18. The Tennis for Champs company is looking into new ways to package tennis balls. The packaging engineer at the company is exploring options for vacuum packing the balls (removing the air from the containers in which they are packaged) so they will retain good bounce. He wonders how much air there is in a standard container of tennis balls. Find the amount of empty space in a cylindrical container that is 18 cm tall and contains three tennis balls 6 cm in diameter. Use 3.14 for pi.

Amount of empty space in the cylinder = 395.64 cm³

Step-by-step explanation:

Since the cylinder contains 3 tennis balls each measuring 6 cm in diameter, we can say that the diameter of the cylinder is 6 cm because the balls are on top of each other. Since we have 18 cm is the height of the cylinder,then;

Formula for volume of a cylinder is;

V = πr²h

We are to use π = 3.14

Thus, V_cylinder = 3.14 × (6/2)² × 18

V_cylinder = 508.68 cm³

Volume of a tennis ball is;

V_tennis ball = (4/3)πr³

V_tennis ball = (4/3) × 3.14 × (6/2)³

V_tennis ball = 37.68 cm³

Thus, volume of 3 tennis balls = 3 × 37.68 = 113.04 cm³

Amount of empty space in the cylinder = V_cylinder - V_3 tennnis ball = 508.68 - 113.04 = 395.64 cm³

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!