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3 years ago

Simplify (6x^-2)^2(0.5x)^4 show work please

Mathematics

2 answers:

tester [92]3 years ago

4 0

(6x^-2)^2(0.5x)^4 = (6^2)(x^-2(2))(1/2)^4(x^4) = 36x^-4(1/16x^4) = (36/16)x^(-4+4) = 9/4

Darina [25.2K]3 years ago

3 0

Answer:

$(6x^{-2})^2(0.5x)^4=\frac{9}{4}$

Step-by-step explanation:

The given equation is:

$(6x^{-2})^2(0.5x)^4$

Simplifying the above given equation, we get

= $(36x^{-4})(\frac{1}{2}x)^4$

= $36x^{-4}(\frac{1}{16})x^4$

= $\frac{36}{16}x^{-4+4}$

= $\frac{36}{16}$

= $\frac{9}{4}$

Thus, $(6x^{-2})^2(0.5x)^4=\frac{9}{4}$

which is the requires simplified form.

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butalik [34]

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Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

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The cross product reduces to

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When you compute the partial derivatives, you'll find that all the components reduce to 0 and

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with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

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$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

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Now

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$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

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