Some nickels and dimes amounting to 2.90. The number of dimes is 1 less than twice the number of nickels. How many coins of each kind?
i think i sorta have a format : d-1(2n)=2.90 argh!
2006-09-26 21:15:41 · 4 answers · asked by Tammy 1 in Education & Reference ➔ Homework Help
try this equation:
x# of coins (no values)
x=nickels
2x-1=number dimes
Value for nickels = .05(x)
Value for Dimes = .10(2x-1)
TOGETHER they equal 2.90
SO
.05(x) + [.10(2x-10] = 2.90 dollars
.05x + .20x-1 = 2.90
.25x-1 = 2.90
.25x=3.90
x = ?
I have to double check that. but I laid it out for you....
2006-09-26 23:21:05 · answer #1 · answered by Anonymous · 0⤊ 0⤋
You have an unknown number of nickels and an unknown number of dimes, and they add up to $2.90, so 0.05n + 0.10d = 2.90. The number of dimes is 1 less than twice the number of nickels, so d = 2n -1. Plug that into the first equation to be 0.05n + 0.2n - 0.10 = 2.90, which you can solve to give you n = 12. Plug in n = 12 to the second equation, and you get d = 23. Thus, you have 12 nickels and 23 dimes. Check it, and it adds up.
2006-09-26 21:26:14 · answer #2 · answered by drunksage 2 · 0⤊ 0⤋
Aw, hell! I can arrive in my mind at 12 nickels and 23 dimes, but now for the hard part...
xn = number nickels
yd = number dimes
So...
yd+1=2xn OR y.1 + 1=2x.05 ---> y.1 (1/2) = x.05
how did i get there? sorry i will work it out with pen and paper and update answer
yeah like the first person said :P
2006-09-26 21:34:21 · answer #3 · answered by quntmphys238 6 · 0⤊ 0⤋
try this formula...
.05x + .1(2x-1) = 2.9
x = number of nickles
i think that should work
sorry i don't feel like doing the math right now
2006-09-26 21:26:53 · answer #4 · answered by Anonymous · 0⤊ 0⤋