OK I understand a bit now :) Thanks
Hmm? I don't quite understand...
So the union of M and N means it is either in M or in N?
And, if you mean this one:
\(\dfrac{7x^2}{3}\cdot \dfrac{9}{14}x\\ =\dfrac{3x^3}{2}\)
Also you may mean this one:
\(\dfrac{7x^2}{3}\cdot \dfrac{9}{14x}\\ =\dfrac{3x}{2}\)
\(7x^{2/3}\cdot \dfrac{9}{14x}\\ =\dfrac{9}{2x^{1/3}}\)
Or you mean this one:
\(7x^{2/3}\cdot \dfrac{9}{14}x\\ =\dfrac{9x^{5/3}}{2}\)
\(\dfrac{8^{26}-8^{25}}{8^{25}8^{26}}\\ =\dfrac{1}{8^{25}}-\dfrac{1}{8^{26}}\)
Wot m8 XD
Now take an attempt on this problem:
\(14^{(1+2+3+4+5+6+...+107)}\pmod {15}=??\)
14^1 mod 15 = 14
14^2 mod 15 = 1
14^3 mod 15 = 14
14^4 mod 15 = 1
.
We can see that 14^(2n+1) mod 15 = 14 and 14^(2n) mod 15 = 1
So that 14^15 mod 15 = 14. :D
Set the whole thing to 0. i.e. f(x) = 0.
\(f(x) = 0 \\ (x-c)^2 + d = 0\\ (x-c)^2 = -d\)
It has no real solutions. Therefore it have no zeroes.
Is it a typing mistake??
Do you mean
\(\displaystyle \int^{\infty}_{-\infty}\dfrac{\sin x}{x}dx\)?
If yes, the answer is π by definition.
I guess it's Mode of X = 3?
