How much paper is used for the label on the can of peas? Round your answer to the nearest tenth of a square inch.

Mathematics

1 answer:

Gekata [30.6K]3 years ago

5 0

Find the circumference of the lid (this is like a side length of a rectangle)

2*pi*2= 12.566in

Then multiply by 3 (the other side length)

12.566*3= 37.7in squared

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Make v the subject of the relation KE = ½mv²

Irina18 [472]

Answer:

$v=\sqrt{\frac{2KE}{m} }$

Step-by-step explanation:

$KE = 1/2mv^2$

$KE/1/2=mv^2$

$2KE=mv^2$

$2KE/m=v^2$

$\sqrt{2KE/m} =v$

7 0

3 years ago

Read 2 more answers

Solve. 5 < x – 2 < 11

lions [1.4K]

The answer is 7 < x < 13.

5 < x - 2 < 11
Add 2 in all sections:
5 + 2 < x - 2 + 2 < 11 + 2
7 < x < 13

3 0

4 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.

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