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Tasya [4]
3 years ago
12

What’s the answer?!!!!!

Mathematics
1 answer:
12345 [234]3 years ago
4 0
C = -5

8c-6=2c-36
8c-2c=-36+6
6c=-30
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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

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

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3 years ago
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Please help!!
crimeas [40]
1)~3x+12=3(x+4)\\\\ 2)~7y-7=7(y-1)\\\\ 3)~5x+30y=5(x+6y)\\\\ 4)~8m+36n=4(2m+9n)\\\\ 5)~6a^2+27=3(2a^2+9)\\\\ 6)~4y^2-24y=4y(y-6)\\\\ 7)~21cd-3d=3d(7d-1)\\\\ 8)~14gh-18h=2h(7g-9)\\\\ 9)~15a^2b-30ab=15ab(a-2)\\\\ 10)~16bc^2+24bc=8bc(2c+3)
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3 years ago
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