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2 years ago

Please can someone help??????

Mathematics

1 answer:

bearhunter [10]2 years ago

6 0

Answer: I’m not sure if I’m right but I think it’s C because there’s only 2 triangles and only one has the same measurement

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What is (4ab^0c^9)^3 ?

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The expression inside parentheses can be simplified by noting that b^0 = 1. Then you have ...

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Intergal of 8sin^4(x)dx

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$\int\limits8sin^4(x)dx =$

Take the constant out $\int\limits a*f(x)dx=a* \int\limitsf(x)dx$

$= 8 \int\limits sin^4(x)dx$

$\int\limits sin^4(x)dx = - \frac{cos(x)sin^3(x)}{4} = \frac{3}{4} \int\limits sin^2(x)dx$

$= 8( -\frac{cos(x)sin^3(x)}{4}+ \frac{3}{4} \int\limits sin^2(x)dx)
$

Use the following identify : $sin^2(x) = \frac{1-cos(2x)}{2}$

$= 8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2}( \int\limits 1 - cos(2x)dx)$

$\int\limits 1dx = x$

$\int\limits cos(2x)dx = \frac{1}{2} sin(2x)$

Apply the sum rule :

$=8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2} (x- \frac{1}{2} sin(2x))$

Simplify :

$8( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))$

Therefore add a constant to the solution:

$= 8 ( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))+C$

hope this helps!

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