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Whitepunk [10]
3 years ago
13

You have an ice cream scoop with a 2-inch diameter. You have an ice cream cone with a 2-inch diameter and a height of 5 inches.

If you place one scoop of ice cream on the cone and let the ice cream melt, will it spill over the cone?
Mathematics
1 answer:
IRISSAK [1]3 years ago
4 0

Answer:

Since the volume of the cone is bigger than the volume of the scoop, it won't spill.

Step-by-step explanation:

In order to calculate if the ice cream will spill we will need to calculate the volume on the ice cream scoop and the volume of the cone. If the volume of the scoop is bigger than the cone it will spill.

volume scoop = (4/3)*pi*r³ = (4/3)*pi*(2/2)³ = 4.1888 in³

volume cone = (pi*r²*h)/3 = [pi*(2/2)²*5]/3 = 5.236 in³

Since the volume of the cone is bigger than the volume of the scoop, it won't spill.

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1/2r - 3 = 3(4-3/2) -----> Simplify the right side.
1/2r - 3 = 3(1/2) -----> Multiply
1/2r - 3 = 3/2 ----> Add 3 to both sides
1/2r = 6/2 + 3/2
1/2r = 9/2 -----> Divide both sides by 1/2 which is the same as multiplying my 2
r = 18/2
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8 0
3 years ago
Approximate the change in the volume of a sphere when its radius changes from r = 30 ft to r=10.1 ft [v(r)=4/3Ï€r^3]. When r cha
ella [17]

Answer: ∆V for r = 10.1 to 10ft

∆V = 40πft^3 = 125.7ft^3

Approximate the change in the volume of a sphere When r changes from 10 ft to 10.1 ft, ΔV=_________

[v(r)=4/3Ï€r^3].

Step-by-step explanation:

Volume of a sphere is given by;

V = 4/3πr^3

Where r is the radius.

Change in Volume with respect to change in radius of a sphere is given by;

dV/dr = 4πr^2

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V'(10) = 400π

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Therefore change in Volume from r = 10 to 10.1 is

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Of by direct substitution

∆V = 4/3π(R^3 - r^3)

Where R = 10.1ft and r = 10ft

∆V = 4/3π(10.1^3 - 10^3)

∆V = 40.4π ~= 40πft^3

And for R = 30ft to r = 10.1ft

∆V = 4/3π(30^3 - 10.1^3)

∆V = 34626.3πft^3

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