I WILL GIVE BRAINLIST IF ALL ANSWERED CORRECTLY Injured runners train on a special track at a rehabilitation center. The track is a square with a half circle on its left and right sides. The area of the square is 128 square feet. What is the length of the track? Use the table to help you answer the questions.

Square 11.02 11.12 11.22 11.32 11.42 11.52
Value 121.0 123.2 125.4 127.7 130.0 132.3
Square 11.62 11.72 11.82 11.92 12.02
Value 134.6 136.9 139.2 141.6 144.0

1. Fill in the blanks to complete the description of the track. (2 points)
The track has sides of the square and the distance around complete circle(s).

2. The length of one side of the square is the square root of its area. Use the table to find the approximate length of one side of the square. Explain how you used the table to find this information. (2 points)

3. Use your answers above to find the total length of the part of the track that is made up of sides of the square. (2 points)

4. The circumference of (distance around) a circle is π times the diameter, or C = πd. A side of the square is the diameter of each half circle.
In your answer to question 1, you gave the number of complete circles included in the track. Use this answer and C = πd to find the approximate length of the circular part of the track to the nearest tenth of a foot. Use 3.14 for π and show your work. (2 points)

5. Find the approximate length of the track, including the straight and circular sections. (2 points)

translate k to n and reflect across the line containing jk.

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).

25% is the answer