Nvm, I think I got it. We can just use the formula a=sqrt(b^2+c^2) where b=8 and c=sqrt(105)
Guest, will you show your reasoning? I've gotten the center of the ellipse is (3, 10), but dont know how to proceed.
thx for the help and support :D
Thank you for the formula, though. I appreciate it and it will help a lot. Will the formula work on all types of problems like this? e.g the 3rd one I posted.
edit: I don't think it will.
Also, sorry about the misrenderance. It was also supposed to be 4 and 3, and not just random blanks :(
Thanks so much for your time and effort. Also, is C in this case the choose function?
I'm so sorry, CPhill. I meant to type the second one not as 1 but 2.
I like liking random comments lol.
Using the quadratic formula, we get: \({-b \pm \sqrt{b^2-4ac} \over 2a}\Leftrightarrow\frac{-\left(-6\right)\pm\sqrt{\left(-6\right)^24\cdot \:1\cdot \:5}}{2\cdot \:1}\Leftrightarrow\frac{6\pm\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1} \Leftrightarrow\frac{6\pm\sqrt{16}}{2\cdot \:1}\Leftrightarrow\frac{6\pm4}{2}=\boxed{5}, \boxed{1}.\)
\(\frac{234}{48}=4.875.\) So, \(\boxed{4}\) times.
ahh, Ok I see. Got it, k+m=22. I get how to do it.
ohh, ok, so would we get k+m=7?
Now that I think about it, why do p+q+r=7?
Ohh, I see. I haven't grasped Vietas yet.
Good job! And, thank you
Elegant response, CPhill!
Ahh, very nice strategy. So concise. Thank you!
