BuilderBoi

Simplify as follows: 

\((3x - 27)^3 + (27x - 3)^3 = (3x + 27x - 30)^3\)

\((3x - 27)^3 + (27x - 3)^3 = (30x - 30)^3\)

\((3(x-9))^3 + (3(9x - 1))^3 = (3(10x - 10))^3\)

\((x-9)^3 + (9x-1)^3 = (10x-10)^3\)

\(-270x^3+2730x^2-2730x+270 = 0\)

\(9x^3 - 91x^2 +91x - 9 = 0\)

Notice that x = 1 is a solution to the equation. Now, divide the left-hand side by (x-1). The result will be quadratic, and you can use the quadratic formula to solve for the other 2 x values. 

Notice that there is only 1 path to get to B from A. 

Then there is 1 way to get to both C and D from B. 

Because you have to get to E from either C or D, there are 1 + 1 = 2 ways to get to E. 

Then there are 2 ways to get to F and G because you have to go through E. 

So then there are 2 + 2 = 4 ways to get to H.

And there is only 1 way to get to I from H, so there are a total of \(\color{brown}\boxed{4}\) paths from A to I.

There are 3 ways to get a \(y^4\) term: \((2y)^4\), \((3y^2) ^2\), and \((2y)^2 \times (3y^2)\)

For the first way, we need 4 \(2y\)'s  and 2 \(-5\)'s, so we have\((2y)^4 \times (-5)^2 = 400y^4\). But we also need to multiply by \({6 \choose 2} = 15\) because there are 15 ways to order them. This makes for \(400y^4 \times 15 = 6000y^4\)

For the second way, we need 2  \(3y^2\)'s and 4 \(-5\)'s, so we have \((3y^2)^2 \times (-5)^4 = 5625y^4\). But like last time, we need to multiply by 15 for the same reason. Thus we have \(5625y^4 \times 15 = 84375y^4\). 

For the final way, we need 2 \(2y\)'s, 1 \(3y^2\)'s, and 3 \(-5\)'s, so we have \((2y)^2 \times (3y^2) \times (-5)^3 = -1500y^3\). But this time, we need to multiply by \({6! \over 3!2!} = 60\). Thus we have \(-1500y^4 \times 60 = -90000y^4\)

So in total, the coefficient of \(y^4\) is \(6,000y^4 + 84,375y^4 - 90,000y^4 = \color{brown}\boxed{375y^4}\)

Let's split this problem into cases: 

Case 1 (1 x 1 isosceles triangle) - There are 36 1 x 1 squares, and each square can have a total of 4 triangles which makes for \(36 \times 4 = 144 \) triangles. 

Case 2 (2 x 2 isosceles triangle) - There are 25 2 x 2 squares, and each square can have a total of 4 triangles, which makes for \(25 \times 4 = 100\) triangles.

Case 3 (3 x 3 isosceles triangle) - There are 16 2 x 2 squares, and each square can have a total of 4 triangles, which makes for \(16\times 4 = 64\) triangles.

Case 4 (4 x 4 isosceles triangle) - There are 9 4 x 4 squares, and each square can have a total of 4 triangles, which makes for \(9 \times 4 = 36\) triangles.

Case 5 (5 x 5 isosceles triangle) - There are 4 5 x 5 squares, and each square can have a total of 4 triangles, which makes for \(4 \times 4 = 16\) triangles.

Case 6 (6 x 6 isosceles triangle) - There is 1 6 x 6 square, and each square can have a total of 4 triangles, which makes for \(1 \times 4 = 4\) triangles.

So, there are \(144 + 100 + 64 + 36 + 16 + 9 + 1 = \color{brown}\boxed{370}\) isosceles right triangles. 

\({x \over x+1} + {3 \over x} = {5 \over 2x}\)

\({x \over x+1} + {6 \over 2x} = {5 \over 2x}\)

\({x \over x+1} = {5 \over 2x}- {6 \over 2x}\)

\({x \over x+1} = {-1 \over 2x}\)

\(2x^2 = -1(x+1)\)

\(2x^2 = -x - 1\)

\(2x^2 + x + 1 = 0\)

\(x = {-1 \pm \sqrt{(1)^2-4(2)(1)} \over 2(2)}\)

\(x = {-1 \pm \sqrt{-7} \over 4}\)

\(x = \color{brown}\boxed{{-1 \pm \sqrt{7}i \over 4}}\)

Rewrite the second equation to \(3a + 2b - mb = k\).

Now, factor out b to get \(3a + (2-m)b = k\). 

For the system of equations to have an infinite number of solutions, the 2 equations must be identical. 

This means that \(2 -m = 2\), so \(\color{brown}\boxed{m = 0}\) and \(\color{brown}\boxed{k = 2}\)

Oh... I get it now. Thanks so much, Melody, Asinus, and Guest!

Why don't you solve it then?

Thanks for the explanation Melody, but why did you reflect AB over x = 2 in particular?

For the quadratic to be factorable, there must be 2 numbers that sum to \(k\) and multiply to \(23 \times -5 =-115\). 

There are 4 pairs of numbers that multiply to \(-115\): \((115, -1)\), \((23, -5)\), \((5, -23)\), \((1, -115)\). 

This means that k can be the sum of these values or \(\color{brown}\boxed{k = -114, -18, 18, 114}\)

What if x is 0? or negative?

Then the distance from \((0, 0)\) to \((1.8, -0.4)\) is \(\sqrt{(1.8)^2 + (0.4)^2} = {\sqrt{85} \over 5} \) and the distance from \((0, 1) \) to \((1.8, -0.4)\) is \(\sqrt{(1.8)^2 + (1.4)^2} = {\sqrt{130} \over 5}\). 

Adding the two gives us \({\sqrt {85 }+ \sqrt {130} \over 5} \approx 4.12426\) but the correct answer was \(\sqrt{17} \approx 4.12311\). 

Sorry, but that's incorrect. The correct answer is actually \(\sqrt{17}\), even though I don't know how to get it. 

Thanks so much for your attempt, though!