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fgiga [73]
3 years ago
5

Kevin and Randy Muise have a jar containing 78 coins, all of which are either quarters or nickels. The total value of the coins

in the jar is $13.90. How many of each type of coin do they have?
Mathematics
1 answer:
gregori [183]3 years ago
5 0
Answer:

50 quarters and 28 nickels

Step-by-step solution:

We have two variables we are solving for. We will call them Q for the number of quarters and N for the number of nickels.

We know that the total number of coins is equal to 78, and we know that the total amount the coins are worth is equal to $13.90. With this information, we can set up a system of equations:

Q + N = 78

0.25Q + 0.05N = 13.90

Now we can solve for either Q or N in the first equation and solve by substitution by plugging the value into the second equation. Let's solve for Q:

Q = 78 - N

Now plug in 78-N for Q:

0.25(78-N) + 0.05N = 13.90

Distribute and combine like terms:

19.50 - 0.25N + 0.05N = 13.90

19.50 - 0.20N = 13.90

-0.20N = -5.60

N = 28

Now we know that there are 28 nickels. We can plug this into one of the original equations to solve for Q:

Q + 28 = 78

Q = 50

There are 50 quarters. We can verify our answer by plugging the values for A and N into the other equation (optional step to check work):

0.25(50) + 0.05(28) = 13.90

12.5 + 1.40 = 13.9

13.9 = 13.9

Our answer is correct.





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Step-by-step explanation:

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The perimeter of a triangle is 41 inches. The three sides of the triangle are all differnt lenghts. The longest side is 10 less
erma4kov [3.2K]

Answer:

Longest\ side=20\ inches\\\\Middle\ side=15\ inches\\\\Shortest\ side=6\ inches

Step-by-step explanation:

Let be "x" the lenght of the longest side, "y" the lenght of the middle side and "z" the lenght of the shortest side.

The perimeter is:

41=x+y+z       [Equation 1]

We know that the longest side is 10 less than twice the middle side length. This can be expressed as:

x=2y-10          [Equation 2]

And the middle side length is three less than three times the shortest side length. This is:

y=3z-3         [Equation 3]

The steps are:

- Solve for "z" from the third equation:

y=3z-3\\\\y+3=3z

z=\frac{y+3}{3}    [Equation 4]

- Substitute [Equation 4] into [Equation 1]

- Substitute [Equation 2] into [Equation 1]

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41=(2y-10)+y+(\frac{y+3}{3})

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41=2y-10+y+\frac{y+3}{3}\\\\41=2y-10+y+\frac{y}{3}+1\\\\41=\frac{10}{3}y-9\\\\41+9=\frac{10}{3}y\\\\\frac{50*3}{10}=y\\\\y=\frac{150}{10}\\\\y=15

- Substitute this value into [Equation 2] and into [Equation 4]:

x=2(15)-10\\\\x=20

z=\frac{(15)+3}{3}\\\\z=6

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