Hi can you me ?question 4 and 5. the second photo shows that you need to do this task in the table. ❤)

It is often easy to work these problems if you can arrange the coins in equal groups. In cases where there are a few more or less of one than another, you can add or subtract the appropriate number of coins to make it come out even.

4. Subtracting 3 quarters ($0.75) Rudolph will have $6.50 and an equal number of nickels and quarters with twice as many dimes. That is, Rudolph can group his coins into groups consisting of 1 quarter, 1 nickel, and 2 dimes, for a total of $.50 per group. 6.50/0.50 = 13 groups will make up the revised total.

Rudolph has 13 nickels, 26 dimes, and 16 quarters.

5. Adding 2 dimes to the total will allow Andrea to group her coins into groups that have 1 quarter, 4 nickels, and 4 dimes. Each group is worth $0.85, and her revised total will be $2.55. That is, she has 2.55/0.85 = 3 groups of coins.

Andrea has 3 quarters, 12 nickels, and 10 dimes.

f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1