BuilderBoi

I don't think it was meant to be that complicated asinus

Simplify as follows: 

\(10 x - y = 2\)

\(-y = 2 - 10x\)

\(y = -2 + 10x\)

The slope is the coefficent of x, or \(\color{brown}\boxed{10}\)

The side will be the length of the segment connecting the two vertices of the triangles. 

This is a triangle with one angle of 120 degrees and 2 sides of 2. 

By the Law of Cosines, the side of the square is \(\sqrt{2^2 + 2^2 - 2 \times 2 \times 2 \times \cos (150)} = {\color{brown}\boxed{\sqrt{8 + 4 \sqrt 3}}} \approx {3.864}\)

\(x^3 - 150 = 6709\)

\(x^3 = 6859\)

\(x = \color{brown}\boxed{19}\)

You're welcome!!

The x-intercept occurs at \((x, 0)\). 

So,  just plug in y = 0 and solve for x.

Let \(m\) be the age of Margaret in 2000, and let \(c \) be the age of Charlie in 2000.

From the first sentence, we know that \(2c = m\)

From the second sentence, we know that \(m + 7 = c + 7 + 7\), meaning \(m = c + 7\), so \(c = 7\) and \(m = 14\)

So, in 2011, Charlie was \(7 + 11 = \color{brown}\boxed{18}\)

Substituting the info gives us a system with 3 equations: 

\(1 = a + b + c\)

\(2 = 4a + 2b + c\)

\(3 = 16a + 4b + c\)

Subtracting the first equation from the second and the second from the third gives us another system: 

\(3a + b = 1\)

\(12a + 2b = 1\)

Solving this system gives us \(a = -{1 \over 6}\) and \(b = 1 {1 \over 2}\)

Substituting this into the first equation gives us \(c = -{1 \over 3}\), so \(a - b - c = -{1 \over 6} - {3 \over 2} + {1 \over 3} = \color{brown}\boxed{-1 {1 \over 3}}\)

Simplify as follows:

 \({a+b} = 5 (a-3b)\)

\(a + b = 5a - 15b\)

\(b = 4a - 15b\)

\(16b = 4a\)

\(4b = a \)

\({\color{brown}\boxed{4}} = {a \over b}\)

There are 4 cases: He plays 3, 4, 5, or 6 rounds of Dance Fever. 

For the first case, he plays 3 rounds of Dance Fever and 9 rounds of Accuracy Ball. There are \({12 \choose 3} = 220\) ways to place the Dance Fever, and the remaining spots are Accuracy Ball.

For the second case, he plays 4 rounds of Dance Fever and 7 rounds of Accuracy Ball. There are \({11 \choose 4} = 330\) ways to place the Dance Fever, and the remaining spots are Accuracy Ball.

For the third case, he plays 5 rounds of Dance Fever and 5 rounds of Accuracy Ball. There are \({10 \choose 5} = 252\) ways to place the Dance Fever, and the remaining spots are Accuracy Ball.

For the final case, he plays 6 rounds of Dance Fever and 3 rounds of Accuracy Ball. There are \({9 \choose 6} = 84\) ways to place the Dance Fever, and the remaining spots are Accuracy Ball.

So, there are \(220 + 330 + 252 + 84 = \color{brown}\boxed{886}\) ways.

I think...

Make a diagram like this: 

The area of the entire square is \((a+b)^2 = a^2 + b^2 + 2ab\)

The area can also be represented as the sum of the area of the 4 triangles and the square in the middle: \(c^2 + 2ab\)

Since they both represent the area of the square, we have the equation \(a^2 + b^2 + 2ab = c^2 + 2ab\)

Subtracting 2ab from both sides gives us \(\color{brown}\boxed{a^2 + b^2 = c^2}\)