Q: How many quarters n three and a half?

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Let the three consecutive integers be n, n + 1 and n + 2. Then (n + 1) + (n + 2) = n/2 + 33 2n + 3 = n/2 + 33 : 4n + 6 = n + 66 : 3n = 60 : n = 20 The three numbers are 20, 21 and 22.

Nickels = 6 Dimes = 2 Quarters = 9 (1/3N) + N + (3+N) = 17 2 1/3N = 14 N = 6

Let the number of nickels, dimes and quarters be n, d, q respectively. Then n +d + q = 30 5n + 10d + 25q = 550 But d = 2n, so: n + 2n + q = 30 => 3n + q = 30 5n + 10(2n) + 25q = 550 => 25n + 25q = 550 => n + q = 22 Which gives two simultaneous equations to solve, resulting in: n = 4, q = 18 So there are 18 quarters (plus 4 nickels and 8 dimes).

5n + 25q = 490; n = 2q so: 10q + 25q = 490 ie 35q = 490 so q = 14 and n must be 28. 14 x 25c = $3.50, 28 x 5c = $1.40, total $4.90. ...and there you have it!

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Q + N = 36;.25Q + .05N = 5.20;.25(36-N) +.05N = 5.20;9- .20N = 5.20;3.8 = .20N;N = 19 and Q = 17

The answer, according to the Wolfram Alpha site, is 0.288788095086602421278899721929230780088911904840685784114741...The input I used is:product(1-1/2^n) from n=1 to infinityIt isn't clear - in this particular page - how this result was obtained.

It is n/2 - 3

One third of three-quarters of a hundred is twenty five. 3/4 of 100 = 75 1/3 of 75 = 25

Call the number of nickels n and the number of quarters q. From the problem statement, q = n + 78 and 25q + 5n = 3240. Substituting the first equation into the second yields 25(n + 78) + 5n = 3240, or 30n = 3240 - 1950 or n = 1290/30 = 43 nickels.

Let the three consecutive integers be n, n + 1 and n + 2. Then (n + 1) + (n + 2) = n/2 + 33 2n + 3 = n/2 + 33 : 4n + 6 = n + 66 : 3n = 60 : n = 20 The three numbers are 20, 21 and 22.

Assuming nickles, dimes, and quarters, there are ten different ways to make change for a half dollar. Just enumerate the combinations... 10 n 8 n 1 d 6 n 2 d 4 n 3 d 2 n 4 d 5 d 5 n 1 q 3 n 1 d 1 q 1 n 2 d 1 q 2 q

Nickels = 6 Dimes = 2 Quarters = 9 (1/3N) + N + (3+N) = 17 2 1/3N = 14 N = 6

Let the number of nickels, dimes and quarters be n, d, q respectively. Then n +d + q = 30 5n + 10d + 25q = 550 But d = 2n, so: n + 2n + q = 30 => 3n + q = 30 5n + 10(2n) + 25q = 550 => 25n + 25q = 550 => n + q = 22 Which gives two simultaneous equations to solve, resulting in: n = 4, q = 18 So there are 18 quarters (plus 4 nickels and 8 dimes).

1/8 = (1/2)3 which is in the form (1/2)n where n is the number of half lives undergone. Therefore the substance has passed three half lives

Three: ' n '

The answer will depend on the value of n.