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.

(x+3)(x+7)

step-by-step explanation:

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all