BuilderBoi

The first digit will represent \(7^3\). The next will represent \(7^2\), then \(7^1\), and finally \(7^0\), so there are \(\color{brown}\boxed{4}\) digits.

As long as the top part \(\sqrt{x-2} \ge 0\) is positive, the function is defined. 

Can you take it from here?

What figure? 

\(400,000 = 4 \times 10^5\)

\((4 \times 10^5)(4 \times 10^5) = (4 \times 4)(10^5 \times 10^5)=16 \times 10^{10}\)

\(16 \times 10^{10} = 1.6 \times 10^{11}\)

\(n = \color{brown}\boxed{11}\)

By the Angle Bisector Theorem, we have \({AY \over AB} = {YC \over CB}\). Plugging in the given info, we have \({6 \over 8} = {3 \over CB}\), meaning \(CB = 4\)

Now, apply the Angle Bisector Theorem again: \({AZ \over AC} = {ZB \over CB}\). Subbing in what we know, we have \({AB \over 9} = {ZB \over 4} \ \ \ (i)\).

For simplicity, let \(AB = x \) and \(ZB = y \). We also know that \(x + y = 8 \ \ \ (ii)\).

Solving this system, we find \(x = AB = {72 \over 13}\) and \(y = ZB = \color{brown}\boxed{32 \over 13}\)

From the second sentence, we know \(o + d = 18\)

From the first sentence, we know \(o = 5d\).

Now, just solve or o and d and see how many games Oliver gave to Darius so they have the same number of games. 

Note that because \(\triangle STU\) is isosceles, \(SN\) is also a median. 

Now, draw the median \(TK\). This intersects at \(X\), and by the properties of medians, we know that \(SX = {2 \over 3}\times SN\).

\(\triangle SNU \) is a right triangle with \(SU = 22\) and \(NU = 8 \div 2 = 4\), meaning \(SN = \sqrt{22^2-4^2} = \sqrt{468} = 6 \sqrt{13}\)

So, \(SX = {2 \over 3} \times 6 \sqrt{13} = \color{brown}\boxed{4 \sqrt {13}}\)

Draw altitude \(CF\). This will intersect at the orthocenter, which is H. This means that \(\triangle AFC \) is a right triangle. 

If we know what \(\angle A\) was, we can figure out what \(\angle HCA \) is, because \(\angle FCA = \angle HCA\) and \(\angle F\) is a right angle. 

Now, note that \(\triangle ADB \) is a right triangle, and since we know \(\angle BAH = 28\) and \(\angle ADB = 90\), \(\angle ABD = 62\).

Also, we know that \(\angle ABH = 24\), meaning \(\angle HBD = \angle ABD - \angle ABH = 62 - 24 = 38\).

Now, \(\triangle BEC\) is a right triangle, so we know that \(\angle BCA = 42\).

This means that \(\angle A = 180 - 62 - 42 = 76\), so \(\angle HCA = \angle FCA = 180 - 90 - 76 = \color{brown}\boxed{14}\)

The side length of square A is \(24 \div 4 = 6\), meaning its area is \(6^2 = 36\). 

Square B's area is \(36 \div 4 = 9\), meaning its side length is 3, so the perimeter is \(3 \times 4 = \color{brown}\boxed{12}\)