MathProblemSolver101

$r^2+s^2 = (r+s)^2 - 2rs$

$r+s = - \frac{4}{3}$

$rs = -4$

\(\begin{align*} r^2 + s^2 &= \left(- \frac{4}{3} \right)^2 - 2 (-4) \\ &= \frac{16}{9} + 8 \\ &= \boxed{\frac{88}{9}} \end{align*}\)

Picture: https://ibb.co/rMsSd3h

Let $a, b, c$ be the radii of $A, B, C$ respectively.

We have: \(\begin{align*} a+b&=14\\ b+c&=18\\ a+c&=16 \end{align*}\)

Add all equations: $2(a+b+c)=48$

Thus: $a+b+c=24$

Subtract $a+b=14, b+c=18, a+c=16$ from $a+b+c=24$ and you will get $c=10, a=6, b=8$

Thus $(a,b,c) = (6,8,10)$

Interesting how this is the length of a Pythagorean triple... probably a coincidence

WLOG, let there be $6+3+1=10$ games played.

There are 6 wins, 3 losses, and 1 tie.

a) $\frac{6}{10} = \frac{3}{5}$

b) $\frac{3}{10}$

c) $\frac{1}{10}$

d) $6 \cdot 5 = 30$ wins, $3 \cdot 5 = 15$ losses, and $1 \cdot 5 = 5$ ties.

e) $6 \cdot 8.2 = 49.2$ wins, $3 \cdot 8.2 = 24.6$ losses, and $1 \cdot 8.2 = 8.2$ ties. This is absurd, so I am positive that (e) is wrongly written.

This problem can be rephrased into: What is the x-value of the vertex of the parabola $t^2 - 12t - 36$?

The formula for $x$ is $- \frac{b}{2a},$ for polynomial $ax^2 + bx + c.$

We have: $-\frac{-12}{2} = \boxed{6}$

You can see that the graph shown here: https://www.desmos.com/calculator/fztszalids is least when $x=6.$

First off, we do offer explanations. Most of those answer-only guest answers are wrong and they are misleading the OP, which the OP kind of deserves due to not paying attention in class. Obviously, you attended a class and that is why you have that as your homework. If you played video games or such, they obviously don't know how to do their homework. I daresay they are doing video games or watching videos after they post the question, waiting for a response. I realise that some people take all the work and copy it and mark it as their own, but that's warez. Cheating. Plagiarism. Even if they don't get caught, realize that if they don't learn, it's their problem, and so they won't have as smooth life as others. You don't need to beat a dead horse.

This looks a lot like catmg's word problem: https://web2.0calc.com/questions/math_66748#r1

@amygdaleon305, nice solution! Next time, consider using the \begin{align*} environment.

Your solution in that environment(you can view the code by double-clicking the TeX and clicking view TeX commands):

\(\begin{align*} x^2+13x+30 &= x^2+3x+10x+30 \\ &= x(x+3) + 10(x+3) \\ &= (x+3)(x+10) \end{align*}\)

$\Rightarrow$ $a = 3$ and $b = 10$

\(\begin{align*} x^2+5x-50 &= x^2+10x-5x-50 \\ &= x(x+10) - 5(x+10) \\ &= (x+10)(x-5) \end{align*}\)

$\Rightarrow c = 5$

$a + b + c = 3 + 10 + 5 = \boxed{18}$

$x^2 + 13x + 30 = (x+10)(x+3)$

$x^2 + 5x - 50 = (x+10)(x-5)$

$10 + 3 + 5 = \boxed{18}$