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kramer
2 years ago
11

HELP ASAP PLEASE AND SHOW WORK!!! Write the vertex form of the following equation. y= x^2+6x-5

Mathematics
1 answer:
valkas [14]2 years ago
4 0

Answer:

<h3>            \boxed{\bold{y=(x+3)^2-14}}</h3>

Step-by-step explanation:

y= x²+6x-5

completing the square: <em>  a²+2ab+b²</em> <em>= (a+b)²</em>  where <em>a²=x² </em>(so <em>a=x</em>) and <em>2ab=6x </em> (so <em>b</em> would be 3):

y=x^2+6x-5\\\\y=\underline{x^2+2\cdot x\cdot3+3^2}-3^2-5\\\\y=(x+3)^2-9-5\\\\\underline{y=(x+3)^2-14}\\\\\\vertex:\ \ (-3,\,-14)

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Question:

Which is equivalent to \sqrt{180x^{11}} after it has been simplified completely?

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\sqrt{180x^{11}} = 6x^{5}\sqrt{5x}

Step-by-step explanation:

Given

\sqrt{180x^{11}}

Required

Simplify

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\sqrt{180x^{11}} = \sqrt{180} * \sqrt{x^{11}}

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So, we have

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Further split the square roots

\sqrt{180x^{11}} = 6*\sqrt{5} *  \sqrt{x^{10}} * \sqrt{x}

From laws of indices; \sqrt{a} = a^{\frac{1}{2}}

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\sqrt{180x^{11}} = 6*\sqrt{5} *  x^{10*\frac{1}{2}} * \sqrt{x}

\sqrt{180x^{11}} = 6*\sqrt{5} *  x^{\frac{10}{2}} * \sqrt{x}

\sqrt{180x^{11}} = 6*\sqrt{5} *  x^{5} * \sqrt{x}

Rearrange Expression

\sqrt{180x^{11}} = 6 *  x^{5} * \sqrt{5} * \sqrt{x}

\sqrt{180x^{11}} = 6x^{5} * \sqrt{5} * \sqrt{x}

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