Kay has 28 coins which includes nickels and dimes if the total is 2.35$, how many of each coin does kay have?

19 dimes and 9 nickels

Step-by-step explanation:

Model the situation using two equations, then solve the system.

let n be number of nickels and d be number of dimes.

n + d = 28     <= This equation is about the number of coins.

0.05n + 0.10d =  2.35   <=This equation is about the worth in dollars.

Rearrange n + d = 28.

n = 28 - d

Now we can substitute this equation into the other one.

Sub n =28-d at the "d" variable for the equation 0.05n + 0.10d =  2.35

0.05n + 0.10d =  2.35

Now there is only one variable.

0.05(28 - d) + 0.10d =  2.35   <=distribute this

0.05(28 - d) + 0.10d =  2.35

1.4 - 0.05d + 0.10d = 2.35   <=combine like terms

1.4 + 0.05d = 2.35    <= begin to isolate d

0.05d = 2.35 - 1.4

0.05d = 0.95

d = 0.95/0.05

d = 19

Now that we know d=19, we can substitute it into any equation.

Choose the equation n + d = 28 because the numbers are easier.

n + d = 28

n + 19 = 28    <=isolate n

n = 28 - 19

n = 9

Since n=9 and d=19, Kay has 9 nickels and 19 dimes.

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead