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2 years ago

10 points!!! What is sqr root 12x^8/ sqr root 3x^2 in simplest form

Mathematics

2 answers:

marta [7]2 years ago

8 0

Answer:

2 x^3

Step-by-step explanation:

sqrt(12 x^8)/ sqrt(3x^2)

combine into 1 term

sqrt(12x^8/3x^2)

when dividing exponents with the same base, we subtract

sqrt(12/3 x^(8-2))

sqrt(4 x^6)

sqrt(4) sqrt(x^6)

2 x^3

boyakko [2]2 years ago

8 0

Answer:

2x^3

Step-by-step explanation:

$\frac{\sqrt{12x^8}}{\sqrt{3x^2}}$

Both top and bottom have square root . We can take square root in common

$\sqrt{\frac{12x^8}{3x^2} }$

simplify the exponent using exponential property

a^m / a^n = a^{m-n}, subtract the exponents

$\sqrt{\frac{12x^8}{3x^2} }$

$\sqrt{4x^6}$

$\sqrt{4} =2$

$\sqrt{x^6} =\sqrt{x^2 \cdot x^2 \cdot x^2} =x^3$

$\sqrt{4x^6}=2x^3$

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ludmilkaskok [199]

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where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

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where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
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We can check this result by evaluating the equivalent surface integral. We have

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so that

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where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

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