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WARRIOR [948]
3 years ago
15

What is the answer to 3z 5z 4-9? every time i answer it differently and i can't get the right answer.?

Mathematics
1 answer:
Sauron [17]3 years ago
5 0
I'm confused as to what the question is
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Please help meeee!!!!
Vedmedyk [2.9K]

Answer:

I think it is 7x+5y=40. Tell me if I'm right because don't think I am.

Step-by-step explanation:

5 0
2 years ago
Help meee plssss!!!!!
lakkis [162]

Circumference of circle is 62.8 inches.

Step-by-step explanation:

Using Formula

Circumference of circle = 2 π r

Substitute the values

Circumference of circle = 2 × 3.14 × 10 inches.

multiply the numbers

Circumference of circle = 62.8 inches

Hence, the circumference of circle is 62.8 inches.

8 0
3 years ago
I need help please!!!
seraphim [82]

Answer: number 1

Step-by-step explanation:

That is how you would write the ratio word for word.

- hope this helped!

7 0
2 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).

By Stoke's theorem, the line integral is equivalent to the surface integral over \mathcal S of the curl of \mathbf f. We have


\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
2 years ago
Need help please and thank you.
AURORKA [14]
X+3=61
x=61-3=58

58+37=95
180-95=85

so x=58 and y=85
5 0
1 year ago
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