All Answers 358099 Answers

$$\\\frac{2^3*2^0}{2}-8*2^{-2}\\\\
=\frac{8*1}{2}-8*\frac{1}{2^{+2}}\\\\
=\frac{4*1}{1}-8*\frac{1}{4}\\\\
=4-2*\frac{1}{1}\\\\
=4-2\\\\
=2$$

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

length x width

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

$$11\frac{1}{4}-10\frac{3}{4}=\frac{1}{2}$$ (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. 

$$\\LHS
=\frac{( ((-5x-3)/(2x-1))-3 )}{( 2((-5x-3)/(2x-1)) + 5)}\\\\
=\left(\frac{(-5x-3)}{(2x-1)}-3 \right )\div \left( 2\times\frac{(-5x-3)}{(2x-1)} + 5\right)\\\\
=\left(\frac{(-5x-3)-3(2x-1)}{(2x-1)} \right )\div \left(\frac{2(-5x-3)+5(2x-1)}{(2x-1)} \right)\\\\
=\left(\frac{(-5x-3)-3(2x-1)}{(2x-1)} \right )\times \left(\frac{(2x-1)}{2(-5x-3)+5(2x-1)} \right)\\\\
=\left(\frac{-5x-3-6x+3}{1} \right )\times \left(\frac{1}{-10x-6+10x-5} \right)\\\\
=\left(\frac{-11x}{1} \right )\times \left(\frac{1}{-11} \right)\\\\
=x\\\\
=RHS \qquad QED$$

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

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

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

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

or because its all multiplication;

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

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

Infinity is an irrational number and so can not be quantified.

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

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

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

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

X square + X square + X square + X square

4 * Xsquare

(2X)square

Correct me if i'm wrong but, isn't is 255 over 225?

Do math homework like you should be doing.

The baker.

By the way, yes, OVER NINE THOUSAND!!!!!!!!!!!

I have chicken for dinner.

It is equvilant to 1 over 9000 or approximately 0.0001111111111111 witht the one repeating.

4 X^2 

4Xsquare

ans: 17/15 or =$${\frac{{\mathtt{17}}}{{\mathtt{15}}}} = {\mathtt{1.133\: \!333\: \!333\: \!333\: \!333\: \!3}}$$ or 1 2/15

Answered here 

http://web2.0calc.com/questions/prove-sin4x-sin2x-cos4x-cos2x

Thanks Chris :)

-1.258 - 0.584

-1.842//

Thanks Sasini but I do not think that is the intended question.

I think the question is

$$5\frac{2}{3}\times\frac{1}{5}$$            can you do that one for us please Sasini ?

5*2/3*1/5

2/3

thatz the answer

10 +9 =19

not 21 answer. Itz wrong answer.......