What is 3/4 inches of 15 ft

1 answer:

6 0

So .. in 1foot, there are 12inches... thus, in 15feet, there are 15*12 inches

how many in 3/4 of an inch? well $\bf \begin{array}{ccllll}
feet&inches\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
15&15\cdot 12\\
x&\frac{3}{4}
\end{array}\implies \cfrac{15}{x}=\cfrac{15\cdot 12}{\frac{3}{4}}\implies \cfrac{15}{x}=\cfrac{\frac{15\cdot 12}{1}}{\frac{3}{4}}
\\\\\\
\cfrac{15}{x}=\cfrac{15\cdot 12}{1}\cdot \cfrac{4}{3}\implies \cfrac{15}{x}=\cfrac{15\cdot 12\cdot 4}{3}$

solve for "x"

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Please help on this math problem

Arte-miy333 [17]

Answer:

1680 in^3

Step-by-step explanation:

SO you got the triangle

24*10=240

240/2=120

now we just need to find the area

There is a formula but I think this is easier

So theres a rectangle with a length of 14

So if you think of the triangle as the base, you just need 14 triangles stacked up to make up the volume

So 120*14=1680

So the volume is 1680

6 0

2 years ago

A conical vessels with base radius 5 cm and height 24cm is full of water this water is emptied into a cylindrical vessel of base

Aliun [14]

Answer:

$2\ cm$

Step-by-step explanation:

$We\ are\ given\ that,\\Radius\ of\ the\ conical\ vessel=5\ cm\\Height\ of\ the\ conical\ vessel=24\ cm\\Hence,\\As\ we\ know\ that,\\Volume\ of\ a\ cone=\frac{1}{3}\pi r^2h\\Hence,\\The\ volume\ of\ the\ conical\ vessel=\frac{1}{3}\pi (5)^224= \frac{1}{3}\pi 25*24=200\pi \\Hence,\\The\ volume\ of\ the\ conical\ vessel=The\ amount\ of\ water\ filled\ in\ it\\Hence,\\The\ amount\ of\ water\ filled\ in\ the\ conical\ vessel=200\pi\\Lets\ consider\ the\ cylindrical\ vessel:\\Volume\ of\ a\ cylinder=\pi r^2h$

$As\ we\ are\ not\ said\ that\ the\ cylindrical\ vessel\ was\ empty,\ let\ its\ initial\\ height\ be\ h_1\\Radius\ of\ the\ Cylindrical\ vessel=10\ cm\\Initial\ height\ of\ the\ cylindrical\ vessel=h_1\\Initial\ volume\ of\ the\ cylindrical\ vessel=\pi *10^2*h_1=100h_1\pi \\Initial\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi \\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=Final\ amount\ of\ water\\ in\ the\ cylindrical\ vessel +Amount\ of\ water\ in\ conical\ vessel$

$Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=100h_1\pi +200\pi \\Now,\\Let\ the\ final\ height\ be\ h_2\\Hence,\\Final\ amount\ of\ water\ in\ the\ cylindrical\ vessel=\pi *10^2*h_2=100h_2\pi \\Hence,\\100h_2\pi =100h_1\pi +200\pi \\Hence,\\100h_2=100h_1+200\\100h_2-100h_1=200\\100(h_2-h_1)=200\\h_2-h_1=2\\Hence,\\As\ the\ rise\ in\ level\ is\ the\ difference\ between\ the\ initial\ and\ final\\ heights:\\The\ rise\ in\ level\ of\ water=2\ cm$