On your wedding day you leave for the church 34 minutes before the ceremony is to begin, which should be plenty of time since the church is only 12.0 miles away. on the way, however, you have to make an unanticipated stop for construction work on the road. as a result, your average speed for the first 17 minutes is only 5.0 mi/h. what average speed do you need for the rest of the trip to get you to the church on time?

Average speed   is 37.35 mi/h

Step-by-step explanation:

given data

leave =  34 minutes before

church distance =  12.0 miles

average speed first 17 minutes =  5.0 mi/h

solution

so we find Total distance travel in first 17 minutes = speed × time

Total distance travel in first 17 minutes = 5 ×

Total distance travel in first 17 minutes = 1.416 mi

and

Distance Remaining = 12 - 1.416 = 10.584 mi

Time Remaining = 34 - 17 min = 17 min

so

remaining distance Average speed  =

Average speed  =

Average speed   is 37.35 mi/h

google dat sis

step-by-step explanation:

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet