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DENIUS [597]
3 years ago
8

There are two vendors at the fair sleeping snow cones for the same price. If the containers are completely filled and then level

ed off across their tops, what is the difference in the amount of snow cone each will hold
Mathematics
1 answer:
aliya0001 [1]3 years ago
4 0

Answer:

d =  787.81cm^3

Step-by-step explanation:

Given

See attachment for the containers

Required

The difference in the amount of snow cone they hold

The amount they hold is determined by calculating the volume of the containers.

The traditional snow cone has the following dimension

Shape: Cone

r=4cm --- radius

h = 13cm --- height

The volume is calculated as:

Volume = \frac{1}{3} \pi r^2h

So, we have:

V_1 = \frac{1}{3} \pi * 4^2 * 13

V_1 = \frac{208}{3} \pi

The snow cone in a cup has the following dimension

Shape: Cylinder

r=8cm --- radius

h = 5cm --- height

The volume is calculated as:

Volume =  \pi r^2h

So, we have:

V_2 =  \pi *8^2 * 5

V_2 =  \pi *320

V_2 =  320\pi

The difference (d) in the amount they hold is:

d = V_2 - V_1

d = 320\pi - \frac{208\pi}{3}

Take LCM

d =  \frac{3*320\pi -208\pi}{3}

d =  \frac{960\pi -208\pi}{3}

d =  \frac{752\pi}{3}

Take \pi = 22/7

d =  \frac{752}{3} * \frac{22}{7}

d =  \frac{752 * 22}{3*7}

d =  \frac{16544}{21}

d =  787.81cm^3

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