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SashulF [63]
3 years ago
15

Simplify the following

0%7B1%7D%7Bx%7D%20%7D" id="TexFormula1" title="\frac{ \frac{1}{x} + 1 } { \frac {1}{x} }" alt="\frac{ \frac{1}{x} + 1 } { \frac {1}{x} }" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Anon25 [30]3 years ago
8 0
Good evening, Riley!

Used properties:

\star~\boxed{\mathsf{a = \dfrac{ax}{x}, \ \ x\neq 0}}\\ \\ \\ \star~\boxed{\mathsf{\dfrac{\frac{a}{b}}{\frac{c}{d}} = \dfrac{ad}{bc}}}

Now, note that:

\mathsf{1 = \dfrac{x}{x},\ x\neq 0}

So:

\mathsf{\dfrac{\frac 1 x + \frac x x }{\frac 1 x} = \dfrac{\frac{x+1}{x}}{\frac 1 x}}=} \mathsf{\dfrac{x}{1}\dfrac{x+1}{x} = \boxed{x + 1}}
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Taking point (0,-2) of graph one and replacing in the equation;

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Taking point (0,2) of graph 2;

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Step-by-step explanation:

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Andreas93 [3]

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Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B5x%2B3" id="TexFormula1" title="x^{2} +5x+3" alt="x^{2} +5x+3" align="absmidd
il63 [147K]
<h3>x²+5x+3+2x²+10x15 =0</h3><h3>x²+2x²+5x+10x+3+15=0</h3><h3>3x²+15x+18=0</h3><h3>3(x²+5x+6) =0 because 3 is common factor</h3><h3>3(x²+3x+2x+6) spill the middle term</h3><h3>3(x(x+3)+2(x+3) take the common factor from term</h3><h3>3(x+2) (x+3)</h3>

<h3>answer is 3(x+2) (x+3)</h3>

please mark this answer as brainlist

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