In row 3 of the periodic table, the third electron shell (the "M" Shell) is being filled in.
Answer:
23/24
Step-by-step explanation:
I found a common denominator of 24, and then multiplied the numbers so that they would equal the common denominator. I then added all of the numbers up, to get an answer of 37/24. Finally, I subtracted 2 12/24 (which is equal to 2 1/2) to get an answer of 23/24. Because that cannot be simplified, that would be your final answer. Hope this helps :)
Answer:
>
Step-by-step explanation:
Divide them, Do 2 divided by 5 and you'll get 0.4 and then do 35 divided by 100 and you'll get 0.35. 0.4 is bigger then 0.35 so the answer will be >.
Answer:
Integration of I= 
=![[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7D%20x%5E%7B%28n%2B1%29%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7Dx%5E%7B%28n%2B1%29%7D%5D)
Step-by-step explanation:
Given integral is I=
Take logx=t
I= 
I= 
Using integration by part,
![I= (t)\int [e^{(n+1)t}]\, dt-\int[\frac{d}{dt}{t}\times\int (e^{(n+1)t})]\\\\I= (t) [\frac{1}{n+1}e^{(n+1)t}]-\int[1\times\frac{1}{n+1}e^{(n+1)t}]\,dt\\\\I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]](https://tex.z-dn.net/?f=I%3D%20%28t%29%5Cint%20%5Be%5E%7B%28n%2B1%29t%7D%5D%5C%2C%20dt-%5Cint%5B%5Cfrac%7Bd%7D%7Bdt%7D%7Bt%7D%5Ctimes%5Cint%20%28e%5E%7B%28n%2B1%29t%7D%29%5D%5C%5C%5C%5CI%3D%20%28t%29%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5Cint%5B1%5Ctimes%5Cfrac%7B1%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D%5C%2Cdt%5C%5C%5C%5CI%3D%5B%5Cfrac%7Bt%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29t%7D%5D)
Writing in terms of x
I=![[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bt%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29t%7D%5D)
I=![[\frac{logx}{n+1}e^{(n+1)logx}]-[\frac{1}{(n+1)^{2}}e^{(n+1)logx}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29logx%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29logx%7D%5D)
I=![[\frac{logx}{n+1}e^{logx^{(n+1)}}]-[\frac{1}{(n+1)^{2}}e^{logx^{(n+1)}}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7De%5E%7Blogx%5E%7B%28n%2B1%29%7D%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7Blogx%5E%7B%28n%2B1%29%7D%7D%5D)
I=
Thus,
Integration of I= =
Answer:
0
Step-by-step explanation:
9-9=0
-9 is basically symbolizing subtraction
