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.