MaxWong

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

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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?