Sam and liam raced each other up and down the hill. Sam's avrage speed up hill was 1 mph, and his avrage speed down hill is 9 mph. Liam ran up and down the hill with the same speed 2 mph. If the path from the bottom of the hill is 1 mile long, how much time did it take the boys to finish-and who is the winner?

a. i. 1.11 h ii. 1 h. b. Liam

Step-by-step explanation:

a.

i. Sam's time

We know distance d = vt where v = speed and t = time.

Since our distance here is the length of the path, d = 1 mile.

For Sam, his speed up the hill is v = 1 mph, so, the time it takes him to go uphill is t = d/v = 1 mi/1 mph = 1 h.

He goes down hill at a speed of v = 9 mph, so the time it takes him to go down hill is t = d/v = 1 mi/9 mph = 1/9 h = 0.11 h

So, the total time it takes him to complete the trip is t = 1 h + 0.11 h = 1.11 h

ii. Liam's time

For Liam, his speed both uphill and downhill is v = 2 mph, so, the time it takes to cover each trip is t = d/v = 1 mi/2 mph = 0.5 h.

So for both uphill and downhill trip, the total time is t' = 2 × 0.5 h = 1 h

b. Since Liam's time t' = 1 h < Sam's time t = 1.11 h, that is t' = 1h < t = 1.11 h, Liam is the winner.

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752