Math

4) Simplify :
(i) (125 ×x^(-3))/(5^(-3)×25 ×x^(-6) )
(ii) (16×10^2×64)/(2^4×4^2 )

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  1. it's better to use * for multiplication, so it does not get confused with x, a variable. So,
    (125*x^(-3))/(5^(-3)*25*x^(-6) )
    = (5^3 x^-3)/(5^-3 * 5^2 x^-6)
    = (5^3 x^-3)/(5*-1 x^-6)
    = 5^(3-(-1)) x^(-3-(-6))
    = 5^4 x^3
    = 625 x^3

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    oobleck
  2. (i) (125 ×x^(-3))/(5^(-3)×25 ×x^(-6) )
    = [5*5*5/x*x*x ] / [ (5*5)/ {5*5*5} / (x*x*x*x*x*x) ]
    = [5*5*5/x*x*x ] * [ {5*5*5} / (x*x*x*x*x*x) ] / (5*5)
    = 5*5*5*5 * (x*x*x)
    = 625 x^3
    your turn now

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    Damon
  3. hint
    16 = 4^2
    64 = 16* 4 = 4^3
    2^4 =( 2^2)^2 = 4^2

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    Damon
  4. (125 ×x^(-3))/(5^(-3)×25 ×x^(-6) )

    write it this way, easier to read:
    (125x^-3)/(5^-3*25*x^-6 ) , the * is often used as multiplication, if an exponent is a monomial you don't need brackets around it.
    = (125/x^3)/((1/125)(25))*x^6
    = 125*5*x^3
    = 625x^3

    or, even simpler ...
    (125x^-3)/(5^-3*25*x^-6 )
    = 125/(25/125) x^(-3+6)
    = 625x^3 or (5^4)(x3)

    looking at the 2nd one, I see all powers of 2 and a 10^2
    (16×10^2×64)/(2^4×4^2 )
    = (2^4)(100)(2^6)/( (2^4)(*2^4))
    = 100(2^10)/2^8
    = 100(2^2) = ...

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    Reiny

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