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madam [21]
3 years ago
15

0%2B%209%20%7D%20%20%7D%20%20%5Cgeqslant%200" id="TexFormula1" title=" \frac{x {}^{2} - 4 }{ \sqrt{x {}^{2} - 6x + 9 } } \geqslant 0" alt=" \frac{x {}^{2} - 4 }{ \sqrt{x {}^{2} - 6x + 9 } } \geqslant 0" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Nimfa-mama [501]3 years ago
7 0

First of all, we can observe that

x^2-6x+9 = (x-3)^2

So the expression becomes

\dfrac{x^2-4}{\sqrt{(x-3)^2}} = \dfrac{x^2-4}{|x-3|}

This means that the expression is defined for every x\neq 3

Now, since the denominator is always positive (when it exists), the fraction can only be positive if the denominator is also positive: we must ask

x^2-4 \geq 0 \iff x\leq -2 \lor x\geq 2

Since we can't accept 3 as an answer, the actual solution set is

(-\infty,-2] \cup [2,3) \cup (3,\infty)

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Put the following in ascending order​
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see explanation

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-------------------------------------------------------

\frac{7}{15} = \frac{7(2)}{15(2)} = \frac{14}{30}

\frac{2}{5} = \frac{2(6)}{5(6)} = \frac{12}{30}

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----------------------------------------------------------------

Placing these in ascending order ( smallest to largest )

\frac{10}{30}, \frac{12}{30}, \frac{14}{30}, \frac{15}{30}, \frac{20}{30}, \frac{27}{30}

----------------------------------------------------------------------

Listing the actual fractions corresponding to the equivalent fractions

\frac{1}{3}, \frac{2}{5}, \frac{7}{15}, \frac{1}{2}, \frac{2}{3}, \frac{9}{10}

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