A large emerald with a mass of 812.04 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. Answer for Blank 1:

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A large emerald with a mass of 812.04 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. Answer for Blank 1:

Mathematics
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i got 294.22
So, 2.76 = 812.04/x 2.76x = 812.04

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wow your good
Haha sorry, miscalculation on my part. That's correct.
mine is correct??
360 no scoped
No, you're fine. But, being that you're rounding to the nearest hundredth, it would be 294.22
okay thats what i got thank you!!
can u check one more real fast?
Find the length of the base of a square pyramid if the volume is 128 cubic inches and has a height of 6 inches. 4 inches 8 inches 16 inches 24 inches
i said 8in
yes you are correct we are anonymous
yes ill help
okay thank you! So 8in is correct
1/3 * x^2 * 6 = 128 see 1/3 * 6 = 2 so, we get 2 x^2 = 128 divided by 2 on both sides, get 2x^2/2 = 128/2 x^2 = 64 x = sqrt(64) = 8 is good
A cylindrical vase has a diameter of 6 inches. At the bottom of the vase, there are 9 marbles, each of diameter 2 inches. The vase is filled with water up to a height of 12 inches. What is the volume of water in the vase? 96π in3 27π in3 65π in3 105π in3
would it be 96 pi3
do you guys know a troller around here?!?!
96pi 3in sorry
hes trolling right know and thats my cuzin whos trolling
8 is your answer
yes can u help me with the other i just posted on here? also??
yea
A cylindrical vase has a diameter of 6 inches. At the bottom of the vase, there are 9 marbles, each of diameter 2 inches. The vase is filled with water up to a height of 12 inches. What is the volume of water in the vase? 96π in3 <---- my answer 27π in3 65π in3 105π in3
tag me ok we are anonymous

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