Each coin in the treasure chest has a radius of 1 inch and a height of 0.0625 inches. If there are 200 coins in the chest, what is the volume of all of the coins?
Answer choices given?
no :-(
ill help
pls help, any help is appreciated
volume of a coin so many times
k holdon
V=πr2 best i can do I know the answer tho so i can check when your done
volume of a circle cylinder, each coin shape, is area of the circle face * height V = pi8r^2 8 h
Volume = pi * r^2 * h
I think he means * instead of 8
haha * is an 8 w/o a shift
V = 3.14 * 1^2 * 0.0625 V = 3.14 * 1 * 0.0625 V = 3.14 * 0.0625 V = 0.19625
Volume = pi * r^2 * h the total is 200*V for all of them
k, that is for one coin, 200 of those is the total they want
39.25
just leave the symbol , for the exact answer
0.0625 * 200 = 39.25 39.25 * pi = 123.245
200 * height is 12.5
where did the 12.5 come from?
you calculated 200*h first how you are doing it 200 * 0. 0625 = 12.5 you mistcalculated that
200 * Volume of Coin = 200 * pi * r^2 * h = 200 * 1^2 * 0.0625 * pi = 12.5*pi
leave it as that, or approximate it using 3.14
12.5 * pi = 39.25
@DanJS if the dimensions of the coins doubled, what would the new volume of the coins be? r=1 ; 1*2 = 2 h = 0.0625 ; 0.0625 * 2 = 0.125
since volume here is V = pi*r^2 *h there are 3 linear terms there, r, r, and h if a linear term is changed to 2 * original a cubic term will change by 2*2*2 * original
to see that a scale in a length , 2*L results in 2^3 * Volume here
a volume is a length cubed V -- L^3 = L*L*L double all of the lengths to 2L 2L * 2L * 2L = 8*L^3 = 8*V double lengths will give 8 times volume
confused..
In general, if you change the scale of lengths by some constant A AREA will change by A*A or A^2 of the original Volume will change by A*A*A or A^3 of the original
Volume is calculated with 3 Length Values V = length^3 = L*L*L if you change the scale of lengths L by that constant A, Volume will be V = (A*L) * (A*L) * (A*L)
so when you double all the lengths, the volume will increase by 2^3 or 8 times larger
can i do 39.25 * 8?
you can see this in this example by calculating the value let each value become double, then calculate the volume the same way, should see it is now 8x the last
\[\large \pi * (2*r)^2 * (2*h) = \pi*2^2 * r^2 + 2*h = (2^3)*\pi*r^2*h \]
the volume is pi*r^2*h this new volume is 2^3 *Volume
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