Natalie has some nickels, Dirk has some dimes, and Quincy has some quarters. Dirk has five more dimes than Quincy has quarters. If Natalie gives Dirk a nickel, Dirk gives Quincy a dime, and Quincy gives Natalie a quarter, they will all have the same amount of money. How many coins did each have originally?
2006-11-14 11:49:13 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Show your work and use only one variable
2006-11-14 12:02:14 · update #1
D = Q + 5
(N-1)*5 + 25 = (D-1)*10 + 5 = (Q-1)*25 + 10
Take the first equation and the second two of the triple below it:
(Q + 5 - 1)*10 + 5 = (Q - 1)*25 + 10
10Q + 45 = 25Q - 15
15Q = 60
Q = 4
D = 9
(N-1)*5 + 25 = (D-1)*10 + 5
5N + 20 = 85
5N = 65
N = 13
2006-11-14 12:05:14 · answer #1 · answered by Dave 6 · 0⤊ 0⤋
For increasing cubic binomials the ordinary formulation is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^one million + 3*a^one million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^one million + 3*x^one million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D
2016-12-17 10:11:13 · answer #2 · answered by nichelle 3 · 0⤊ 0⤋
4 dimes, 3 nickels, and 2 quarters.
Need to know how to solve it?
2006-11-14 11:59:11 · answer #3 · answered by Anonymous · 0⤊ 1⤋
we use variables
N=0.05
D=0.10
Q=0.25
Dirk: 5*x D
Quincy: x Q
Natalie: yN -1N
Dirk: 5xD +1N
Dirk: 5xD +1N -1D
Quincy: xQ +1D
Quincy: xQ +1D - 1Q
Nancy: yN -1N +1Q
Nancy = Dirk = Quincy (use the last equation for each)
yN-1N+1Q=5xD+1N-1D=xQ+1D-1Q <- solve for the knowns
(y-1)*0.05+0.25=(5x-1)*0.25+0.05=(x-1)*0.25+0.05
First solve the x's since there are two sides with x's
0.25(5x-1)+0.05=0.25(x-1)+0.05
once you know x subsitute into other equation
0.05(y-1)+0.25=0.25(5x-1)+0.05 solve for y
substitue in and decode.
Have fun!
2006-11-14 12:01:31 · answer #4 · answered by unlv_engineer 2 · 0⤊ 0⤋