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!