Answer:
15
Step-by-step explanation:
Let n, d, q represent the numbers of nickels, dimes, and quarters. The problem statement tells us ...
n +d +q = 37
n = d +4
q = n +2
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Rearranging the second equation gives ...
d = n -4
Substituting that into the first, we get ...
n + (n -4) +q = 37
2n +q = 41 . . . . . . . add 4 and simplify
Rearranging the third original equation gives ...
n = q -2
Substituting into the equation we just made, we get ...
2(q -2) +q = 41
3q = 45 . . . . . . . . add 4 and simplify
q = 15 . . . . . . . . . divide by 3
Joe has 15 quarters.
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<em>Check</em>
The number of nickels is 2 fewer, so is 13. The number of dimes is 4 fewer than that, so is 9. The total number of coins is 15 + 13 + 9 = 37, as required.
The answer is 168 logs.
1st row: 15 logs
2nd row: 18 logs
3rd row: 21 logs
(so, each row has 3 more logs)
4th row: 24 logs
5th row: 27 logs
6th row: 30 logs
7th row: 33 logs
The sum of logs from each row is: 15 + 18 + 21 + 24 + 27 + 30 + 33 = 168 logs
-7.14 should be your answer!
Answer:
3003 units^3.
Step-by-step explanation:
Just multiply the dimensions :
11 * 13 * 21
= 143 * 21
= 3003 units^3.
