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AveGali [126]
3 years ago
6

Can someone please simplify the complex fraction?

Mathematics
1 answer:
dmitriy555 [2]3 years ago
8 0
\bf \cfrac{\qquad \frac{x+3}{4x^2-16}\qquad }{\frac{2x^2+10x+12}{2x-4}}\implies \cfrac{\frac{x+3}{4(x^2-4)}}{\frac{\underline{2}(x^2+5x+6)}{\underline{2}(x-2)}}\implies \cfrac{\frac{x+3}{4(x^2-2^2)}}{\frac{x^2+5x+6}{x-2}}\implies \cfrac{\frac{x+3}{4(x-2)(x+2)}}{\frac{(x+3)(x+2)}{x-2}}

\bf \cfrac{\underline{x+3}}{4\underline{(x-2)}(x+2)}\cdot \cfrac{\underline{x-2}}{\underline{(x+3)}(x+2)}\implies \cfrac{1}{4(x+2)}\cdot \cfrac{1}{x+2}
\\\\\\
\cfrac{1}{4(x+2)(x+2)}\implies \cfrac{1}{4(x^2+4x+4)}\implies \cfrac{1}{4x^2+16x+16}
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<h3>How to find the image after the two reflections?</h3>

A reflection across a line moves a point perpendicularly towards that line and translates it to the other side of the line, such that the distance between the point and the line doesn't change.

Here the original point is z(1, 1)

And the first reflection is across line x = 2.

This is a vertical line, then it only will change the x-value.

Now, the distance between the x-value of our point and the line is:

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