MaxWong

That does not have a meaning.

14 over 56 x 100%

= 1/4 x 100% <-- as 14/56 = 1/4

= 25%.

Any convolution is defined as
$(f*g)(t)=\displaystyle\int^{\infty}_{-\infty}f(x)g(t-x)dx=\int^{\infty}_{-\infty}g(x)f(t-x)dx$

Trying to do it... Never done this type of question before.

$(f*h)(3)=\displaystyle\int^{\infty}_{-\infty}f(x)h(3-x)dx=\int^{\infty}_{\infty}h(x)f(3-x)dx$

But f(x) is only defined in {-1<=x<=1|x$\in\mathbb{Z}$}

h(x) is only defined in {1<=x<=5|x$\in\mathbb{Z}$}

So Idk what does it mean by integrating f(x)h(3-x) from infinity to negative infinity... How on Mars could I do that?

But u = e^(-st) so:

$\dfrac{u}{s^2}(\ln u-1)+k\\ =\dfrac{e^{-st}}{s^2}(-st-1)+k$

I'd guess "8"s are actually "i"s and 5 is "s" and w is "n"

Just guessing

Yes. 

A triangle with a 90-degree-angle is called a "right triangle"(Am) or a "right-angled triangle"(Br)

It's not like that...

$\boxed{a-b=-(b-a)}$

That's because $\boxed{-(-a)=a}$

So 14 - 32 = -(32 - 14) 

You do the stuff in the bracket: = -18 :D

32 - 14 is elementary math :)

Mean of these numbers = 20

Median of these numbers = 12

Mode of these numbers = 12

Range is from 2 to 52.

Definitely, you did something wrong.

$a=1+\sqrt{7a-7}\\ a-1=\sqrt{7a-7}\\ 7a-7=(a-1)^2\\ a^2 - 2a + 1 = 7a - 7\\ a^2 - 9a + 8 = 0\\ (a-1)(a-8)=0\\ a=1\text{ or }8$