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CaHeK987 [17]
3 years ago
13

(w^5 z^2 )(−9w^2 z^5)

Mathematics
1 answer:
Lady_Fox [76]3 years ago
5 0
We have that
(w^5*z^2)*(-9*w^2*z^<span><span>5)</span> 

</span><span>Multiplying exponential expressions
we know that
</span>w^5<span> multiplied by </span>w^<span>2 = w</span>^<span>(5 + 2) = w</span>^7
and
z^2<span> multiplied by </span>z^<span>5 = z</span>^<span>(2 + 5) = z</span>^7

therefore
(w^5*z^2)*(-9*w^2*z^5) =-9*( w^7)*( z^7)
-9*( w^7)*( z^7)=-(3^2)*( w^7)*( z^7)

the answer is
-(3^2)*( w^7)*( z^7

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4 0
3 years ago
Evaluate the following integral using trigonometric substitution
serg [7]

Answer:

The result of the integral is:

\arcsin{(\frac{x}{3})} + C

Step-by-step explanation:

We are given the following integral:

\int \frac{dx}{\sqrt{9-x^2}}

Trigonometric substitution:

We have the term in the following format: a^2 - x^2, in which a = 3.

In this case, the substitution is given by:

x = a\sin{\theta}

So

dx = a\cos{\theta}d\theta

In this question:

a = 3

x = 3\sin{\theta}

dx = 3\cos{\theta}d\theta

So

\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}

We have the following trigonometric identity:

\sin^{2}{\theta} + \cos^{2}{\theta} = 1

So

1 - \sin^{2}{\theta} = \cos^{2}{\theta}

Replacing into the integral:

\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C

Coming back to x:

We have that:

x = 3\sin{\theta}

So

\sin{\theta} = \frac{x}{3}

Applying the arcsine(inverse sine) function to both sides, we get that:

\theta = \arcsin{(\frac{x}{3})}

The result of the integral is:

\arcsin{(\frac{x}{3})} + C

8 0
2 years ago
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