Melody

Question 5 is right.  

Question 11   There may be a 'normal' method for doing this, I am not a physics specialist.

$$magnitude\\
=\sqrt{7.6^2+5.2^2}\\
=9.20869\\
=9.2$$

$$\\tan\theta=\frac{5.2}{7.6}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.2}}}{{\mathtt{7.6}}}}\right)} = {\mathtt{34.380\: \!344\: \!723\: \!845^{\circ}}}$$

direction = 180+34.38 = 214.38 degrees

So the answer is B      9.2 and 214.38 degrees with the positive x axis.

If you do not understand what I have done then make sure you ask me to explain more.  

How about

$$\frac{1}{3}=\frac{3(8-7)(6-5)(4-2-1)}{9}$$

Ninja you should include all these calculator instrutions in one of your threads! Are you already doing that?

They are really good 

$$\\(x+2)(x+1)-2(x+5)\\
=x(x+1)+2(x+1)-2(x+5)\\
=x^2+x+2x+2-2x-10\\
=x^2+x-8\\$$

$$\begin{array}{rlll}
2x-5y&=&3\; &(1)\\\\\\
\frac{-2}{5}x + y &=& \frac{3}{10}\qquad &(2a)\\\\
5(\frac{-2}{5}x + y) &=&5( \frac{3}{10})\qquad\\\\
-2x + 5y &=&1.5\qquad &(2b)\\\\
2x - 5y &=&-1.5\qquad &(2c)\\\\
\end{array}$$

It now becomes clear that the 2 equations are parallel lines hence there is no simultaneous solution!

I shouold have recognised this from the beginning but better late than never!

-20+18

and that is it

mult by 2 then add the zeros.

$${\frac{{\mathtt{3\,043}}}{{\mathtt{897\,960}}}} = {\mathtt{0.003\: \!388\: \!792\: \!373\: \!825\: \!1}}$$

I just put it straight on the web2 calc.

$${\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\mathtt{8}} = -{\mathtt{60.740\: \!078\: \!950\: \!105\: \!126\: \!8}}$$