BuilderBoi

Let \(\angle IAB = x\) and \(\angle IBA = y\)

We know that \(x + y + 106 = 180\), meaning \(x + y = 74\)

We also know that \(\angle C = 180 - 2x - 2y = 180 - 2(x+y)\)

So, \(\angle C = 180 - 2 \times 74 = \color{brown}\boxed{32}\)

Simplify: 

\(a(x+b(x+3)) = 2(x+6)\)

\(a(x+bx+3b) = 2x + 12\)

\(ax+abx+3ab = 2x + 12\)

Equate coefficicents:

\(ab = 4\)

\(a + ab = 2\)

Solving this system, we find \(a = -2\) and \(b = -2\) so \(a + b = \color{brown}\boxed{-4}\)

Here's a hint: The 3 numbers can only sum to a square if the square is divisible by 3

\(-7 = f(2)\)

  \(-7 = f(5-3)\)

\(\color{brown}\boxed{(5, -7)}\)

Multiply the numerators: \(12x^{3 \over 2} \times 5 \times 2 \times x^{1 \over 2}\times 6x^{32} = 720x^{34}\)

Multiply the denominators: \(3x^{24} \times 2x^{5 \over 2} \times 3x^{7 \over 2} = 18x^{30}\)

So, \({720x^{34} \over 18x^{30}} = \color{brown}\boxed{40x^4}\), just as proyaop found 

14 fl oz per ____? 

Because \(x \in (-\infty, 2)\cup(8,\infty)\), we know when x = 2, y = 0 and when x = 8, y = 0. 

So, the quadratic is \((x-2)(x-8) = \color{brown}\boxed{x^2 - 10x + 16}\)

There is a way to solve this without finding the roots of the quadratic. 

The equation simplifies to \(0 = x^2 -6x + 11\). The sum of the roots is \(-{b \over a} = 6\) and the product is \({c \over a} = 11\)

The new quadratic would be \((x-a^2)(x-b^2)\), which simplifies to \(x^2 - (a^2 + b^2)x + a^2b^2\)

Note that \(a^2 + b^2 = (a+b)^2 - 2ab = 36 - 2 \times 11 = 14\) and \(a^2 b^2 = (a\times b)^2 = 11^2 = 121\).

So the quadratic is \(\color{brown}\boxed{x^2 - 14x + 121}\)

How many days are there in a week?

Now, multiply that by 0.5.

Here's my approach: 

Each of the triplets receives 0 stickers: Then there are 13 stars and 2 bars so there \({15 \choose 2} = 105\) ways to distribute the stickers.

Each of the triplets receives 1 sticker:   Then there are 10 stars and 2 bars for \({12 \choose 2} = 66\) ways to distribute the stickers.

Each of the triplets receives 2 stickers: Then there are 7 stars and 2 bars for \({9 \choose 2} = 36\) ways to distribute the stickers.

Each of the triplets receives 3 stickers: Then there are 4 stars and 2 bars for \({6 \choose 2} = 15\) ways to distribute the stickers. 

Each of the triplets receives 4 stickers: Then there are 1 star and 2 bars for \({3 \choose 2} = 3\) ways to distribute the stickers. 

So, there are \(105 + 66 + 36 + 15 + 3 = \color{brown}\boxed{225}\) ways to distribute the stickers.