thelizzybeth

Add both equations up and the y's cancel out. So we have:

2x = 18

x = 9

Plug this into our first equation:

9 + y = 15

y = 6

So (9, 6)

I say this for every geometry problem I solve, but I really wish I had a diagram. In order for my solution to make sense, please draw out what I wrote below while reading it. :)

Extend line BA to the other side of the circle. Let's call the point where it touches the circle F. Inscribed angle CBF, as given by the problem, is x. So arc CF is 2x because a central angle/arc equals twice the inscribed angle.

Arc CE is 6x because ∠CDE=3x.

Since∠BAE=90, and line BAF is straight, then EAF is also ∠90.

arc CF + arc CE = arc EAF

2x + 6x = 90

8x = 90

x = 11.25

I'm not sure if my solution is correct because I didn't utilize ∠BCD=2x.

Put the point (4, -3) in.

1-k*4 = 9

4k = -8

k = -2

No worries! :)

28! means 28 factorial which is 28*27*26...*1. 

27! is 27*26*25...*1.

So we can take out 27!. 

Thank you! I appreciate the kind words :)

100 can be expressed as 25*4. 

We can multiply this by 6 to get 25*24. 

So the smallest positive multiple of 100 is 100*6=600

Since both have 27 factorial, we can factorize to get 27! (1+28) = 27! * 29. 

The largest prime that divides 27! + 28! is 29. 

The question asks that the two R's are adjacent and the two O's are adjacent. RR and OO don't have to be adjacent to each other. It could be xooxxrrx

I think you intepreted it as "The two R's and the two O's are adjacent". Then you would be correct.

The total number of ways to arrange TRAPDOOR is \(\frac{8!}{2!*2!}\). 8 choices for the first letter, 7 choices for the second etc. We divide by 2!*2! because there are repeating cases with the two R's and the two O's. 

Now onto the cases that the two R's are adjacent and the two O's are adjacent. 

Think of the two R's as a "block" or a "bundle". Think of the two O's the same way.

So now we find the number of ways to arrange 6 letters (The 4 distinct letters T, A, P, D & the 2 "bundles" of R and O). This is just 6!. We don't have to divide by anything because they are all distinct.

\(\frac{6!}{\frac{8!}{2!*2!}}\)= \(\frac{1}{14}\)