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Mathematics

1 answer:

g100num [7]2 years ago

5 0

The slope of the line representing the linear graph is a rate of 3.2 meters per second.

<h3 /><h3>Linear equation</h3>

A linear equation is in the form:

y = mx + b

where y, x are variables, m is the slope of the line and m is the y intercept.

Let y represent the distance in meters and x represent the time in seconds. Hence:

Using the points (20, 64) and (60, 192)

$Slope = \frac{y_2-y_1}{x_2-x_1} =\frac{192-64}{60-20} =3.2$

The slope of the line representing the linear graph is a rate of 3.2 meters per second.

Find out more on linear equation at: brainly.com/question/14323743

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3 years ago

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Find the flux of the vector field F = 〈e-z,4z,6xy) across the curved sides of the surface S = {(x,y,z): z= cos y, lys π, 0sxs4}

Len [333]

I'll go ahead and assume you meant to say that <em>S</em> is the surface given by

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This immediately gives us a parameterization for the surface,

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The upward-pointing normal vector to this surface is then

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Then the flux of $\vec F(x,y,z) = \left\langle e^{-z}, 4z, 6xy\right\rangle$ across <em>S</em> is

$\displaystyle \iint_S \vec F(x,y,z)\cdot\mathrm d\vec s = \int_0^4\int_0^\pi \vec F(x,y,\cos(y))\cdot\vec n\,\mathrm dy\,\mathrm dx \\\\ = \int_0^4\int_0^\pi \left\langle e^{-\cos(y)},4\cos(y),6xy\right\rangle \cdot \left\langle0,\sin(y),1\right\rangle \,\mathrm dy\,\mathrm dx \\\\ = \int_0^4\int_0^\pi (4\sin(y)\cos(y)+6xy)\,\mathrm dy\,\mathrm dx \\\\ = 2 \int_0^4\int_0^\pi (\sin(2y) + 3xy)\,\mathrm dy\,\mathrm dx = \boxed{24\pi^2}$

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2 years ago

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