zacismyname

The lines are parralel. This means they have the same gradient. 

First we find the gradient of RS.

The formula for gradient is: m = (y{2}-y{1})/(x{2}-x{1}).

 y{2}=0, y{1}=3, x{2}=-1, x{1}=5 

$${\mathtt{m}} = {\frac{\left({\mathtt{0}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{5}}\right)}}$$

$${\mathtt{m}} = {\frac{\left(-{\mathtt{3}}\right)}{\left(-{\mathtt{6}}\right)}}$$

$${\mathtt{m}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$

Now we can sub our gradient into the equation for line PQ's gradient.

m = (y{2}-y{1})/(x{2}-x{1})

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\frac{\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}}$$ (keep in mind the y corresponds to point Q)

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\frac{\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left(-{\mathtt{5}}\right)}}$$ Now we solve for y.

$$-{\mathtt{5}} = {\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)$$

$$-{\mathtt{5}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}$$

$$-{\mathtt{7}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{y}}$$

$${\mathtt{y}} = -{\mathtt{3.5}}$$

We can check this by substituting y into the equation. Does (-3.5 --1)/(-5) = 1/2? 

This is our y coordinate for point Q.

So point Q = (-2, -3.5)

Hope this helps :)

length times width or length^2

$$11\frac{1}{4}-?=\frac{1}{2}$$

$$11\frac{1}{4}-10\frac{3}{4}=\frac{1}{2}$$ (for addition simply add $$-10\frac{3}{4}$$)

$$11\frac{1}{4}/?=\frac{1}{2}$$

$$\frac{11\frac{1}{4}}{22\frac{1}{2}}=\frac{1}{2}$$

$$11\frac{1}{4}\times?=\frac{1}{2}$$

$$11\frac{1}{4}\times\frac{2}{45}=\frac{1}{2}$$      (I found this by saying 11.25*x = 0.5  therefore 0.5/11.25 = x = 2/45)

I'm not sure if this is what you were asking but I hope it helped anyway. 

$$\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}}\right){\mathtt{\,\times\,}}\left({\mathtt{4.3}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{6}}}\right)$$

$${\mathtt{200\,000}}{\mathtt{\,\times\,}}{\mathtt{4\,300\,000}}$$ 

$${\mathtt{860\,000\,000\,000}}$$

$${\mathtt{8.6}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{11}}}$$

or because its all multiplication;

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{4.3}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{6}}}$$

$${\mathtt{8.6}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{11}}}$$

$$\left({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)$$

$$\left({\frac{{\mathtt{17}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right)$$

$${\frac{\left({\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{1}}\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}}$$

$${\frac{{\mathtt{17}}}{{\mathtt{15}}}}$$ or  $${\mathtt{1}}$$$${\frac{{\mathtt{2}}}{{\mathtt{15}}}}$$

The formula for the volume of a pyramid is $${\mathtt{V}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{base}}{\mathtt{\,\times\,}}{\mathtt{height}}$$

so $${\mathtt{V}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}\left({\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{21}}\right){\mathtt{\,\times\,}}{\mathtt{15}}$$         (The brackets aren't needed I just wanted to separate the base)

$${\frac{{\mathtt{0}}}{{\mathtt{0}}}} = {\mathtt{C}}$$

$${\mathtt{0}} = {\mathtt{0}}{\mathtt{\,\times\,}}{\mathtt{C}}$$

$${\mathtt{0}} = {\mathtt{0}}$$

Therefore C can be any number. It is "indeterminate". I didn't make this up its from a book called "50 Mathematical Ideas You Really Need to Know" by Tony Crilly. 

A stomach ache?

Some people have dedicated a lot of time to pi, hoping that one day a pattern of numbers will repeat. This would allow them to write it as a fraction but there would still be no last digit. Sadly for them it still remains irrational.

1 and itself . . . ?