tertre

I have finally solved the problem. Oh, you changed the problem! So, is the answer \(\boxed{-\frac{137}{60}\quad}.\)

Now, the LATEX: \(\left(\frac{\left(-\frac{3}{5}+\frac{1}{10}\cdot \:3\right)}{\left(\frac{\frac{9}{20}}{\left(-\frac{1}{4}-\frac{1}{5}\right)}\right)}\right)-\frac{3}{4}-\frac{\frac{2}{5}}{\frac{\left(\frac{1}{20}-\frac{1}{4}\right)}{\left(-\frac{5}{6}-\frac{1}{13}\left(\frac{5}{4}-\frac{1}{6}\right)\right)}}\)

All problems can be handy by learning the Principle of Inclusion-Exclusion formula. Look at CPhill's solution for more detail!

2. Remember that OR=addition, so we have 6/20+2/20=8/20 or 2/5.

Heron's: \(\sqrt{s(s-a)(s-b)(s-c)}\), where \(s\) is the semiperimeter and \(a,b,c\) are the sides. The semiperimeter can be found by \(\frac{A+B+C}{2}=\frac{4+6+8}{2}=\frac{18}{2}=9.\) Now, we have \(\sqrt{9(9-4)(9-6)(9-8)}=\sqrt{9*5*3*1}=\sqrt{135}=3\sqrt{15}.\)

Last one, then I have to go.

We have \(y=x^2-6x+8\) and \(y=-2x+5\) . We know that both equations are the same, since we have found \(y\) in terms of \(x.\) Therefore, \(x^2-6x+8=-2x+5\) . This time, by factoring we get \(x=1, x=3,\) so when \(x=1\), \(y\) equals   \(y=-2(1)+5=-2+5=3\) and when \(x=3, \) \(y\) equals \(y=-2(3)+5, y=-6+5, y=-1.\) Thus, the ordered pair is \(\boxed{(1,3), (3,-1)}.\)

We have \(9x, 10y, -9y, 10\), so our like terms are  \(10y, -9y\) or the first option. (A). 

I think I can convert this into LaTex. Does it look like this \(\left(\frac{\left(-\frac{3}{5}+\frac{1}{10}\cdot \:3\right)}{\left(\frac{\frac{9}{20}}{\left(-\frac{1}{4}-\frac{1}{5}\right)}\right)}\right)-\frac{3}{4}-\frac{\frac{2}{5}}{\frac{\left(\frac{1}{20}-\frac{1}{4}\right)}{\left(-\frac{5}{6}-\frac{1}{13}\left(\frac{5}{4}-\frac{1}{6}\right)\right)}}\) . Sorry, it's a bit hard to read. First, try to start from the bottom.

I think you mean: \(y=x^2-5x-6\) and \(x+y=-3.\) Taking \(x\) to the other side in our second equation, we have \(y=-3-x\) . This means \(-x^2-5x-6=-3-x\), so by using the quadratic formula, we get \(x=-3,\:x=-1.\) . 

From here, first by plugging \(x=-3\) in the second equation to solve for \(y\), we get      \(​​-3+y=-3, y=-3+3, y=0.\) Consequently, by plugging in \(x=-1\)\(\), we get    \(-1+y=-3, y=-3+1, y=-2.\) Finally, our two ordered pairs are \(\boxed{(-3,0), (-1,-2)}\) , so (A) or the first option. 

All of angles are interior angles, right?

So, \(x+(2x+4)+60=180, 3x+4=120, 3x=116, x=\frac{116}{3}.\)

joe paid=x

mary paid=x

m=j+2, joe paid: 2/5j-1 and mary paid 1/3(j+2). Now, we have an equation: 2/5j-1=1/3j+2/3, 1/15j=1 2/3, so j=5/3*15, j=25.

So, Mary paid 25+2=$27. Thus, the answer is $25+$27=$52.