chrissy

for a number to be divisible by 5, it must end in either 5 or 0, which means that \(C\) is either 5 or 0. for a number to be divisible by 7... well, we'll deal with that later. and for it to be divisible by 9, the sum of the digits must be divisible by 9. since \(5+7+6=21\), \(A+B+C\) is equal to 6, or a multiple of 9 plus 6. 

if \(C\) is 0, then \(A+B\) is equal to 6 or a mutliple of 6 plus 1, and if \(C\) is 5, \(A+B\) is equal to 1 or a multiple of 9 plus 1. this is the point where i begin to bash, case by case, and where there is almost definitely a more graceful solution. 

if \(A+B=6\), then \(A\) and \(B\) can be \((0, 6), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1),\) or \((6, 0)\). we check if each of these make the original number divisible by 7 (using a calculator in my case), and we have \((4, 2, 5)\) as an ordered triplet.

if \(A+B=6+1(9)=15\), then \(A\) and \(B\) can be \((6, 9), (7, 8), (8, 7),\) or \((9, 6)\). once again, we check if any of these will make the original number divisible by 7, but this time, none of the values for \(A\) and \(B\) work.

if \(C = 5\), then \(A\) and \(B\) can be 1 or 10. \(A+B=1\) yields no working ordered triplets and \(A+B=10\) yields \((4, 6, 0)\).

all in all, the only possible values for the number are \(\boxed{579425}\) and \(\boxed{579460}\).

note: this solution, once again, is extremely messy, and i would love it if someone else could offer some suggestions for a more graceful answer. also, when i mention ordered pairs, i mean in the form \((A, B)\), and when i mention ordered triplets, i mean in the form \((A, B, C)\).

to the original poster: if you could post the official solution once you get it (or if it exists), that would be amazing. thanks for the awesome problem!

can you reply with the question with the numbers in? i think the problem formatted wonky somewhere among internet dimensions :|

\(\frac{3}{1.2} = \frac{30}{12} = \boxed{\frac{5}{2}}\)
to check, we can do the same thing with the \(y\) values:
\(\frac{9}{3.6} = \frac{90}{36} = \frac{15}{6} = \frac{5}{2} \)
yup, the answer is right :D

posted my answer to the original post :)

we first solve for \(f(10)\):

\(\begin{align*} f(10) &= 3(10)^2 - 2(10) + 4 \\ &= 300 - 20 + 4 \\ &= 284 \end{align*}\)

then, we solve for \(g(10)\) in terms of \(k\):

\(\begin{align*} g(10) &= 10^2 - 10k - 6 \\ &= 100 - 10k - 6 \\ &= 94 - 10k \end{align*}\)

plugging in these two values into \(f(10) - g(10) = 10\), we get

\(\begin{align*} 284 - (94 - 10k) &= 10 \\ 284 - 94 + 10k &= 10 \\ 190 + 10k &= 10 \\ 10k &= -180 \\ k &= \boxed{-18}. \end{align*} \)

i'm not quite sure i'm interpreting the problem right, but i'm going to solve the problem assuming that the problem is asking you to find the leg length of a 45-45-90 triangle, where one leg is on the ground and another leg is the height of the treehouse, with the 90-foot rope being the hypotenuse.

we know that the hypotenuse of a 45-45-90 triangle with leg length \(x\) is \(x\sqrt2\), so we know that the leg length of a 45-45-90 triangle with hypotenuse \(y\) is \(\frac{y}{\sqrt2} = \frac{y\sqrt2}{2}\), so plugging in \(y = 90\) (since the rope is the hypotenuse), we get 

\(\frac{y\sqrt2}{2} = \frac{90\sqrt2}{2} = \boxed{45\sqrt2}.\)

just a quick solution for anyone else confused by the question : )
 

the circumfrence of a circle with a radius of \(2\) is \(2\pi r = 4\pi\). the perimeter of a square with side length \(x\) would be \(4x\), so we have 

\(\begin{align*} 4\pi &= 4x \\ x &= \boxed{\pi}. \end{align*}\)

the graph of \(y = f(x)\) has the line \(x=5\) as an axis of symmetry. the graph also passes through the point \((8, -7)\). find anohter point that must lie on the graph of \(y = f(x)\).
 

since \(x=5\) is the line of symmetry, we know that the point \((8, -7)\) reflected over \(x=5\) must also be on the graph of \(y = f(x)\). \((8, -7)\) is 3 units away from \(x=5\) on the \(y\)-axis, so the point 

\((8 - 2(8-5), -7) = (8 - 6, -7) = \boxed{(2, -7)}\)

must also be on the graph ​of \(y = f(x)\).

omg yeah, i've been trying to register and i tried different devices, different browsers, etc, and i keep getting an internal server error. and this has been going on for like 7-8 weeks now for me.