6. The time t required to drive a certain distance varies inversely with the speed r. If it takes 2 hours to drive the distance at 40 miles per hour, how long will it take to drive the same distance at 45 miles per hour?

A. 80 hours

B. 900 hours

C. about 1.78 hours

D. 2.25 hours

C. about 1.78 hours

Step-by-step explanation:

1.78 hours

Step-by-step explanation:

If t varies inversely with the speed, the formula for that is  t= \frac{k}{r} .  If t = 2 and r = 40, then we can solve for k, the constant of variation.  Multiply both sides by 40 to get  k = 80.  Now we need to find the value for t when r = 45.  Using the k value we just solved for, we will fill in the equation as  t= \frac{80}{45}   for a t value of about 1.78 hours

about 1.78 hours

t = k/r

2 = k/67

k = 134 (which happens to be the distance)

t = 134/r

t = 134/55

t = 2.436 hours

The formula for this is t = k/r, read "t varies inversely as r". If the info they give us is 4 hours and 35 miles per hour, we find that k = 140. Now use that 140 in another equation to solve for t: t = 140/45, and t = 3.1 hours

t = k/r\

4 hours and 35 miles per hour,

we find that k = 140.

Now use that 140 in another equation to solve for t:

t = 140/45, and

t = 3.1 hours

the answer is b. 60. each interior angle in 120, so the exterior angle must be 60 in order to add up to 180

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).