trumpstinks

\(QED = (11-5i)(3-2i)(2i)\)

We can split this up into two parts. Let's first compute \((11-5i)(3-2i)\), and then multiply it again by \(2i\).

\((11-5i)(3-2i) = 11(3-2i)-5i(3-2i) = 33-22i-15i+10i^2 = 33-37i-10 = 23-37i.\)

\(2i\cdot(23-37i) = 46i-74i^2 = \boxed{74+46i}\)

jimkey17, you had the right idea, but you made a mistake in your work. It should be choose 5, not 4. The real answer is 11, as 11 choose 4 > 365, but 10 choose 4 isn't. (Note: I solved this problem on AoPS, and 11 is correct, so you need have no doubts in the solution)  

I already tried that response a long time ago. It was wrong. I'm confused.

Since \(p\), \(q\), and \(r\) must all be positive, \(p = 1\) because if it was any higher, \(q\) and \(r\) would need to be negative to make equation true. So we have \(1+\frac{1}{q+\frac{1}{r}}=\frac{25}{19}\). So \(\frac{1}{q+\frac{1}{r}}=\frac{6}{19}= \frac{6}{6q+\frac{6}{r}}\), so \(19=6q+\frac{6}{r}\). The only possibility of \(q\) is \(\boxed{3}\), with \(r\) being \(6\).

thanks!! I thought it was only integers ;(

The answer is obviously 39, which is not the hypotenuse in any pythagorean triples.

The average speed equals \(\frac{total distance}{total time}\). The total time is the time going uphill + the time going downhill. This is \(\frac{100}{30}+\frac{50}{60} = \frac{125}{30}=\frac{25}{6}\). So the fraction\(\frac{total distance}{total time} = \frac{150}{\frac{25}{6}} = 150/25*6=\boxed{36}\)

You can use guess and check and find the sum of squares that satisfy the inequality, because of the Pythagorean Theorum, a^2+b^2 = c^2. In this case, c, the hypotenuse is equal to sqrt(n), so a^2+b^2 = n. By using guess and check, we find that the leg lengths of 3 and 6 satisfy this, 3^2 + 6^2 = n, so n is \(\boxed{45}\)

First post:

The two letters that appear twice in the word BRILLIANT are I and L. The probability that I choose an I is \(\frac{1}{9}\), and the probability that I pick another I right after is 1/8 (as I already picked a letter, so there is one less to pick from). So the probability of picking two I's is \(\frac{1}{9}*\frac{1}{8}=\frac{1}{72}\), and the same goes for L too, so the probability of picking the same two letters in a row is \(\frac{2}{72}=\boxed{\frac{1}{36}}\)

Edited:

The two letters that appear twice in the word BRILLIANT are I and L. The probability that I choose an I is \(\frac{2}{9}\) (2 I's to pick from) , and the probability that I pick another I right after is 1/8 (as there is only one "I" left to pick, and there are now only 8 letter to pick from). So the probability of picking two I's is \(\frac{2}{9}*\frac{1}{8}=\frac{2}{72}=\frac{1}{36}\), and the same goes for L too, so the probability of picking the same two letters in a row is \(\frac{1}{36}+\frac{1}{36}=\boxed{\frac{1}{18}}\)

There isn't an actual strategic way I know of. For 1, I would have x = 0, y = 0, and z = 1, or x = 1, y = 0, z = 0, etc., for 2 I would have two of the three variables be 1, and so on. Like I said before, there isn't exactly a strategic way, and you might run into some radicals that aren't integers, like with four.

Next time put (-1) in parentheses, so that there aren't two operators next to each other, which makes things very messy! But, I think I understand your expression is -1-(-1)+(-1x)-(-1)+5x-1-2-5x-1-2. This simplifies to -1+1-x+1+5x-3-5x-3 = -x+1+5x-3-5x-3 = -x-5. The solution is -x-5.

Arc IF is 180 degrees. Since arc CH and GF are congruent, arcs CH, BG, HI, and GF are all congruent, as this is in a circle. 180 degrees - 60 degrees = 120 degrees = CH + BG + HI + GF. Since they are all congruent, we can say that all of the arcs equal BG, so 120 degrees = 4BG, so 30 degrees = BG. The answer is a.