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How do you divide #(x^2-9)/(4x+12)div(x-3)/6#?

Answer 1

Savannah Anderson

First, notice that #x^2-9 = x^2-3^2 = (x-3)(x+3)# This comes from a standard identity: #(a^2 - b^2) = (a-b)(a+b)#

So #(x^2 - 9)/(x-3) = ((x-3)(x+3))/(x-3) = (x+3)#

Hence

#(x^2-9)/(4x+12)-:(x-3)/6#

#=(x+3)/(4x+12)-:1/6 =(6(x+3))/(4(x+3)) = 6/4 = 3/2#

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Answer 2

Isaiah Anthony

#((x - 3)(x + 3))/(2(x + 6)) . (6)/(x - 3)# = # (6(x + 3))/(2(x + 6))#

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Answer 3

Eva Adamson

To divide the expression (x^2-9)/(4x+12) by (x-3)/6, you can follow these steps:

Simplify the expression (x^2-9) by factoring it as a difference of squares: (x^2-9) = (x+3)(x-3).
Simplify the expression (4x+12) by factoring out the common factor of 4: (4x+12) = 4(x+3).
Rewrite the division expression as a multiplication by taking the reciprocal of the second fraction: (x^2-9)/(4x+12) ÷ (x-3)/6 = (x^2-9)/(4x+12) * 6/(x-3).
Cancel out common factors between the numerators and denominators: (x^2-9)/(4x+12) * 6/(x-3) = [(x+3)(x-3)]/[4(x+3)] * 6/(x-3).
Simplify further by canceling out common factors: [(x+3)(x-3)]/[4(x+3)] * 6/(x-3) = (x-3)/4.

Therefore, the simplified expression is (x-3)/4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.