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Well, it's been a while since I've answered a question on this website, and what better way to celebrate with a geometry question :)

So, we don't have a right triangle, so we're gonna have to settle with Law of Cosines. 

Also, note that we have 

$(cos PMQ) = (- cos PMR )$

Now, let's apply Law of Cosines and add subsitute using equation above. 

$PQ^2 = QM^2 + PM^2 - 2(QM * PM) * (-cos PMR) \\ PR^2 = RM^2 + PM^2 - 2 (RM * PM) * ( cos PMR)$

Simplify

$5^2 = 5.5^2 + PM^2 + 2(5.5 * PM)cos(PMR) \\ 8^2 = 5.5^2 + PM^2 - 2(5,5 * PM) cos (PMR)$

Now, we're stuck...but we can add these two equations to keep progressing. 

$5^2 + 8^2 = 2 * 5.5^2 + 2PM^2 \\ 89 = 60.5 + 2PM^2 \\ 28.5 / 2 = PM^2\\ 14.25 = PM^2 \\ \sqrt {14.25} = PM ≈ 3.77$

And we're done!

Also, side note, this is gonna be my 1000 answer on this website which is a weird flex but whatever. 

I might be gone for another while, but hopefully be back in the summer. 

~NTS

The angles theta that work are pi/3 and pi/4.  Hope this helps!

$If\ BD = 8,\ BC = 3,\ CE = 1,\ AC = 19\ and\ AB = 13,\ then\ what\ is\ DE?$

$f_{AB}(x)=\sqrt{13^2-x^2}\\ f_{BC}(x)=\sqrt{3^2-(x-19)^2}\\ 169-x^2=9-x^2+38x-361\\ 38x=169-9+361=521\\ x_B=13.7105\\ y_B=\color{red }\sqrt{-18.9...}$

The triangle ABC, with sides 3, 19, 13 is not possible.

 !

First, we can simplify f(x): $f(x)=-3x^2+6x+27$. Then, we factor out -3 for $f(x)=-3(x^2-2x-9)=-3(x-1)^2+30$.

 From this equation, we get that the vertex of f(x) is (x, y) = (1, 30).

Ok, I found the answer.

We can use polynomial long division to simplify: 

$\frac{5x^2+20x+33}{x^2+4x+7}=5-\frac{2}{x^2+4x+7}$

Then, h(x)<5, and we can find the minimum by completing the square on the quadratic.

$x^2+4x+7=(x+2)^2+3$,

so the minimum is 3. Hence, the least possible value of h(x) is $\frac{13}{3}$, and the range is $\boxed{h(x)\in[\frac{13}{3},5)}$

I solved this problem before.  The answer is 10*sqrt(5).  Hope this helps!

Quotient is x^2 - 11x + 14, remainder is 33x + 5.

The answer is 38.

It says the area of the squares created by the side lengths of the triangle are shown below. Does this mean that 9, 12, 15 are NOT the side lengths of the triangle? its a bit confusing when you dont have an image. 

IF 9, 12, 15 ARE the side lengths of the triangle, it must be a right triangle. So in that case, the key is probably wrong. 

IF 9, 12, 15 are not the side lengths of the triangle, that would mean its either b or d. Now based on the information given, I'm assuming 9, 12, 15 are the AREAS of squares, and the squares each share a side length with the triangle. So that means each side length would be the square root of 9, 12, 15. So when you apply pythagoras, 9 + 12 is not equal to 15, so it should be b. 

Even Function Properties: An even function satisfies the condition f(x) = f(-x).

Odd Function Properties: An odd function satisfies the condition f(x) = -f(-x).

Sum of Even Functions: The sum of even functions is always an even function.

Product of Even and Odd Functions:

Even * Even = Even

Even * Odd = Odd

Odd * Odd = Even

Sum of Even and Odd Functions: The sum of an even and an odd function is neither even nor odd, unless the terms cancel out. 

Guys, I ended up figuring it out :) 

For those who want to know the answer:

It's n+2 choose 2. That's because you can let n be a number of pieces of candy, and let x, y, and z be the number of candies you give each of 3 kids. Now, applying the Hockey Stick Identity, the answer would be n+3-1 choose 3-1, which is n+2 choose 2. 

Completing the square in $x$ and $y$, we get

$\begin{align*} -4 (x + 1)^2 + (y - 1)^2 = 6. \end{align*}$

We can re-write this as

$\begin{align*} \frac{(y - 1)^2}{6} - \frac{(x + 1)^2}{3/2} = 1. \end{align*}$

The center of the hyperbola is at $\boxed{(-1,1)}$