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moongazer

  • 5 years ago

How do you equate this? 9^2+x^2=4^2+(10-x)^2 Please explain clearly that anyone could understand it easily. (Guaranteed medal for a good answer)

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  1. anonymous
    • 5 years ago
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    Combine the x^2 together on one side and the constants on another side. Set them equal to each other. x^2 - (10-x)^2 = 4^2 - 9^2

  2. anonymous
    • 5 years ago
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    Expanding it, you get x^2 - 100 +20x - x^2 = 4^2 - 9^2 The x^2 , cancels out. You are left with -100 + 20x = 4^2 - 9^2 20x = 4^2 - 9^2 + 100 = 35 x = 35/20

  3. moongazer
    • 5 years ago
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    Thanks!

  4. anonymous
    • 5 years ago
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    9^2+x^2=4^2+(10-x)^2 Sort out the constants first, so 9^2 = 81 and 4^2 = 16 Rewrite 81 + x^2 = 16 +(10-x)^2 Expand (10-x)^2 which gives 100 -20x + x^2 Rewrite 81 + x^2 = 16 +100 -20x + x^2 = 116 -20x + x^2 Subtract x^2 from both sides 81 = 116 -20x Subtract 116 from both sides -35 = -20x divide both sides by -5 7 = 4x Divide both sides by 4 7/4 = x

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