BuilderBoi

Write as system: 

x = cost of first hotel

y = cost of second hotel

0.075x + 0.095y = 542.25

y = x - 1000

SOlving, we find \(\color{brown}\boxed{x=3,750, y = 2,750}\)

I think it is \(11 \choose 7\). There are 7 ways to choose the boys with orange caps, and the rest have yellow caps. 

Thus, there is \(\color{brown}\boxed{330}\) ways. (Note, \({11 \choose 7} = {11 \choose 4}\), so the answer would be the same either way. 

Here's a better way to get the same answer. 

For groups of 2 including  ₯, there is \(4 \choose 1 \) choices, because 1 of them is already taken. 

For groups of 3 including  ₯, there is \(4 \choose 2 \) choices, because 1 of them is already taken. 

For groups of 4 including  ₯, there is \(4 \choose 3\) choices, because 1 of them is already taken. 

For groups of 5 including  \(₯\), there is \(4 \choose 4\) choices, because 1 of them is already taken. 

Adding everything up, there is \(\color{brown}\boxed{15}\) subsets that contain ₯.

I will number these from right to left, starting with 1, going up to 5. The rightmost one is 1, the leftmost is 5. 

2 1

2 3

2 4

2 5 

2 1 3 

2 1 4 

2 1 5 

2 3 4 

2 3 5 

2 4 5

2 1 3 4 

2 1 3 5 

2 1 4 5 

2 3 4 5 

2 1 3 4 5 

Thus, there is a total of \(\color{brown}\boxed{15}\) subsets with the ₯ symbol. 

It has to be in the form ABCBA. There are 5 choices for A, 5 for B, and 5 for C, making a total of \(\color{brown}\boxed{125}\) choices.

There are 2 cases: The second digit is a 2 or it is a 4

If the second digit is a 2: \(9 \times 1 \times 10 \times 5 = 450\)

If the second digit is a 4: \(9 \times 1 \times 10 \times 9 = 810\)

Add these together for a total of \(\color{brown}\boxed{1,260}\) ways

HAPPY ANIVERSARY CPHILL!!!

You made it so easy to learn new math skills and topics!!!

Because it's a right triangle, and the remaining angles are equal, this is a 45-45-90 triangle, with a width and heigth of 6, making the area \(6 \times 6 \div 2 = \color{brown}\boxed{18}\)

Write as an equation: 

\({2 \over 5} ={ x + 1000 \over 2x+ 1000}\)

Solving for x, we find John originally had \(\color{brown}\boxed{-3,000}\) dollars. 

Here's another way to prove it's impossible:

\(k \times 1984 \times 2 = k \times 2^7 \times 31\)

Using this logic, you can create the factors: 1, 2, 4, 8, 16, 31, 32, 62, 64, 124, 128, 248, 496, 992, 1984, and 3968.

Because we want 21 factors, we need the number to be a perfect square. The smallest way to do this is by multiplying \(3968 \times 31 \times 2\). This yields 27 factors, way too big. Any bigger and you get more factors. 

Thus, this problem is impossible, like geno said (Note: If this problem was just \(k \times 1984\), we could multiply by 31, and get 21 factors. )