Please help me and show your work!!

1 answer:

4 0

$\bf \cfrac{\sqrt{250x^{16}}}{\sqrt{2x}}\qquad 
\begin{cases}
250=2\cdot 5\cdot 5\cdot 5\\
\qquad 2\cdot 5\cdot 5^2\\
x^{16}=x^{8\cdot 2}\\
\qquad (x^8)^2
\end{cases}\implies \cfrac{\sqrt{2\cdot 5\cdot 5^2\cdot (x^8)^2}}{\sqrt{2\cdot x}}$

$\bf \cfrac{5x^8\sqrt{2\cdot 5}}{\sqrt{2}\cdot \sqrt{x}}\implies \cfrac{5x^8\underline{\sqrt{2}}\cdot \sqrt{5}}{\underline{\sqrt{2}}\cdot \sqrt{x}}\implies \cfrac{5x^8\sqrt{5}}{\sqrt{x}}\impliedby 
\begin{array}{llll}
\textit{and rationalizing the}\\
\textit{denominator}
\end{array}$

$\bf \cfrac{5x^8\sqrt{5}}{\sqrt{x}}\cdot \cfrac{\sqrt{x}}{\sqrt{x}}\implies \cfrac{5x^8\sqrt{5}\cdot \sqrt{x}}{(\sqrt{x})^2}\implies \cfrac{5x^8\sqrt{5x}}{x}\implies 5x^8x^{-1}\sqrt{5x}
\\\\\\
5x^{8-1}\sqrt{5x}\implies 5x^7\sqrt{5x}$

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write an equation for the line that is perpendicular to the given line and passes through the given point y-3=8/3(x+2);(-2,3)

Stolb23 [73]

Answer:

$y-3=-\frac{3}{8} (x+2)$

Step-by-step explanation:

We can write the equation of a line in 3 different forms including slope intercept, point-slope, and standard depending on the information we have. We have a point (8,1) and a slope from the equation. We will chose point-slope since we have a point and slope.

Point slope: $y-y_1=m(x-x_1)$

$m\neq \frac{8}{3}$ in our new equation because it is perpendicular to it. This means we will need to change it into its negative reciprocal which is $m=-\frac{3}{8}$ .