Nathan walked on an asphalt pathway for 12 miles. he walked the 12 miles back to his car on a gravel road through the forest. on the asphalt he walked 2 miles per hour faster than on the gravel. the walk on the gravel took one hour longer than the walk on the asphalt. how fast did he walk on the gravel?

4 miles/hour

Step-by-step explanation:

Distance walked on asphalt = Da = 12 miles

Distance walked on gravel = Dg= 12miles

time spent on asphalt = ta = ?

time spent on gravel = tg = ?

However, time spent on gravel was 1 hour longer than on asphalt

Therefore,

(tg)hrs - (ta)hrs = 1 hr

tg = ta + 1 ... Equation 1

Magnitude of velocity or speed

Vg = speed on gravel

Va = speed on asphalt

However, he walked 2 miles per hour faster on asphalt than on gravel

Therefore,

Va = Vg + 2 ... Equation 2

Note that all velocity values will carry the miles/hr unit

However,

Velocity = distance/time

Da/ta = (Dg/tg + 2) miles/hour

12/ta = [(12/tg) + 2] Equation 3

But tg = ta + 1 (from equation 1)

We substitute for tg in equation 3

12/ta = [(12/ta+1) +2] Equation 4

Taking the LCM of the the fraction, we have

12/ta = [12+2(ta+1)]/ta+1

Opening the inner bracket, we have

12/ta = (12+2ta+2)/ta+1

Cross multiplying, we have

12ta+2ta2+2ta = 12ta+12

We can eliminate 12ta as it appears on both sides

2ta2 + 2ta - 12= 0

This has become q quadratic equation, we factorize into;

(2ta - 4) (ta+3) = 0

2ta - 4 = 0

2ta = 4

ta= 4/2

ta = 2

Since tg = ta + 1

tg = 2+1

= 3hours

Vg = 12/tg

= 12/3 = 4 miles/hour

His speed on the gravel was 4miles/hour

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

5

step-by-step explanation: