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3 years ago

What is the formula of a radius circle ?

Mathematics

2 answers:

UkoKoshka [18]3 years ago

5 0

I assume you happen to mean, what is the formula for a circle.

The formula is:

A= $\pi r^{2}$

Bas_tet [7]3 years ago

3 0

r= C/2 x 3.14 would be the answer to this

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Determine the slope of the line shown on the graph

vovangra [49]

Answer:

The Answer Is:

B. The Slope is 0

Hope This Helps!

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3 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.

4 0

3 years ago

What is the minum of two colors needed?

Alexus [3.1K]

4 colour needed for this…i think

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2 years ago

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mixer [17]

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2 years ago

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zepelin [54]

The first one would be 120 the second one would be 68

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3 years ago

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