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kiruha [24]
3 years ago
12

7 (Picture) CONVERGENT AND DIVERGENT SERIES PLEASE HELP!!

Mathematics
1 answer:
Elena-2011 [213]3 years ago
7 0

Answer:

Option a. convergent: A

Step-by-step explanation:

a0=3/2

a1=9/8

a2=27/32

a1/a0=(9/8)/(3/2)=(9/8)*(2/3)→a1/a0=3/4

a2/a1=(27/32)/(9/8)=(27/32)*(8/9)→a2/a1=3/4

r=a1/a0=a2/a1→r=3/4

The abosute value of r= !r! = !3/4! = 3/4 = 0.75<1, the series is convergent

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If there is such a scalar function <em>f</em>, then

\dfrac{\partial f}{\partial x}=4y^2

\dfrac{\partial f}{\partial y}=8xy+4e^{4z}

\dfrac{\partial f}{\partial z}=16ye^{4z}

Integrate both sides of the first equation with respect to <em>x</em> :

f(x,y,z)=4xy^2+g(y,z)

Differentiate both sides with respect to <em>y</em> :

\dfrac{\partial f}{\partial y}=8xy+4e^{4z}=8xy+\dfrac{\partial g}{\partial y}

\implies\dfrac{\partial g}{\partial y}=4e^{4z}

Integrate both sides with respect to <em>y</em> :

g(y,z)=4ye^{4z}+h(z)

Plug this into the equation above with <em>f</em> , then differentiate both sides with respect to <em>z</em> :

f(x,y,z)=4xy^2+4ye^{4z}+h(z)

\dfrac{\partial f}{\partial z}=16ye^{4z}=16ye^{4z}+\dfrac{\mathrm dh}{\mathrm dz}

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Integrate both sides with respect to <em>z</em> :

h(z)=C

So we end up with

\boxed{f(x,y,z)=4xy^2+4ye^{4z}+C}

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