Kevin and randy muise have a jar containing 69 coins, all of which are either quarters or nickels. the total value of the coins in the jar is $nbsp 9.05. how many of each type of coin do they have?

Let us assume number of quarters = q.

Let us assume number of nickels = n.

Total number of coins = 69.

We can setup first eqation,

Number of quarters + number of nickels = 69.

q + n = 69 equation(1).

Total value of all the coins = $9.05

We can setup second equation as,

0.25*number of quarters + 0.05*number of nickels  - $9.05.

0.25q + 0.05n =9.05   equation(2).

We got two equations of a system of equation.

Let us apply substitution mehod of solving system of linear equations.

Solving first equation q+n=69 for n.

Adding q on both sides of the equation, we get

q-q+n = 69-q.

n= 69-q.

We need substitute n=69-q in second equation 0.25q + 0.05n =9.05.

0.25q +0.05(69-q) = 9.05

Distributing 0.05 over (69-q).

0.25q +0.05*69 -0.05*q =9.05.

0.25q +3.45 -0.05q =9.05

Combining like terms 0.025q -0.05q, we get

0.20q +3.45 =9.05

Subtracting 3.45 from both sides.

0.20q +3.45-3.45 =9.05-3.45

0.20q = 5.6

Dividing both sides by 0.20

0.20q/0.20 = 5.6/0.20

q = 28.

Plugging q=28 in equation(1)

q+n =69    => 28 +n =69

Subtracting 28 from both sides, we get

28-28 +n =69-28.

n = 41.

Therefore, number of quarters 28 and number of nickels 41.

step-by-step explanation: i dont

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation: