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3 years ago

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A rectangular prism has a length of 212 centimeters, a width of 212 centimeters, and a height of 5 centimeters. Justin has a sto

rage container for the prism that has a volume of 35 cubic centimeters. What is the difference between the volume of the prism and the volume of the storage container? Enter your answer in the box as a simplified mixed number or a decimal.

1 answer:

BlackZzzverrR [31]3 years ago

3 0

Answer:

$3.75\ cm^{3}$ or $3\frac{3}{4}\ cm^{3}$

Step-by-step explanation:

step 1

Find the volume of the rectangular prism

we know that

The volume of a rectangular prism is

$V=LWH$

In this problem we have

$L=2\frac{1}{2}\ cm=\frac{2*2+1}{2}=\frac{5}{2}\ cm$

$W=2\frac{1}{2}\ cm=\frac{2*2+1}{2}=\frac{5}{2}\ cm$

$H=5\ cm$

substitute in the formula

$V=(\frac{5}{2})(\frac{5}{2})(5)=\frac{125}{4}=31.25\ cm^{3}$

step 2

Find the difference between the volume of the prism and the volume of the storage container

$35\ cm^{3}-31.25\ cm^{3}=3.75\ cm^{3}$

$3.75=3\frac{3}{4}\ cm^{3}$

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<h3>How to apply limits and derivatives to the study of particle motion</h3>

a) To determine the limit for $t = 2$ , we need to apply the following two <em>algebraic</em> substitutions:

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The limit of the position of particle $Q$ when time approaches 2 is $-\pi$ . $\blacksquare$

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$v_{Q}(t) = \frac{f'(t)\cdot g(t)-f(t)\cdot g'(t)}{g(t)^{2}}$ (3)

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$v_{Q}(t) = \frac{\pi \cdot \cos \pi t \cdot (2-t)-\sin \pi t}{(2-t)^{2}}$

$v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$

The velocity of particle $Q$ is $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$ for all $t \ne 2$ . $\blacksquare$

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<h3>Remark</h3>

The statement is incomplete and poorly formatted. Correct form is shown below:

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