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.