3 years ago

Explainnnnnnnn plzzzzz!

Mathematics

1 answer:

miskamm [114]3 years ago

7 0

Answer:

i and iv...................

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gayaneshka [121]

1.09 that's the answer ! I just wrote extra stuff for the sake of it . x= 1.09

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3 years ago

Which two triangles are congruent? Complete the congruence statement.

shtirl [24]

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tsu and hgi

Step-by-step explanation:

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Sarah's math grade average is 93.3333(repeating)%. What is the repeating decimal part of her grade as a fraction?

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2 years ago

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What does x equal in this equation -5x + 8 = -2

AnnyKZ [126]

Answer:

x = 2

Step-by-step explanation:

-5x + 8 = -2

-5x + 8 - 8 = -2 -8

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2 years ago

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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.

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

So, this integral is divergent.

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3 years ago