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valentinak56 [21]
3 years ago
11

Each cone of the hourglass has a height of 12 millimeters. The total height of the sand within the top portion of the hourglass

is 47 millimeters. The radius of both the cylinder and cone is 4 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 dripped to the bottom of the hourglass?
6.4
62.4
8.5
56.0
Mathematics
1 answer:
kolbaska11 [484]3 years ago
8 0

The solution would be like this for this specific problem:

Volume of a cylinder = pi * r^2 * h 

Volume of a cone = 1/3 * pi * r^2 * h 

Total Height = 47

Height of the cone = 12

Height of the cylinder = 35

If the top half is filled with sand, then:

volume (sand) = pi * 4^2 * 36

volume (cone) =  1/3 * pi * 4^2 * 12

Total volume = 1960.353816 cubic millimeters

353816 / (10 * pi) = 62.4 seconds.

It will take 62.4 seconds until all of the sand has dripped to the bottom of the hourglass.

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Answer:

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Step-by-step explanation:

The ratio of Milky Way bars to 3-Musketeers is 3:4 if there are 48 3-muskateers 48/4=12 so 3x12 is 36.

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Help PLZZ
Alexus [3.1K]
<h3 /><h3>\huge \rm༆ Answer ༄</h3>

To find slope we use the formula ~

  • \sf \dfrac{y2 - y1}{x2 - x1}

[ Ratio of the difference of y - coordinates of the points and the difference of x - coordinates of the points ]

let's find the slope of line passing through the given points ~

<h3>1. (10 , -9) and (-4 , -14) </h3>

  • \sf \dfrac{ - 9 - ( - 14)}{10 - ( - 4)}

  • \sf \dfrac{ - 9 + 14}{10 + 4}

  • \sf \dfrac{5}{14}
<h3>2. (16 , 6) and (-4 , -14)</h3>

  • \sf \dfrac{6 - ( - 14)}{16 - ( - 4)}

  • \sf \dfrac{6 + 14}{16 + 4}

  • \sf \dfrac{20}{20}

  • \sf1
<h3>3. (9 , -4) and (5 , -20)</h3>

  • \sf \dfrac{ - 4 - ( - 20)}{9 - 5}

  • \sf \dfrac{ - 4 + 20}{4}

  • \sf \dfrac{16}{4}

  • \sf4

<h3>4. (-13 , 20) and (-18 , 20)</h3>

  • \sf \dfrac{20 - 20}{ - 13 - ( - 18)}

  • \sf\dfrac{0}{ - 13 + 18}

  • 0

I hope it helps ~

___________________________________

꧁ \: \: \large \rm{kaul} \: ꧂

3 0
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