Jack ran at 880 ft./min. for the first 10 min. of the 3 mi. race before slowing his pace to finish the race with a total time of 20 min. what was his pace (in feet per minute) for the second part of the race?

Jack ran 704 ft/min during the second half of the race.

Step-by-step explanation:

Here the distance traveled was 3 miles, and the total elapsed time was 20 minutes.  Recall that distance = rate * time, or rate = distance / time.

Find the pace (speed, rate) for the second part of the race.

Here, the total distance was 3 miles and the elapsed time 20 minutes.  Jack's average speed was thus 3/20 miles/minute.  Converting this to feet per minute, we get:

3 mi          5280 ft

* = 792 ft/min    (Jack's average speed)

20 min        1 mi

Recall that the formula for finding the average of x and y is (x + y) / 2.

(1/2)(880 ft/min  +  p) = 792 ft/min; find p, which represents Jack's pace for the second part of the race.

If (1/2)(880 ft/min  +  p) = 792 ft/min, and we mult. both sides by 2, we get:

880 ft/min + p = 1584 ft/min.

To solve this for p, subtract 880 ft/min from both sides, obtaining:

p = 704 ft/min

Jack ran 704 ft/min during the second half of the race.

c

step-by-step explanation:

c

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