Melody

do you get 11% each year or 11% at the end of 10 years. ?

The question is not clear.   

One universal symbol for exponent is ^

5^6

looks like this

$${{\mathtt{5}}}^{{\mathtt{6}}} = {\mathtt{15\,625}}$$

what is 460sqm as a percentage of 174,320sqm?

460/174320 *100   then put a percent sign.  

On the web2 calc it usually comes out as both. 

it does not matter what the original population is.

$$\\1.03=1(1+r)^{10}\\\\
1.03=(1+r)^{10}\\\\
log(1.03)=log(1+r)^{10}\\\\
log(1.03)=10log(1+r)}\\\\
10^{\left(\frac{log(1.03)}{10}\right)}=10^{log(1+r)}}\\\\
10^{\left(\frac{log(1.03)}{10}\right)}=1+r\\\\
r=10^{\left(\frac{log(1.03)}{10}\right)}-1\\\\$$

$${{\mathtt{10}}}^{\left({\frac{{log}_{10}\left({\mathtt{1.03}}\right)}{{\mathtt{10}}}}\right)}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0.002\: \!960\: \!253\: \!145\: \!653\: \!4}}$$

so the yearly rate of increase is approximately    0.296%

P (NOT rolling a particular number in n trials) = $$\left(\frac{5}{6}\right)^n$$

so

P(Rolling a particular number in n trials ) =      $$1-\left(\frac{5}{6}\right)^n$$

$${\mathtt{1.064}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{19}}}{\mathtt{\,-\,}}{\mathtt{8.074}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{20}}}$$

$$\\10^{-20}(1.064*10-8.074)\\
=10^{-20}(10.64-8.074)\\
=2.566\times 10^{-20}$$

It is a joke wildhorse.  Well i thought it was funny anyway :))

Using Vectors

\vec{V_1}

$$\vec{v_1}$$

Binomials

\displaystyle\binom{5}{i}

$$\displaystyle\binom{5}{i}$$

\displaystyle \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{5-1}} *(-1)^i*\binom{5}{i}* (5-i)^3 \hspace {15}| \hspace {15} \Text {N=5 \; k=3} \\\

$$\displaystyle \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{5-1}} *(-1)^i*\binom{5}{i}* (5-i)^3 \hspace{15pt}| \hspace{15pt} \Text {N=5 \; k=3} \\\$$

Partial Derivatives

\dfrac{\partial z}{\partial x}

$$\dfrac{\partial z}{\partial x} =0.3x^{-0.7}y^{0.5}$$

you are right Chris,

I can at time be difficult to avoid Kool-Aid