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2 years ago

Can you please help me??

Mathematics

1 answer:

lord [1]2 years ago

4 0

15 minutes

1 hour divided by 4 is 15

hope that helped

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HELP PLEASEEEE it’s due in 10 minutes

lana [24]

Answer:

24.187

Step-by-step explanation:

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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.

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

So, this integral is divergent.

4 0

3 years ago

Question from my math test

frutty [35]

Answer:

The manager skipped June

Step-by-step explanation:

In the first bar we can see the sales are from

January - February

Then,

March - April.

But then, the store manager goes from

May - July,

and totally skipped June, therefore making it look like the sales have increased.

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3 years ago

Geometry please help asap!

m_a_m_a [10]

Step-by-step explanation:

AB = 3

by similaritiy criterion

CE/ CB= ED/AB

4/6= 2/AB

AB = 3

plz mark my answer as brainlist plzzzz also vote me ..if you find it useful.

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3 years ago

Please help me in with these two questions Geometry.

horrorfan [7]

Answer:

#1 is A #2 is A

Step-by-step explanation:

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2 years ago