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drek231 [11]
3 years ago
5

Aran moved a cone-shaped pile of sand that had a height of 4 ft and a radius of 3.2 ft. He used all of the sand to fill a cylind

rical pit with a radius of 6 ft. How high did the sand reach in the pit? Use 3.14 to approximate pi and round your answer to one decimal place.
Mathematics
1 answer:
Yuliya22 [10]3 years ago
5 0
So volue of a cone=area of circle base times heigt times 1/3

height=4
area of base=pi times r^2
area of abse=aprox 3.14 times 3.2^2=32.1536
32.1536 times 4 times 1/3=32.1536 times 4/3=42.87 the amount of sand is 42.9 ft^3
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Your answer is: 20

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hope this helps!
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3 years ago
What is an explicit formula for the following sequence 1,5,9,13,17 ? plssss helppp mee
morpeh [17]

Answer:

plus 4 every time

Step-by-step explanation:

1+4=5+4=9+4=13+4=17

7 0
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Write the equation for the statement the sum of three times x and 13 is 25 ​
Sonja [21]

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3x+13=25

Step-by-step explanation:

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How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title
kotegsom [21]
Let t=\sqrt y, so that t^2=y, t^4=y^2, and \mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}. Then

\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}

Now let y=\sqrt2\tan z, so that \mathrm dy=\sqrt2\sec^2z\,\mathrm dz. Then

\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C

Transform back to y to get

\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C

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3 0
3 years ago
Use the intercepts from the graph below to determine the equation of the function.
Tems11 [23]

(-3,0) and (5, -2)

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8y = -2x - 6

2x + 8y = - 6

Answer is

D. 2x + 8y = -6

5 0
3 years ago
Read 2 more answers
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