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Phoenix [80]
3 years ago
6

If $\sqrt{5+n}=7$, then what is the value of $n$?

Mathematics
2 answers:
alina1380 [7]3 years ago
3 0

Answer:

44

Step-by-step explanation:

We can solve the equation by simplifying it.

$\sqrt{5+n}=7$, let's square both sides.

5+n = 49. Now lets subtract 5 from both sides.

n = 44.

Hope this helped!

Scorpion4ik [409]3 years ago
3 0

Answer:

  44

Step-by-step explanation:

  \sqrt{5+n}=7\\\\5+n=7^2=49\\\\n=49-5\\\\\boxed{n=44}

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so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

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\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

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<h2>x³ + x² - 3x - 3</h2>
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Step-by-step explanation:

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