3 years ago

What is |6| = |-10| = if you know pleas tell me

Mathematics

2 answers:

scZoUnD [109]3 years ago

8 0

Answer:

False

Step-by-step explanation:

|6| = +6

|-10| = +10

kifflom [539]3 years ago

4 0

Answer:

the answer is false

Step-by-step explanation:

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PLEASE HELP WITH GEOMETRY!!!

svetoff [14.1K]

Answer:

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Step-by-step explanation:

7 0

3 years ago

Determine whether each integral is convergent or divergent. If it is convergent evaluate it. (a) integral from 1^(infinity) e^(-

AVprozaik [17]

Answer:

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.

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