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3 years ago

HELP PLEASE I NEED IT LIKE NOW ILL GIVE BRAINLEST

Mathematics

1 answer:

Mazyrski [523]3 years ago

6 0

Answer:

Step-by-step explanation:

Since the scale factor is , we know that the bigger shape went to the smaller shape.

If we know that the smaller shape's side, 12, is of the bigger one, we can make the equation

To solve for x, we can divide both sides by .

We can multiply by the reciprocal:

Hope this helped!

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Problem 24

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$\displaystyle \int \csc(x)\left(\sin(x)+\cot(x)\right)dx\\\\\displaystyle \int \frac{1}{\sin(x)}\left(\sin(x)+\frac{\cos(x)}{\sin(x)}\right)dx\\\\\displaystyle \int \frac{1}{\sin(x)}*\sin(x)+\frac{1}{\sin(x)}*\frac{\cos(x)}{\sin(x)}dx\\\\\displaystyle \int 1+\frac{\cos(x)}{\sin^2(x)}dx\\\\\displaystyle \int 1 dx+\int \frac{\cos(x)}{\sin^2(x)}dx\\\\\displaystyle x+C_1+\int \frac{1}{u^2}du \ \text{ ... where } u = \sin(x)\\\\\displaystyle x+C_1+\int u^{-2}du\\\\$

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$\displaystyle x+C_1+\frac{1}{1+(-2)}u^{-2+1}+C_2\\\\\displaystyle x+C_1+\frac{1}{-1}u^{-1}+C_2\\\\\displaystyle x+C_1-u^{-1}+C_2\\\\\displaystyle x+C_1-\frac{1}{u}+C_2\\\\\displaystyle x+C_1-\frac{1}{\sin(x)}+C_2\\\\\displaystyle x-\frac{1}{\sin(x)}+C_1+C_2\\\\\displaystyle x-\frac{1}{\sin(x)}+C\\\\\displaystyle x-\csc(x)+C\\\\$

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Don't forget about the plus C constant

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Fortunately, there aren't as many steps for this problem.

$\displaystyle \int \frac{dy}{\csc(y)}\\\\\displaystyle \int \frac{1}{\csc(y)}dy\\\\\displaystyle \int \sin(y)dy\\\\\displaystyle -\cos(y)+C\\\\$

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