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anonymous

  • 4 years ago

simplify (πd²/4)-(π[d+1]²/4)

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  1. anonymous
    • 4 years ago
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    Factor pi/4. Then, it is a difference of squares. Simplify.

  2. anonymous
    • 4 years ago
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    wouldn't it be π/4 (d^2-d^2+1)?

  3. anonymous
    • 4 years ago
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    No. It is pi/4(d^2 - (d+1)^2)

  4. anonymous
    • 4 years ago
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    so how do i solve that? :S

  5. anonymous
    • 4 years ago
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    a^2 - b^2 = (a-b)(a+b)

  6. anonymous
    • 4 years ago
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    In the above, a = d, b = d+1.

  7. anonymous
    • 4 years ago
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    i got π(2d-1)/4 .... is that right? because the back of my book says the answer is π(2d+1)/4

  8. anonymous
    • 4 years ago
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    pi/4 (d^2 - (d+1)^2) = pi/4 (d + d + 1)(d - (d-1)) = pi/4 (2d+1)(d-d+1) = pi/4 (2d+1)

  9. anonymous
    • 4 years ago
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    how is it pi/4 (d + d + 1)(d - (d-1)) ?

  10. anonymous
    • 4 years ago
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    Forget pi/4 since that is already factored out. Focus on (d^2 - (d+1)^2). Like I said, a^2 - b^2 = (a+b)(a-b). Here, you have d^2 - (d+1)^2 = (d+d+1)(d-(d+1))

  11. anonymous
    • 4 years ago
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    OOOOOH ! im subtracting it.. instead of multiplying (x my bad

  12. anonymous
    • 4 years ago
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    wait but i still get 2d-1 :S

  13. anonymous
    • 4 years ago
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    Actually, you get -(2d+1)

  14. anonymous
    • 4 years ago
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    ): it says the answer is pi/4 2d+1

  15. anonymous
    • 4 years ago
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    I see the answer as the same, but with a minus sign. But, may be missing something.

  16. anonymous
    • 4 years ago
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    oh well i'll just keep it like this for now and ask my teacher tomorrow. thanks

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