Andrea made a replica of the Empire State Building. The surface area of the model is 45 feet2. What is the surface area of the actual Empire State Building in ft.2 if the scale factor is 44?
Enter your answer as a whole number, with the appropriate unit, like this: 42 ft.^2

87,120 ft²

1. Let's review the information given to us to answer the question correctly:

Surface area of the Empire State building replica = 45 square feet

Scale factor = 44

2. What is the surface area of the actual Empire State Building in square feet?

For answering the question, we will make the following calculations:

45 square feet = 9 feet * 5 feet or,

45 square feet = 10 feet * 4.5 feet, or,

45 square feet = 20 feet * 2.25 feet.

Now, let's apply the scale factor to all of these probable lengths of the sides of the surface area of the replica, as follows:

(9 *44) * (5 * 44) = 396 * 220 = 87,120 ft²

(10 * 44) * (4.5 * 44) = 440 * 198 = 87,120 ft²

(20 * 44) * (2.25 * 44) = 880 * 99 = 87,120 ft²

As we proved, no matter what are the lengths of the sides of the surface area of the replica. after applying the scale factor, the result is the same: 87,120 ft²

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).