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!