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3 years ago

What is the expression in radical form? (5x^3 y^2) 2/8

Mathematics

1 answer:

Gre4nikov [31]3 years ago

8 0

Answer: (y^{2}x^{3}⋅5)/4

Step-by-step explanation:

Convert to Radical Form

$(5x^3y^2)\frac{2}{8}$

Reduce the expression $\frac{2}{8}$ by cancelling the common factors.

Factor 2 out of 2.

$5x^3y^2\frac{2(1)}{8}$

Factor 2 out of 8.

$5x^3y^2 \frac{2(1)}{2(4)}$

Cancel the common factor.

$5x^3y^2 \frac{1}{4}$

Combine 5 and $\frac{1}{4}$ .

$x^3y^2\frac{5}{4}$

Combine $x^{3}$ and $\frac{5}{4}$ .

$y^2\frac{x^3(5)}{4}$

Combine $y^{2}$ and $\frac{x^3*5}{4}$

$\frac{y^2x^3*5}{4}$

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4. Describe the proportion method for solving a percentage problem:

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Step-by-step explanation:

Consider the provided information.

For the proportion method first set up the equation like this:

$\frac{Part}{Whole}=\frac{Percentage}{100}$

Perform the cross multiplication and then solve for the missing part.

For example:

Find 80 percentage of 10.

Step 1: Set up the equation.

$\frac{Part}{10}=\frac{80}{100}$

$\frac{Part}{10}=\frac{4}{5}$

Step 2: Perform the cross multiplication

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Step 3: Solve for the missing part.

$Part=4 \times 2$

$Part=8$

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3 years ago

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Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

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with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

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$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$

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