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Grace [21]
3 years ago
15

The vertex of this parabola is at (-5, -2). When the x-value is -4, the

Mathematics
2 answers:
ale4655 [162]3 years ago
4 0
Standard form of a parabola is (y-k) =a(x-h)² or y= a(x-h)² +k

The vertex of a parabola is v(h,k).

We have already h & k ( given) ==>v(-5, -2), we still have to calculate "a", the coefficient of the squared expression:
1st info: replace h & k by their respective values: y=a(x+5)² -2
2nd info: when x= -4, y= 2. Replace y & x  in the above equation so that to find a:  2 =a(-4 + 5)² -2==> 2=a(1)² - 2==> a=4 coef. of sqaared expression

dusya [7]3 years ago
3 0
I am pretty sure that it's
A. 1
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Hey there! I'm happy to help!

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Have a wonderful day! :D

7 0
3 years ago
Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin
saw5 [17]
Treat \mathcal C as the boundary of the region \mathcal S, where \mathcal S is the part of the surface z=2xy bounded by \mathcal C. We write

\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

with \mathbf f=(y+7\sin x,z^2+9\cos y,x^3).

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\nabla\times\mathbf f=(-2z,-3x^2,-1)

so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv


where \mathbf s(u,v) is a vector-valued function that parameterizes \mathcal S. In this case, we can take

\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

with 0\le u\le1 and 0\le v\le2\pi. Then

\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

and the integral becomes

\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi<span />
4 0
3 years ago
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DerKrebs [107]
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3 0
3 years ago
Solve for x. (7x+4)+(8x-4)
Iteru [2.4K]

Answer:

x=15

Step-by-step explanation:

thats what i got

Hope this helps!

7 0
3 years ago
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Black_prince [1.1K]

Answer:

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7 0
3 years ago
Read 2 more answers
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