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enot [183]
3 years ago
11

7Bax-ay%7D%20%7D%20-%5Csqrt%7Bx%5E%7B3%7D-x%5E2y%20%7D%20%7D" id="TexFormula1" title="2y\sqrt{x-y} +x\sqrt{\frac{1}{x-y} } -a\sqrt{\frac{a}{ax-ay} } -\sqrt{x^{3}-x^2y } }" alt="2y\sqrt{x-y} +x\sqrt{\frac{1}{x-y} } -a\sqrt{\frac{a}{ax-ay} } -\sqrt{x^{3}-x^2y } }" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:

(2y-x)\sqrt{x-y} +(x-a)\sqrt{\frac{1}{x-y} }

Step-by-step explanation:

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Find the perimeter of the square. (hint : make one side = to the other side to find y. Remember, you are finding the perimeter)
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Step-by-step explanation:

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3 years ago
Joe has 37 coins consisting of nickels, dimes, and quarters. There are four more nickels than dimes and two more quarters than n
astraxan [27]

Answer:

  15

Step-by-step explanation:

Let n, d, q represent the numbers of nickels, dimes, and quarters. The problem statement tells us ...

  n +d +q = 37

  n = d +4

  q = n +2

___

Rearranging the second equation gives ...

  d = n -4

Substituting that into the first, we get ...

  n + (n -4) +q = 37

  2n +q = 41 . . . . . . . add 4 and simplify

Rearranging the third original equation gives ...

  n = q -2

Substituting into the equation we just made, we get ...

  2(q -2) +q = 41

  3q = 45 . . . . . . . . add 4 and simplify

  q = 15 . . . . . . . . . divide by 3

Joe has 15 quarters.

_____

<em>Check</em>

The number of nickels is 2 fewer, so is 13. The number of dimes is 4 fewer than that, so is 9. The total number of coins is 15 + 13 + 9 = 37, as required.

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3 years ago
(5/8 x - 10) - (1/8 x - 12)
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Step-by-step explanation:

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Step-by-step explanation:

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