find three consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third (o
1 answer:
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So hmmm there are 5pennies in a nickel, and 10pennies in a dime and 25pennies in a quarter
now, in $6.10, there are 610 pennies
n = amount of nickels coins
d = amount of dime coins
q = amount of quarter coins
so... we know, there are 5*n or 5n pennies in the nickels, and 10*d or 10d pennies in the dimes and 25*q or 25q in the quarter coins, and we know their sum is 610 pennies total, thus
5n + 10d + 25q = 610
now, there are 4 more "n" then "d", so whatever "d" is, "n" is 4 more than that
n = d + 4
and twice as many "q" than "n", so, whatever "n" is, then
q = 2n
solve for "d", to see how many dimes are there
what about the nickels? well, n = d + 4
what about the quarters? well, q = 2d + 8
now, in $6.10, there are 610 pennies
n = amount of nickels coins
d = amount of dime coins
q = amount of quarter coins
so... we know, there are 5*n or 5n pennies in the nickels, and 10*d or 10d pennies in the dimes and 25*q or 25q in the quarter coins, and we know their sum is 610 pennies total, thus
5n + 10d + 25q = 610
now, there are 4 more "n" then "d", so whatever "d" is, "n" is 4 more than that
n = d + 4
and twice as many "q" than "n", so, whatever "n" is, then
q = 2n
solve for "d", to see how many dimes are there
what about the nickels? well, n = d + 4
what about the quarters? well, q = 2d + 8
Answer:
$7897.50
Step-by-step explanation:
using both compound interest and simple interest, they will need to pay $7897.50 by the end of a year
