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3 years ago

-12x +8y=-32 in slope-intercept form

Mathematics

2 answers:

Sergeeva-Olga [200]3 years ago

4 0

Answer:

y=3/2x-4

Step-by-step explanation:

First we add 12x to both sides

8y=12x-32

Now divide both sides by 8

y=3/2x-4

Tatiana [17]3 years ago

3 0

The slope-intercept form: y = mx + b

$-12x+8y=-32\qquad|\text{add 12x to both sides}\\\\8y=12x-32\qquad|\text{divide both sides by 8}\\\\y=\dfrac{12}{8}x-\dfrac{32}{8}\\\\\boxed{y=\dfrac{3}{2}x-4}$

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le Gianna just started a running plan where she runs 20 miles the first week and then increases the number of miles she runs by

yuradex [85]

Answer:

Gianna ran 40 miles over the course of 17 weeks.

Step-by-step explanation:

20 miles in the first week.

5% increase each week.

17 weeks.

5% * 20 = 1

1 mile increase each week.

17 * 1 = 17

17 mile increase over the course of 17 weeks.

20 + 17 = 37

37 miles over the course of 17 weeks.

Round 37 to the nearest whole number.

37 ^ 40

4 0

3 years ago

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

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

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

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

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

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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3 years ago

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Yuri [45]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

6th out 10 questions

kicyunya [14]

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