Mathematics, 17.07.2020 01:01 ingle75

You have 20 coins in your pocket that are either quarter or dimes. They total $3.95. How many quarters do you have? A.16 B.13 C.12 D.10 E.15 I WILL GIVE THE FIRST PERSON BRAINLIEST, BUT U HAVE TO BE RIGHT.

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Answer from: dazzlecaptain212

B. 13 quarters

Step-by-step explanation:

Let:

quarters = x = $0.25

dimes = y = $0.10

Get the set of equation:

x + y = 20

0.10y + 0.25x = 3.95

First, solve for x. Use the first equation. Isolate the x by subtracting y from both sides:

x + y (-y) = 20 (-y)

x = 20 - y

Plug in 20 - y for x in the second equation:

0.10y + 0.25(2020 - y) = 3.95

Simplify. Remember to follow PEMDAS. Isolate the variable, y.

First, multiply 0.25 to all terms within the parenthesis:

0.10y + (0.25 * 20) + (0.25 * -y) = 3.95

0.10y + 5 - 0.25y = 3.95

Combine like terms:

(-0.25y + 0.10y) + 5 = 3.95

(-0.15y) + 5 = 3.95

Isolate the variable, y. Do the opposite of PEMDAS. First, subtract 5 from both sides:

(-0.15y) + 5 (-5) = 3.95 (-5)

(-0.15y) = 3.95 - 5

-0.15y = -1.05

Isolate the variable, y. Divide -0.15 from both sides:

(-0.15y)/-0.15 = (-1.05)/-0.15

y = 1.05/0.15 = 7

y = 7

There are 7 dimes.

Plug in 7 to one of the equations in the operation for y (dimes):

x + y = 20

x + (7) = 20

Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Subtract 7 from both sides:

x + 7 (-7) = 20 (-7)

x = 20 - 7

x = 13

There are 13 quarters.

Check: Plug in 13 for x, and 7 for y in the second equation:

0.10y + 0.25x = 3.95

0.10(7) + 0.25(13) = 3.95

0.70 + 3.25 = 3.95

3.95 = 3.95

Note that you are solving only for quarters. In this case, your answer will be B. 13.

Answer from: ahhhhhhhh2806

B. 13 Quarters

Step-by-step explanation:

So since we don't have an amt. specified like e.g there are 10 more quarters than dimes, we can only rely on guess and check.

First we can try 10 quarters --> 10 x 25 = 250 --> 395 - 250 = 145 = 14.5 dimes

=> 10 is not equal to 14.5 so we know that the amt. of quarters must be more than 10 and must be an odd amt. => so now we can try 13

13 x 25 = 325 --> 395 - 325 = 70 => 70/10= 7 --> 13 + 7 = 20

Thus, we have our answer of 13

Hope this helps!

Answer from: Quest

no, because the sides are not congruent

Is the following shape a square? how do you know?

Answer from: Quest

well we know there are 4 clubs in 52 cards so it’s replaced so it should be 3/51

step-by-step explanation:

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