lokiisnotdead

hi and welcome to the site! :)
i have an account on aops, and I'm pretty sure there's a lot of ppl here who have an account there too since 90% of all questions on this forum are aops questions lol

please fix your code! :)

hi guest!

so if 1 tower is 18m and another is 30m it would form a trapezoid, something like this:

*this is not drawn to scale*

as you can see, it forms a right triangle, with legs of 16 and 30-18=12!

so, the missing length is \(16^2+12^2=y^2\)

\(400=y^2\)

\(\boxed{y=20}\)

i hope this helped you! :)

hey guest!
 

so we can express x as 2y. so the first equation becomes 2y+1=y-8.

2y+1=y-8

2y+9=y

y+9=0

y=-9

cause a constant term is just an integer, and there isn't a constant term here

the answer is actually 0 since there is no constant term

Each term is of the form
\(\binom{9}{k} (2z)^{9-k}\Big(-\frac{1}{\sqrt{z}}\Big)^k = \binom{9}{k}(-1)^k 2^{9 - k} \frac{z^{9-k}}{z^{k/2}}.\)

Therefore, the constant term occurs when 9-k=k/2 , or k=6. Putting k=6 in our expression above, the constant term is

\(\binom{9}{6} (-1)^6 2^{9 - 6} = \boxed{672}.\)

yea Melody was right! thanks for helping :)

So for this question you can just cross multiply! 

\((1+\frac{1}{x}) \cdot x=1\)

\(x+1=1\)

\(x=0\)

but, this isn't possible since 0 cannot be a denominator, which means that there is no solution.