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2 years ago

Please help quarter ends tomorrow and I don’t know what to do I’m failing all my classes

Mathematics

1 answer:

Nutka1998 [239]2 years ago

3 0

Step-by-step explanation:

x = -10, y = 10

y= -4x+4

m1×m2 = -1

-4 × m2 = -1

m2 = 1/4

y-y1 = m(x-x1)

y-10 = 1/4(x+10)

y-10 = 1/4x +10/4

(×4)

4y-40=x+10

4y=x+50

x-4y+50=0

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What is the axis of symmetry for f(x) = 2x2 − 4x + 5?

Romashka-Z-Leto [24]

Answer:

Axis of symmetry is

C) x = 1

Step-by-step explanation:

we are given equation as

$2x^2 -4x+5$

we can use axis of symmetry formula

Let's assume we are given quadratic equation as

$y=ax^2+bx+c$

Axis of symmetry is

$x=-\frac{b}{2a}$

now, we can compare both equations

and we get

$a=2,b=-4$

now, we can find axis of symmetry

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3 years ago

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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

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

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

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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WARRIOR [948]

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kirill115 [55]

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