Sarah has a collection of nickels, dimes, and quarters worth 15.75. she has 10 more didmes than nickels and twice as many quarters as dimes. how many coins of each kind does she have?

Let n, d, q represent the numbers of nickels, dimes, and quarters Sarah has. The problem statement gives rise to three equations:

... 5n +10d +25q = 1575 . . . . . . the collection's value in cents

... n + 10 = d . . . . . . . . . . . . . . . . the number of dimes is 10 more than the number of nickels

... q = 2d . . . . . . . . . . . . . . . . . . . she has twice as many quarters as dimes

Solving the second equation for n and substituting into the first equation, we have ...

... 5(d -10) + 10d + 25(2d) = 1575

... 65d -50 = 1575

... 65d = 1625

... d = 1625/65 = 25

... n = d -10 = 15

... q = 2d = 50

Sarah has 15 nickels, 25 dimes, and 50 quarters.

huh? im pretty sure it is in 199

step-by-step explanation:

cut 138 centimeters off

step-by-step explanation:

250-112 = 138