I need help I don't understand this

Mathematics

1 answer:

Lady_Fox [76]3 years ago

4 0

<h3><u>Answer</u> :</h3>

It is clear from the diagram that,

ㄥ6 + ㄥ8 = 180

⇒ (2x - 5) + (x + 5) = 180

⇒ (2x + x) + (5 - 5) = 180

⇒ 3x + 0 = 180

⇒ x = 180/3

⇒ x = 60°

◈ From the geometry of figure,

ㄥ3 = ㄥ6

⇒ ㄥ6 = (2x - 5)

⇒ ㄥ6 = (2×60 - 5)

⇒ ㄥ6 = 120 - 5

⇒ <u>ㄥ6 = 115°</u>

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a) So, this integral is convergent.

b) So, this integral is divergent.

c) So, this integral is divergent.

Step-by-step explanation:

We calculate the next integrals:

$\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\$

So, this integral is convergent.

$\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\$

So, this integral is divergent.

$\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\$