NEED SOMEONE WITH BIG BRAINS
Imagine we have a piece of string 25,000 miles long, just long enough to exactly circle the globe at the equator. We take the string and fit it snugly around the equator, over oceans, deserts and jungles. When we have completed our task, we find that the string is just one yard too long. To overcome the error, we decide to tie the ends together and to distribute the extra 36 inches evenly over the entire 25,000 miles, figuring that it will never be noticed. How far do you think the string will stand off from the ground at each point, merely by virtue of the fact that it is 36 inches (3 feet) too long? Explain your answer.

just an extra one foot

Step-by-step explanation:

Given that the length of the piece of string = 25,000 miles

 Since the string is long enough to exactly circle the globe at the equator over oceans, deserts and jungles. We can say that the string could completely circle a sphere

Assuming that the earth is a perfect sphere.

The circumference of the circle = 25,000 mile.

But note that the circumference of a circle is directly proportional to the radius. If you double the radius, the circumference will also be doubled. Half the circumference, half the radius. Increase the circumference by 30%, radius will also be increased by 30% and so on.

In this question, the radius is being increased by 3 feet, so we need to know that as a fraction of the original radius.

Circumference = 2πr

Let's also convert miles to feet

25000 × 5280 = 2(3.143)r

r = 132 × 10^6/2π

r = 21008452.5 feet

3 feet - that's how far off the ground we're lifting the string out of 21008452.5 feet is close to one part in 25000 miles. The string was 132000000 feet long to start with, so we need just an extra 1 foot.

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).

option a is correct.

step-by-step explanation:

scale factor: it is used to find proportional measurements.

proportions means having the same ratio.

scale factor is the ratio of the model measurement to the actual measurement in a simplest form.

or we can say that the ratio of any two corresponding lengths in two similar geometric figures

if the scale factor is more than 1, then it is an enlargement,

if the scale factor less than 1, then it is a reduction.

given the two geometric figure:

by definition of proportions, we have