tertre

2. We have $y=2+\frac{1}{4}x$

and $3x-5y=-3$

Plug in the value of $y$ for $x$, so $3x-5(2+\frac{1}{4}x)=-3$ . Simplifying, we get $3x-10-\frac{5}{4}x=-3$ , so     $\frac{7}{4}x=7, x=4$ and the value of $y$ equals . $2+1=3$ . Thus, $\boxed{(x,y)=(4,3)}.$

3. I can't really see the LATEX, but $\frac{\left(x-1\right)\left(x+4\right)}{2x-1}=\boxed{\frac{x^2+3x-4}{2x-1}}.$

5+(n-1)(4)=-111, (n-1)(4)=-116, n-1=-29, so n=30, 30 terms. 

$112.65 rounds to $113

$29.75 rounds to $30

$48.33 rounds to $48

So, 113+30+48=$191, which is very close to $190, the third option.

12. Taking the x's and the constants to the other side, we get $y=\frac{-6-x}{3x-1}.$

Okay, here we go: Let's start by finding the midpoint of $s_1.$ We have $(\frac{7+\sqrt{2}}{2}, 6)$ and $(\frac{9-\sqrt{2}}{2}, 4).$ Now, we take the midpoint of these two numbers, yielding us with $\boxed{(4, 5)}.$

Remember, when a trapezoid is orthodiagonal, meaning having a 90 degrees angle; a right trapezoid, the sum of the two bases divided by 2 is the altitude. In this case, we have the two bases as 8 and 12, so $\frac{8+12}{2}=10$ and by using the formula for the area of a trapezoid, we get $\frac{1}{2}*20*10=\boxed{100}.$

Hope this helped! 

Since the trapezoid is orthodiagonal or right, we have $a=\frac{8+12}{2}=10,$ where $a$ is the altitude. Thus, the area is 100.

180(n-2), 180*3=540. So, 540/5=108 and the answer is 108.

Thank you, CPhill!