AsianMathGuy

We can use Heron's formula to first find the area of the triangle, \(\sqrt{s(s-a)(s-b)(s-c)}\). Since the semiperimeter is \(17\), we have 

\(\sqrt{17(17-15)(17-9)(17-10)} = \sqrt{1904} = 4\sqrt{119}\)

as the area of triangle \(ABC\). The shortest altitude of a triangle has the longest side as its base, so we have 

\(\frac{15h}{2} = 4\sqrt{119}\),

where \(h\) is the altitude. Solving for \(h\) yields \(h = \boxed{\frac{8\sqrt{119}}{15}}\).

^{(I'm not sure if there's a way to solve this without using Heron's, but if there is, please let me know! Thanks.)}

Simplifying we get \(x^2 + 12x - 4 = 0\). Using the quadratic formula we find the roots to be \(-6 + 2\sqrt{10}\) and \(-6 - 2\sqrt{10}\). Taking their squares, we get \(36 + 4\sqrt{10} + 40\) and \(36 - 4\sqrt{10} + 40\). Adding them together, we get \(\boxed{152}\)

Expanding \((x-2)(x+2)\) gives \(x^2 - 4\). In order to get it into the form \(Ax^2+Bx+1\) where the constant term is \(1\), we can divide the original quadratic by \(-4\) to get \(-\frac{1}{4}x^2 + 1\), so \(A+B = -\frac{1}{4} + 0 = \boxed{-\frac{1}{4}}\)

Or did you mean to find the units digit of \(13^{287}\)? It's a bit more complicated but still doable.

The units digit of both numbers are \(3\) and \(7\), so \(3\cdot7 = 21\), where the units digit is \(1\). The units digit of \(13\cdot287\) is \(\boxed{1}\)

The slope of the line is \(\frac{150-17}{82-3} = \frac{133}{79}\) .(This cannot be further simplified). Starting from point \((3,17)\), the next point with integer coords will be \((82, 150)\), which is the second endpoint. So there are only \(\boxed{2}\) lattice points in the line, that being the two endpoints.

If we simplify it we get \(x^2 + 8x + 9\). The discriminant is \(b^2 - 4ac\) so we get \(64 - 4\cdot9 = 28\)

\(28\) is the discriminant.