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3 years ago

Anusha needs to find a rectangular container large enough to hold a volume of 130 in.. Which container should sh

Mathematics

1 answer:

Alinara [238K]3 years ago

4 0

Answer:

"a container with the dimensions 3 by 5 by 10 inches"

Step-by-step explanation:

To find volume of rectangular box, we multiply the length, width, and height. Or if given in 3 dimensions form, we multiply all 3.

Lets find volume of each of the answer choices:

a container with the dimensions 2 by 5 by 10 inches = 2 * 5 * 10 = 100

a container with the dimensions 3 by 4 by 10 inches = 3 * 4 * 10 = 120

a container with the dimensions 3 by 5 by 10 inches = 3 * 5 * 10 = 150

a container with the dimensions 4 by 2 by 10 inches = 4 * 2 * 10 = 80

We want a container LARGE ENOUGH to contain a volume of 130 cubic inches. Which of the containers has a volume GREATER than 130?

Yes, only the 3rd option. It is 150 cubic inches in volume.

So, this is the answer:

a container with the dimensions 3 by 5 by 10 inches

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kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

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$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

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3 years ago