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the complete answer in the attached file (response greater than 5000 characters))

the complete question below with their respective answers

For reasonable distances, a certain jogger can maintain an average speed of 6 miles per hour while running on level ground. The jogger decides to go to a local park and use one of the paths there for a workout routine one day each week. This path is a gently sloping one that winds its way to the top of a hill.

1.The jogger can run at an average speed of 5.5 miles per hour up the slope and 6.5 miles per hour going down the slope. The jogger decides to cover 2 miles by going uphill for 1 mile and then returning 1 mile back down the hill.

a.How long does it take the jogger to run 1 mile uphill?

Let

r> average speed

d > distance in miles

t > is the time in hours

we know that

r=d/t > t=d/r

for r=5.5 mph and d=1 mile

t=1/5.5 hours

the answer Part 1a) is (1/5.5) hours

b.How long does it take the jogger to run 1 mile downhill?

we know that

r=d/t > t=d/r

for r=6.5 mph and d=1 mile

t=1/6.5 hours

the answer Part 1b) is (1/6.5) hours

c.Use your answers to a and b to determine how long, in hours, the full trip will take (1 mile uphill and 1 mile downhill). Give an exact answer expressed as a fraction in simplest terms and then give a decimal approximation correct to three decimal places.

Reduced Fraction: Decimal Approximation:

(1/5.5)+(1/6.5)=(6.5+5.5)/(5.5*6.5)=12/35.75 >multiply by 100 both members 1200/3575

Divide by 25 both members

48/143=0.336 hours

The answer part 1c) is

Reduced Fraction: 48/143 hours

Decimal Approximation: 0.336 hours

2.On a steeper slope, the jogger can run at an average speed of 5.3 miles per hour up the slope and 6.7 miles per hour going down the slope.

a.What is the jogger’s uphill time in hours?

Let

r> average speed

d > distance in miles

t > is the time in hours

we know that

r=d/t > t=d/r

for r=5.3 mph and d=1 mile

t=1/5.3 hours

the answer Part 2a) is (1/5.3) hours

b.What is the jogger’s downhill time in hours?

Let

r> average speed

d > distance in miles

t > is the time in hours

we know that

r=d/t > t=d/r

for r=6.7 mph and d=1 mile

t=1/6.7 hours

the answer Part 2b) is (1/6.7) hours

c.Use your answers to a and b to determine how long, in hours, the full trip will take (1 mile uphill and 1 mile downhill). Give an exact answer expressed as a fraction in simplest terms and then give a decimal approximation correct to three decimal places.

Reduced Fraction: Decimal Approximation:

(1/5.3)+(1/6.7)=(6.7+5.3)/(5.3*6.7)=12/35.51 >multiply by 100 both members 1200/3551=0.338 hours

The answer part 2c) is

Reduced Fraction: 1200/3551 hours

Decimal Approximation: 0.338 hours

3.Now let’s generalize.

a.Running uphill the jogger runs c mph slower than 6 mph. Write an

expression representing a speed of c mph slower than 6 mph.

we know that

r=(6-c) mph

the answer Part 3a) is r=(6-c) mph

b.Write an algebraic expression that represents the time it takes to

run 1 mile uphill at a speed that is c mph slower than 6 mph.

we know that

r=d/t > t=d/r

for r=(6-c) mph and d=1 mile

t=1/(6-c) hours

the answer Part 3b) is t=1/(6-c) hours

c.Running downhill the jogger runs c mph faster than 6 mph. Write an

expression represents a speed of c mph faster than 6 mph.

we know that

r=(6+c) mph

the answer Part 3c) is r=(6+c) mph

d.Write an algebraic expression that represents the time it takes to

run 1 mile downhill at a speed that is c mph faster than 6 mph.

we know that

r=d/t > t=d/r

for r=(6+c) mph and d=1 mile

t=1/(6+c) hours

the answer Part 3d) is t=1/(6+c) hours

e.Use your answers to b and d to write an algebraic expression for the total time, in hours, that it takes the jogger to cover 2 miles by going uphill for 1 mile and then returning 1 mile back down the hill.

(1/(6-c))+(1/(6+c))

The answer part 3e) is (1/(6-c))+(1/(6+c))

f.Simplify your answer to part e into a single algebraic fraction. (Remember to find a common denominator first.)

(1/(6-c))+(1/(6+c))=((6+c)+(6-c))/(6*6-c*c)=12/(36-c^2)

The answer part 3f) is 12/(36-c^2) hours

g.What is the value of c for #1? (How much slower does she run uphill?)

Use this c to test your answer to part f. (Check that it gives the correct answer for #1c.)

we know that

(6-c)=5.5

c=6-5.5

c=0.5 mph

substitute the value of c in [12/(36-c^2) ]

12/(36-c^2)

12/(36-0.5^2)

12/(36-0.25)

12/35.75=48/143 hours > is correct

h.What is the value of c for #2? (How much slower does she run uphill? )

Use this c to test your answer to part f. (Check that it gives the correct answer for #2.)

we know that

(6-c)=5.3

c=6-5.3

c=0.7 mph

substitute the value of c in [12/(36-c^2) ]

12/(36-c^2)

12/(36-0.7^2)

12/(36-0.49)

12/35.51=1200/3551 hours > is correct