The jones family drove to kankakee in 5 hours and returned in 4 hours. their rate on the return trip was 9 mph faster than their rate going. find their rate going?

36 mph

Step-by-step explanation:

Let

D be the distance covered in on trip

x be the rate on return

y be the going rate

if the Jones family drove to Kankakee in 5 hours and returned in 4 hours.

y = D/5 and x = D/4

D = 5y and D = 4x

5y = 4x Equation 1

Their rate on the return trip was 9 mph faster than their rate going

x = y + 9  Equation 2

substitute into equation 1

5y = 4(y + 9)

5y = 4y + 36

collect like terms

5y - 4y = 36

y = 36 mph

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

b, i believe, but not sure.