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lys-0071 [83]
2 years ago
11

HELP 20 POINTS!!! Does the infinite geometric series diverge or converge? Explain.

Mathematics
2 answers:
Nitella [24]2 years ago
8 0

Answer:

c. it converges, it has a sum

Thepotemich [5.8K]2 years ago
4 0

Answer:

It converges; it has a sum, I did the test and got 100%

Step-by-step explanation:

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Antonio had his hair cut at the salon. After he returned home, he realized he had accidentally tipped the stylist 114% of the co
stepan [7]
=>C

1.14*20.95=23.883
5 0
3 years ago
Picture of reflected trapezoid above the line
asambeis [7]

Answer:

which line!

x or y axis?

first image is over the x axis

second is over the y

Step-by-step explanation:

HOPE THIS HELPS, BRAINLIEAST PLZZ

3 0
2 years ago
Evaluate the surface integral:S
rjkz [21]
Assuming S does not include the plane z=0, we can parameterize the region in spherical coordinates using

\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle

where 0\le u\le2\pi and 0\le v\le\dfrac\pi/2. We then have

x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v
(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v

Then the surface integral is equivalent to

\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du

We have

\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle
\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle
\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle
\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v

So the surface integral is equivalent to

\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du
=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv
=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw

where w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv.

=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}
=\dfrac{243}2\pi
4 0
2 years ago
Determine the values of the constants b and c so that the function given below is differentiable. f(x)={2xbx2+cxx≤1x>1
Lera25 [3.4K]
Assuming the function is

f(x)=\begin{cases}2x&\text{for }x\le1\\bx^2+cx&\text{for }x>1\end{cases}

For f(x) to be differentiable, it necessarily has to be continuous. For this condition to be met, we need

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)
\iff\displaystyle\lim_{x\to1}2x=\lim_{x\to1}(bx^2+cx)
\iff2=b+c

For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have

f'(x)=\begin{cases}2&\text{for }x1\end{cases}

\displaystyle\lim_{x\to1^-}2=\lim_{x\to1^+}(2bx+c)
\iff2=2b+c

Now we solve for b and c:

\begin{cases}b+c=2\\2b+c=2\end{cases}\implies b=0,c=2
5 0
3 years ago
The function f(x) is shown in this graph.
Minchanka [31]
Where’s the graph?????? The slope of g(x) is -2
8 0
3 years ago
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