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Mathematics 8 Online
OpenStudy (anonymous):

An hourglass consists of two sets of congruent composite figures on either end. Each composite figure is made up of a cone and a cylinder, as shown below: Each cone of the hourglass has a height of 18 millimeters. The total height of the sand within the top portion of the hourglass is 54 millimeters. The radius of both cylinder and cone is 8 millimeters. Sand drips from the top of the hourglass to the bottom at a rate of 10π cubic millimeters per second. How many seconds will it take until all of the sand has

OpenStudy (anonymous):

68.3 38.4 268.8 230.4

OpenStudy (anonymous):

I think its 268.8

OpenStudy (anonymous):

@Marcorie @misssunshinexxoxo @jdoe0001 @dan815

OpenStudy (anonymous):

is their a pic?

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

OpenStudy (anonymous):

@mathmate can you please help him

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

sorry im not really sure try asking someone else i tried it and got the wrong answer

OpenStudy (anonymous):

Thats okay :)

OpenStudy (anonymous):

@Anon.

OpenStudy (anonymous):

@slade

OpenStudy (anonymous):

i agree with u @CaptainLlama49

OpenStudy (anonymous):

Cool I just might have it right

OpenStudy (astrophysics):

Well you have the volume of the cylinder which is \[V = \pi r^2 h\] and cone is \[V_c = 1/3 \pi r^2h\] we are given that the height of the hourglass is 18 mm, and the total height within the top portion of the hour glass is 54 mm. It also tells us the radius of the cylinder and the cone is 8mm. So we just need to find the total volume, using the formulas above and once we have that we can just divide by \[10 \pi~ mm^3\] to figure out the time it will take the sand to be at the bottom I assume?

OpenStudy (anonymous):

yes

OpenStudy (astrophysics):

Yeah, so you should be able to figure out the rest

OpenStudy (anonymous):

I got 268.8

OpenStudy (anonymous):

Is that correct

OpenStudy (astrophysics):

Well lets find the volume of the cone and cylinder first, what did you get?

OpenStudy (anonymous):

268.8

OpenStudy (anonymous):

@Astrophysics

OpenStudy (astrophysics):

Well I don't really feel like doing the whole problem, I gave you the process you can plug in the numbers and check.

OpenStudy (anonymous):

100% YES!

OpenStudy (astrophysics):

Haha, so it was right? :)

OpenStudy (anonymous):

Yep

OpenStudy (astrophysics):

Nice work

OpenStudy (anonymous):

Thank you :)

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