When 10 is added to the product of 5 and a number, the result is 50. What is the number?

Mathematics

2 answers:

ser-zykov [4K]4 years ago

7 0

8 because 10+5(8) =50. I subtracted 10 from 50 then divided 40 by 5 to get 8

dalvyx [7]4 years ago

5 0

It's 35 because 10 plus 5 is 15. You the subtract it by 50 and you get 35.

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Perform the subtraction -2 - 15=

Natali5045456 [20]

the answer is -17 :)

7 0

3 years ago

Read 2 more answers

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]
$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\frac{x^{2n+1}}{2n+1}\bigg|_{x=0}^{x=t}$
$=\displaystyle8\sum_{n\ge0}\frac{(-1)^n}{2^{2n}(2n+1)}\lim_{t\to2^-}t^{2n+1}$
$=\displaystyle16\sum_{n\ge0}\frac{(-1)^n}{2n+1}$
$=16-\dfrac{16}3+\dfrac{16}5-\dfrac{16}7+\dfrac{16}9+\cdots$