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Elden [556K]
3 years ago
5

What is 2(n+5)=-2 showing work

Mathematics
2 answers:
olganol [36]3 years ago
8 0
Simplify:
2n+10=-2

Subtract 10 from both sides
2n+10-10=-2-10
2n=-12

Divide both sides by 2
2n/2=-12/2

n=-6
BabaBlast [244]3 years ago
4 0
2(n + 5) = -2

First, our goal to this problem is to get the variable (n) and an unknown number on each side. To start off, we can divide both sides by 2.
n + 5 =  -\frac{2}{2}

Second, we can go ahead and cancel 2 out, which will replace it with a -1.
n + 1 = -1

Third, our next step will to be to subtract 5 from each side.
n = -1 - 5

Fourth, our last step is to simplify -1 - 5. This is quite simple. -1 - 5 = -6, so -6 is what the variable (n) equals. 
n = -6

Answer: \fbox {n = -6}
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Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about th
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Step-by-step explanation:

See the graph of the region attached.

To find the intersection between the line and the parabola we set the equation equal to each other, and solve that quadratic equation by factorization:

x^2=9-8x\\x^2+8x-9=0\\(x+9)(x-1)=0\\x=-9,x=1

Since the region is the one for x\ge 0 then the intersection we are interested on is x=1 as it can also be seen in the graph.

Then we set the integral using shell method for revolving about the y-axis:

\displaystyle\int_a^b 2\pi\,x\,h(x)\,dx

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Then the integral becomes:

\displaystyle\int_0^1 2\pi\,x\,(8x-9-x^2)\,dx

Notice the limits of the integral are the x-axis (x=0) and the intersection of the parabola and the line that we found before (x=1)

Now, solving the integral:

Start by factoring the 2\pi and distributing the x:

=\displaystyle2\pi \int_0^19x-8x^2-x^3dx

Then use the basic rule to integrate:

=\displaystyle2\pi \left[\frac{9x^2}{2}-\frac{8x^2}{3}-\frac{x^4}{4}\right|_0^1

Then evaluate the antiderivative in the limits and subtract:

=\displaystyle2\pi\left[\frac{9}{2}-\frac{8}{3}-\frac{1}{4}-0\right]=\frac{19\pi}{6}

So the volume of the solid is \displaystyle\frac{19\pi}{6}

4 0
3 years ago
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