Add: (g2 - 494 + 5g + 9) + (-393 +352 - 6) Complete the steps to find the sum. What is the sum?
O -794 + 493 – 3g+ + 5g - 3
1. Rewrite terms that are subtracted as addition of the 0-494-393 + 4g2 + 5g + 3
opposite.
g2 + (494) + 5g +9+ (-393) + 3y2 + (-6)
0 494 + 4g2 + 148 - 6
0 -394 + 149 - 6
2. Group like terms.
3. Combine like terms.
4. Write the resulting polynomial in standard form.

15 quarters
10 dimes
5 nickels

five more dimes than nickels means we know there is at least 5 dimes, or 50 cents
if we subtract this from $5, we have $4.50.
With those 5 dimes gone, the number of dimes equals the number of nickels. If there are 3 times as many quarters as nickels, there are also 3 times as many quarters as dimes. So for every 3 quarters (0.75), there is one dime (0.10) and one nickel (0.05).
0.75 + 0.10 + 0.05 = 0.90
you find the number of remaining coins by dividing the total minus the 5 dimes (4.50) by 0.90.
4.50 ÷ 0.90 = 5
this is how many times 3 quarters, one dime, and one nickel go into 4.50
so, the answer to how many of each coin is:
3 quarters × 5 = 15 quarters
one dime × 5 = 5 plus the original 5 we took out = 10 dimes
one nickel × 5 = 5 nickels

to check
5.00 = 15(0.25) + 10(0.10) + 5(0.05)
5.00 = 3.75 + 1.00 + 0.25
5.00 = 5.00 ✔