shape of the pond is shaped somewhat like a triangle. He claims that he can approximate the perimeter of the pond without doing any measurement of the actual pond or on the diagram. What assumption would Paul need to make in order to justify his claim? B. Based on the assumption in part A, determine the approximate length of side c of the pond. Explain your reasoning in at least two sentences. C. Determine the approximate perimeter of the pond. Explain your reasoning

 A) the curves are ellipses

 B) 161 yards

 C) 366 yards

Step-by-step explanation:

A) The general shape of the side marked c has the appearance of two quarter-ellipses. The major axis is approximately 125 yards, and the minor axis is approximately 80 yards. Paul could assume that the curves are quarter-ellipses.

B) For an ellipse with this eccentricity, its perimeter is within 2% of that of a circle with a diameter equal to the average of these dimensions.*

So, the approximate length of side c is ...

c ≈ π(80 +125)/4 ≈ 161 . . . . yards; length of side c  

C) The perimeter of the pond is the sum of the side lengths, so will be ...  

80 yds + 125 yds + 161 yds = 366 yds . . . pond perimeter  

Indian mathematician Ramanujan developed an approximate formula for the circumference of an ellipse, which would otherwise need to be computed using an elliptic integral. it is ...

C = π(a+b)(1 + 3λ²/(10 +√(4 -3λ²))    where λ = (a -b)/(a +b) and a, b are the semi-axis lengths.

For this ellipse, λ = 9/41, so the last factor is about 1.01208313. Using this factor, the circumference is expected to be good to about 7 significant digits. Continuing the calculation, we find  C ≈ 325.904 . . . . circumference of an ellipse 125 yd long by 80 yd wide

This is twice the length we imagine for c, so c ≈ 163 yards. As we said, the approximation used above is within 2% of this value.

The attachment shows the shape if the curve is actually an ellipse.

no he doesnt because "-16' is negatiive

step-by-step explanation:

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)

=====================================

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

=====================================

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).