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FrozenT [24]
2 years ago
5

The diameter of a cylindrical construction pipe is 4ft If the pipe is 17ft long, what is its volume? Use the value 3.14 for /pi,

and round your answer to the nearest whole number. Be sure to include the correct unit in your answer.
Mathematics
1 answer:
Leya [2.2K]2 years ago
5 0

Answer:

213.52 cubic ft

Step-by-step explanation:

Diameter of pipe = 4ft

radius for circle is half of diameter

thus radius of pipe = 1/2 of 4ft = 2 ft

length of pipe = 17 feet

volume of cylinder is given by \pi r^2l

where r is the radius and l is the length of cylinder

.

thus volume of pipe is calculated below using above values

volume \ of \ pipe = \pi r^2l\\volume \ of \ pipe = \3.14* 2^2*17 \\volume \ of \ pipe =  213.52

volume of pipe is 213.52 cubic ft.

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myrzilka [38]
Answer to first part 1: First you obtain number of intervals you would like to have by getting the square root of the total number of observations. In this case, square root of 14 = 3.74; round up and use 4.0.
Answer to Second part: 2:  The interval width is then obtained by dividing the range by the number of intervals. For this case, interval width = (42-9)/4 = 8.25. This can be rounded up to the convenient evenfigure of 10.0.



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5 0
3 years ago
1. Leo made a container to store his camping gear. The container is in the shape of a rectangular prism. The container has mater
AleksandrR [38]
<span>(a) What is the capacity of the container?
Volume = length * width * height
V = 4ft * 2ft * 3ft
V = 24 ft</span>³
<span>
(b) How much material was used to make the container?
Surface area = sum of 6 faces of the prism
4 ft x 3 ft = 12 ft</span>² → 12 ft² x 2 = 24 ft²
4 ft x 2 ft = 8 ft² → 8 ft² x 2 = 16 ft²
2 ft x 3 ft = 6 ft² → 6 ft² x 2 = 12 ft²
<span>
Surface area = 24 ft</span>² + 16 ft² + 12 ft² = 52 ft²
5 0
3 years ago
Please explain how to find the answer
irina1246 [14]

Answer:

\rm x=\frac{46}{3}

\rm y=-\frac{25}{24}

Step-by-step explanation:

\huge\begin{array} {|c|} \hline\left. \begin{cases} {   \rm\frac{ 1  }{ 4  }  x+ \frac{ 4  }{ 5  }  y =  3  } \\ {   \rm\frac{ 5  }{ 16  }  x- \frac{ 1  }{ 5  }  y=5  }  \end{cases} \right. \\  \hline \end{array}

\\ \Large \overline{ \rm -   \: Steps \:  using  \: substitution \: - }

\rm \frac{1}{4}x+\frac{4}{5}y=3

\rm \frac{1}{4}x=-\frac{4}{5}y+3

\rm x=4\left(-\frac{4}{5}y+3\right)

\rm x=-\frac{16}{5}y+12

┈

\rm \frac{5}{16}\left(-\frac{16}{5}y+12\right)-\frac{1}{5}y=5

\rm -y+\frac{15}{4}-\frac{1}{5}y=5

\rm -\frac{6}{5}y+\frac{15}{4}=5

\rm -\frac{6}{5}y=\frac{5}{4}

\bold{ y=-\frac{25}{24}}

┈

\rm x=-\frac{16}{5}\left(-\frac{25}{24}\right)+12

\rm x=\frac{10}{3}+12

\bold{ x=\frac{46}{3}}

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