BuilderBoi

a) The area of the triangle is \(33 \times 56 \div 2 = 924\)

The hypotenuse of the triangle is \(\sqrt{33^3 + 56^2} = 65\). 

So, the length of m is \(924 \div 56 \times 2 = {\color{brown}\boxed{1848 \over 65}} \approx 28.4307\)

\(x + y+ xy = 39\)

\(x( 1 + y) + y = 39\)

\(x(1+y) + 1 + y = 39 + 1\)

\(x(1+y) + 1(1+y) = 40\)

\((x+1)(y+1)=40\)

The only pairs that work are \((0,39)\), \((1, 19)\), \((3, 9)\), and \((4, 7)\)

The product of sums and products are 0, 380, 324, and 308. 

So, the largest possible value is \(\color{brown}\boxed{380}\)

\( x\in (-\infty, -2)\cup(3,\infty)\)

So we have \((-2)^2 -2b + c = 0\), meaning \(-2b + c = -4\)

We also have \((3)^2 + 3b + c = 0\), meaning \(3b + c = -9\)

Solving the system gives us \(b = -1\) and \(c = -6\).

Thus, \(b + c = -1 - 6 = \color{brown}\boxed{-7}\)

\(3x+\frac{2}{x}-4=2(x+9)-(7-\frac{1}{x})\)

\(3x+\frac{2}{x}-4=2x + 18-7+\frac{1}{x}\)

\(3x + {1 \over x} - 4 = 2x + 11\)

\(x + {1 \over x} - 4 = 11\)

\(x + {1 \over x} = \color{brown}\boxed{15}\)

Note that \(x^2 + y^2 = (x+y)^2 - 2xy\)

So, \(17^2 - 2xy = 145\)

Now, solve for xy.