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sp2606 [1]
3 years ago
10

What value of n makes the equation true? (n – 27.3 = 15.9)

Mathematics
2 answers:
Softa [21]3 years ago
7 0
Add 27.3 to both sides

Lady_Fox [76]3 years ago
7 0
N=15,9+27,3
n=43,2
the answer is D
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Using the given points, determine Δy.<br><br> (-3, -5) and (0, 10)
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Can you please help me?
ASHA 777 [7]

Answer:

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3 years ago
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Rom4ik [11]

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Now, lets evaluate the same integral using power series. first, find the power series for the function f(x = \frac{32}{x^2+4}. t
BlackZzzverrR [31]
No idea what the previous part of the problem is, but you have

f(x)=\dfrac{32}{x^2+4}=\dfrac8{1-\left(-\frac{x^2}4\right)}=\displaystyle8\sum_{n\ge0}\left(-\frac{x^2}4\right)^n
f(x)=\displaystyle8\sum_{n\ge0}\left(-\dfrac14\right)^nx^{2n}

which is valid for \left|-\dfrac{x^2}4\right|, or |x|. So the integral from 0 to 2 is

\displaystyle\int_0^2f(x)\,\mathrm dx=\int_0^28\sum_{n\ge0}\left(-\frac14\right)^nx^{2n}\,\mathrm dx
=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\int_0^2x^{2n}\,\mathrm dx

Note that since the power series only converges on the interval if x is strictly less than 2, which means we have to treat this as an improper integral.

=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\int_0^tx^{2n}\,\mathrm dx[/tex]
=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\frac{x^{2n+1}}{2n+1}\bigg|_{x=0}^{x=t}
=\displaystyle8\sum_{n\ge0}\frac{(-1)^n}{2^{2n}(2n+1)}\lim_{t\to2^-}t^{2n+1}
=\displaystyle16\sum_{n\ge0}\frac{(-1)^n}{2n+1}
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