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2 years ago

HELP AHHHHHHHHHHHHHHHH

Mathematics

1 answer:

allochka39001 [22]2 years ago

8 0

Answer:

y = 3x+4

Step-by-step explanation:

slope formula is changes of y over the changes of x

$\large \boxed{m = \frac{y_2-y_1}{x_2-x_1} }$

m is slope in y = mx+b.

Substitute the coordinate points in.

$\large{m = \frac{7 + 2}{1 + 2} } \\ \large{m = \frac{9}{3}} \\ \large{m = 3}$

Since we know the slope, we rewrite by substituting m = 3 in the slope-intercept form.

$\large{y = mx + b} \\ \large{y = 3x + b}$

Then we either substitute one of given coordinate points to find the value of b. I will substitute (1,7) in the equation.

$\large{7 = 3(1)+ b} \\ \large{7 = 3 + b} \\ \large{7 - 3 = b} \\ \large{4 = b}$

Then rewrite the equation from y = 3x+b by substituting the value of b.

$\large{y = 3x + b} \\ \large{y = 3x + 4}$

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Answer:

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Step-by-step explanation:

Given an arbitrary ellipse (x²/a²) + (y²/b²) = 1 (a, b > 0)

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α(t) = (-a*Sin(t), b*Cos (t)) where t ∈ [0, 2π]

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If t = 3π/2

α(3π/2) = (-a*Sin(3π/2), b*Cos (3π/2)) = (a, 0)

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tresset_1 [31]

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Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

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$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

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$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

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