A box contains $7.45 in nickels, dimes, and quarters. There are 43 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coins of each kind are there?

Number of nickels = 6

Number of dimes = 14

Number of quarters = 23

Step-by-step explanation:

Total number of coins = 43

Let number of nickels = n

Let number of dimes = d

Let number of quarters = q

Money in n nickels =

Money in d dimes =

Money in q quarters =

Total money = $7.45

The sum of the numbers of nickels and dimes is 3 less than the number of quarters.

i.e.

Let us put the value  in (1):

Putting the value of q in (1) and (2):

n+d = 20 .... (3)

(4) - (3):

d = 14

Now, from (3):

n = 6

Number of nickels = 6

Number of dimes = 14

Number of quarters = 23

1. credits

2. dual - aspect

hope i : )

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."