3 questions

#14 use distance formula:                                    

You have #14 wrong.  It's asking for distance, and you have the midpoint

yeah i didnt, understand that question either

so if anyone could help with that too 

would be much appreciated 

OK...I'm doing 15, 16 and 17....if that's what you need....

(15)...using the Law of Cosines we have

(PQ)^2 = 5^2 + 6^2 - 2(5)(6) cos(100) =

(PQ)^2 = 25 + 36 -60cos(100) 

(PQ)^2 = 61 - 60cos(100)          take the square root of both sides

PQ = [61 - 60cos(100)]^(1/2) = about 8.45 cm.

(16)  Note that if we join the two "bottom" vertices, we have a central angle that is twice the measure of 43 = 86. This is equal to the measure of the minor arc that this central angle intercepts. Then "x" represents the measure of the major arc of the remaining part of the circle = 360-86 = 274.

(17)  (4)/(x+3) + (5)/(x-2) = 10

So..multiplying both sides by  (x+3)(x-2) we have

4(x-2) + 5(x+3) = 10(x+3)(x-2)    ....simplify......

4x-8 + 5x + 15  =10(x^2 +x -6)

9x +7 = 10x^2 +10x  -60   subtract 9x + 7 from both sides

0 = 10x^2 +x -67  this isn't factorable...using  our site's solver we get...

$${\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{67}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{2\,681}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{20}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{2\,681}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{20}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{2.638\: \!918\: \!693\: \!199\: \!923\: \!6}}\\
{\mathtt{x}} = {\mathtt{2.538\: \!918\: \!693\: \!199\: \!923\: \!6}}\\
\end{array} \right\}$$

Yuck....that's not very pretty, is it??? (I also solved the original equation with a solver and got the same answers...!!!)

Also...your answer to (14) is incorrect..I think it should be "5"

Thanks, TaviSage.......