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Mathematics 8 Online
OpenStudy (shaw74):

Find the length of the base of a square pyramid if the volume is 48 cubic inches and has a height of 9 inches. Use 3.14 for pi, and round your answer to the nearest hundredth. 4 inches 8 inches 16 inches 24 inches

OpenStudy (shaw74):

can anybody help me?

jimthompson5910 (jim_thompson5910):

Use the formula V = (1/3)*s^2*h In this case, V = 48 s = unknown (you are solving for s) h = 9

OpenStudy (shaw74):

what is s

OpenStudy (shaw74):

so do i multiply 1/3 times 9? equals 3 now what

jimthompson5910 (jim_thompson5910):

V = (1/3)*s^2*h 48 = (1/3)*s^2*9 48 = (1/3)*9*s^2 48 = 3*s^2 isolate and solve for s

OpenStudy (shaw74):

16

jimthompson5910 (jim_thompson5910):

s^2 = 16 so s = ???

OpenStudy (shaw74):

16

OpenStudy (shaw74):

s = 16 i thought

jimthompson5910 (jim_thompson5910):

nope

jimthompson5910 (jim_thompson5910):

s^2 = 16 you have to take the square root of both sides to isolate s

jimthompson5910 (jim_thompson5910):

since the square root undoes squaring

OpenStudy (shaw74):

256

jimthompson5910 (jim_thompson5910):

square ROOT

OpenStudy (shaw74):

4

jimthompson5910 (jim_thompson5910):

yep \[\Large s^2 = 16\] \[\Large \sqrt{s^2} = \sqrt{16}\] \[\Large s = 4\]

OpenStudy (shaw74):

i got it before you said it haha yay

jimthompson5910 (jim_thompson5910):

lol nice work

OpenStudy (shaw74):

thanks man

OpenStudy (shaw74):

Find the volume of a cone with a diameter of 5 inches and a height that is 3 times the radius. Use 3.14 for pi, and round your answer to the nearest hundredth. 49.06 in3 78.51in3 147.19 in3 392.50 in3

OpenStudy (shaw74):

can you help with this

jimthompson5910 (jim_thompson5910):

diameter of 5 inches radius = ??

OpenStudy (shaw74):

49.06

OpenStudy (shaw74):

A large emerald with a mass of 378.24 grams was recently discovered in a mine. If the density of the emerald is 2.76grams over centimeters cubed, what is the volume? Round to the nearest hundredth when necessary and only enter numerical values, which can include a decimal point.

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