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lara [203]
4 years ago
12

A spherical balloon has a 32-in. diameter when it is fully inflated. Half of the air is let out of the balloon. Assume that the

balloon remains a sphere.
a. Find the volume of the fully-inflated balloon.
b. Find the volume of the half-inflated balloon.
c. What is the radius of the half-inflated balloon?
Mathematics
1 answer:
Anna11 [10]4 years ago
6 0

Answer:

Volume of Fully-inflated balloon: 17148.58 inches^3

Volume of half-inflated balloon: 8574.29 inches^3

Radius of half-inflated balloon: 12.7 inches

Step-by-step explanation:

Given

Diameter of inflated balloon=d=32 inches

As we have to use the radius of the balloon to calculate the volume then,

r=d/2

r=32/2

r=16 inches

Then balloon can be sphere when it has gas in it.  

Volume of fully-inflated balloon:

Vhf=4/3 πr^3

=4/3*3.14*(16)^3

=12.56/3*4096

=51445.76/3

=17148.58 inches^3

Volume of Half-Inflated balloon:

As there will be half of the gas left, so the volume of half-inflated balloon will be the half of fully-inflated balloons volume.

Vh=Vf/2

=17148.58/2

=8574.29 inches^3

Radius of Half-inflated balloon:

As we know the volume of half inflated balloon, we will use the formula for volume to calculate the radius.

Vh=4/3 πr^3

8574.29=4/3*3.14*r^3

8574.29*3/4*1/3.14=r^3

r^3=2047.999

∛r= ∛2047.999

So,

r=12.69 ≈12.7 inches

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F(x)=x^3+ax^2+bx+3 where a and b are constants. bx Given that when f (x) is divided by (x+2) the remainder is 7, (a) show that 2
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Part I - First synthetic division

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Part II - Second Synthetic Division

We draw another synthetic division table, this time with (x - 1), so the number on the left hand side will be +1

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             1      (A+1)      A+B+1       A+B+4 
                            
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Part III - Solving for A and B with our two simultaneous equations

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Learn more on midpoint of a line here: brainly.com/question/5566419

#SPJ1

3 0
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