Justin recently drove to visit his parents who live 96 miles away. On his way there his average speed was 20 miles per hour faster than on his way home (he ran into some bad weather). If Justin spent a total of 4 hours driving, find the two rates (in mph). Round your answer to two decimal places, if needed. Justin's average speed to his parents' house: mph

Justin's average speed from his parents' house: mph

Justin's average speed to his parents' house: x= 40mph

Justin's average speed from his parents' house: x+20=60 mph

Step-by-step explanation:

Let the average speed be x +20  

Then average speed from his parents' x

And the distance is S=96

Then we know :

Justin's average speed to his parents' house:  x= 40mph

Justin's average speed from his parents' house:  x+20=60 mph

Outward bound::

rate = x+9 ; dist = 100 m ; time = 100/(x+9) hrs

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Homeward

rate = x mph ; dist = 100 m ; time = 100/x hrs

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Equation:

time + time = 5 hrs

100/(x+9) + 100/x = 5 hrs

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100x + 100(x+9) = 5x(x+9)

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200x + 900 = 5x^2 + 45x

5x^2 - 155x - 900 = 0

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x^2 - 31x - 180 = 0

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x = 36 mph

x+9 = 45 mph

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so x=40km

and average speed=x+20=60mph

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).