BuilderBoi

Notice that the number of ways to get to point C is the sum of the number of ways to get to point A and point B. We can then use this logic on all the other points until we get to point I. 

There is 1 way to get to point A and 1 way to get to point B, so there are 2 ways to get to point C. 

There are 2 ways to get to point C and 1 way to get to point B so there are 3 ways to get to point D. 

There are 3 ways to get to point D and 2 ways to get to point C so there are 5 ways to get to point E. 

There are 5 ways to get to point E and 3 ways to get to point D so there are 8 ways to get to point F. 

There are 8 ways to get to point F and 5 ways to get to point E so there are 13 ways to get to point G. 

There are 13 ways to get to point G and 8 ways to get to point F so there are 21 ways to get to point H. 

There are 21 ways to get to point H and 13 ways to get to point G so there are $\color{brown}\boxed{34}$ ways to get to point D. 

To get a constant term, we need the z's to cancel out. Notice that we can't eliminate the z with less than 4 $2z$'s and ${1 \over z \sqrt z}$'s. The closest we can do is $(2z)^3 \times ({1 \over z \sqrt z})^2$, but that uses 5, so it doesn't work. 

Thus, the only constant term will be the 1 multiplied by itself 4 times, which is $1^4 = \color{brown}\boxed{1}$

Let $x = \log_b ({2 \over 3})$ and $y = \log_b ({2 \over 3})$

Then $b^x = {2 \over 3}$ and $b^y = {3 \over 2}$

Multiplying the two gives us $b^x \times b^y = b^{x + y} = 1$

Now, notice that the only way the expression equals 1 is if $x + y = \color{brown}\boxed{0}$.

Note: In general, $\log_b({x}) + \log_b({y}) = \log_b({xy})$

$\sqrt{7! + 6!}$

$\sqrt{6!(7 + 1)}$

$\sqrt{720 \times 8}$

$\sqrt{2^7 \times 3^2 \times 5}$

$24 \sqrt{2 \times 5}$

$\color{brown}\boxed{24 \sqrt{10}}$

$(2x+10)^4(x+5)^4$

$(2(x+5))^4(x+5)^4$

$16(x+5)^4(x+5)^4$

$\color{brown}\boxed{16(x+5)^8}$

Simplify the left-hand side to $x^2 + 28x - 21$.

Now, notice that the right-hand side is just vertex form. This means that a is the coefficient of the $x^2$ term (in this case it's 1) and that $(h, k)$ is the vertex. 

The vertex is located at $x = -{b \over 2a} = -14$. Substituting this in gives us $(-14)^2 + 28(-14) - 21 = -217$. So, the vertex is $(-14, -217)$.

Thus, the ordered triple is just $\color{brown}\boxed{(1, -14, -217)}$

Let's divide this problem up into cases:

Case 1 (2 1's, 2 2's, 3 4's) - There are ${7! \over 2! 2! 3!} = 210$ cases.

Case 2 (1 1's, 3 2's, 3 4's) - There are ${7! \over 3! 3!} = 140$ cases.

Case 3 (1 1's, 2 2's, 4 4's) - There are ${7! \over 2! 4!} = 105$ cases.

Case 3 (1 1's, 2 2's, 3 4's, 1 3, 5, or 6) - There are ${7! \over 2! 3!} \times 3 = 1260$ cases.

So, there are $210 + 140 + 105 + 1260 = \color{brown}\boxed{1,715}$ sequences.