2
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5 0
4 years ago
I need help again don't get it
Dmitry [639]

Answer:

x=45

Step-by-step explanation:

70+65=135

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3 0
3 years ago
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