HELPMEEEEEEEEEEEEE

Call this "x" : \(\cfrac{9}{9 + \cfrac{9}{9 + \cfrac{9}{9 + \dotsb}}} \)

Which simplifies to: \(\frac{9}{9+x}\)

From this, we can set up the equations:
\(x=\frac{9}{9+x}\)

\(x^2+9x=9\)

\(x^2+9x-9=0\)---> This quadratic equation had 2 solutions, one positive and one negative, for obvious reasons we can rule out the negative solution. This means that the positive solution is our answer:

\(x=\frac{-9+3\sqrt{13}}{2}\)

I appreciate your time!


*Note: I'm not 100% sure about my answer. If anyone sees anything wrong with my thinking process, make sure to point it out!

Factoring the quadratic: (x+1)(x+4)
The values of x that make this equation 0 are -1 and -4  which are the roots of this quadratic.

"The positive five-digit integers that use each of the digits 1, 2, 3, 4 and 5 exactly once are ordered from least to greatest. What is the integer in the list?"

I think you are missing a part of the question. The question doesn't specify which integer in the list it wants (i.e.) the 5th integer or the 24th integer. Check if there is any missing information that can help us solve the question.

I appreciate your time!

5+5+55+5+55+5+55+5+55+5+55+5+55 = 365

This is a link to a question we cannot answer due to the fact that we need an account to log-in. If you still require help, try typing out the whole question so we can view it.

I appreciate your time!

120 / (2-12) = 120 / -10 = -12.

1)
\(x^2=9\)

\(\sqrt{x^2}=\sqrt{9}\)

\(x=±3\)

2)
\(x^2=-5\)

\(\sqrt{x^2}=\sqrt{-5}\)

\(x=±\sqrt{5}i\)

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7!= 5040
...

From this, you can see the units digit from 5! and beyond is 0. So the units digit of the sum of the factorials, is just the sum of the units digits of 1! - 4! which is 1 +2 + 6 + 4 = 13. Since we only need the units digit, the answer is 3.