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Zepler [3.9K]
4 years ago
7

Please help me answer the following question. Thanks in advance.

Mathematics
2 answers:
aivan3 [116]4 years ago
8 0

Answer:

11.2ft

Step-by-step explanation:

14 is 2/3 of 21 so the scale factor is 2/3.

16.8x2/3= 11.2

Artist 52 [7]4 years ago
5 0
Answer is 11.2
14/21 is 2/3
So, scale factor is 2/3
16.8*2/3 is 11.2
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Solve x:<br> (On the attachment). Please help! Homework due in for tomorrow! Xx
emmasim [6.3K]

Hello!

Explanation:

Distributive property: → a(b+c)=ab+ac

First do expand.

4(x+1)=4x+4

Then you subtract by 4 from both sides of an equation.

4x+4-4=2x+4-4

Simplify by equation.

4x=2x

Next, subtract by 2x from both sides of an equation.

4x-2x=2x-2x

Simplify.

2x=0

You can also divide by 2 from both sides of an equation.

\frac{2x}{2}= \frac{0}{2}

Finally, simplify by equation and dividing into the groups.

0/2=0

Answer: → \boxed{x=0}

Hope this helps!

6 0
4 years ago
22 ounces increased to 25 ounces
Zielflug [23.3K]
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8 0
3 years ago
PLEASE HELP!!!
Vilka [71]

Answer:

B)

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8 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.
7 0
3 years ago
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