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

i am currently in the exact same situation rn. algebra2/adv algebra is killing

step-by-step explanation:

answer: a and c are correct

step-by-step explanation:

the +5 is the vertical translation

the added negative symbol in front is the reflection over the y-axis