Can U Solve Y= -x^2 5x 6

How do I solve (x+5) ^5=5^5?

[math](x+5)^5=5^5[/math][math]\dfrac{(x+5)^5}{5^5}=1[/math]Let [math]y = \frac{x+5}{5}[/math][math]y^5=1[/math]That would give us the 5th roots of unity.[math]y \in \{1, \cos \frac{2\pi}{5} + i\sin \frac{2\pi}{5}, \cos \frac{4\pi}{5} + i\sin \frac{2\pi}{5}, \cos \frac{6\pi}{5} + i\sin \frac{6\pi}{5}, \cos \frac{8\pi}{5} + i\sin \frac{8\pi}{5}\}[/math]Solving the substitution for [math]x[/math], we have:[math]x=5y-5[/math]Therefore, our answers are:[math]x \in \{0, 5\cos \frac{2\pi}{5} - 5 + 5i\sin \frac{2\pi}{5}, 5\cos \frac{4\pi}{5} - 5 + 5i\sin \frac{4\pi}{5}, 5\cos \frac{6\pi}{5} - 5 + 5i\sin \frac{6\pi}{5}, 5\cos \frac{8\pi}{5} - 5 + 5i\sin \frac{8\pi}{5}\}[/math]Which are approximately:[math]x \in \{0, -3.454915028+4.755282581i, -9.045084972+2.938926261i, -9.045084972-2.938926261i, -3.454915028-4.755282581i\}[/math]

How do I solve inequalities?

I’m horrible at words. Like, really horrible at words. So: I’m going to do the work and explain to you what I’m doing and hopefully that’ll help. If I get decent responses that probably means that I still know how to do these. The basic gist is that yes, you’re on the right track. It is possible for there to be a gap within a domain.[math]-1\leq\frac{6}{x}\leq1[/math] // initial problem[math]\abs{\frac{6}{x}}\leq1 [/math]// simplification by treating the bound as a singular inequality for the sake of minimizing the number of steps taken at a time[math]\abs{\frac{1}{x}}\leq\frac{1}{6}[/math] // div 6 both sidesNow, we split it back up because here’s where things get weird.[math]\frac{-1}{6} \leq \frac{1}{x} \leq \frac{1}{6}[/math] // undo absolute valueHere, we have two inequalities:[math]\frac{-1}{6} \leq \frac{1}{x}[/math] and [math]\frac{1}{x} \leq \frac{1}{6}[/math].Now, we flip all fractions. Regardless of the signs involved, if [math]n\geq m[/math], then [math]\frac{1}{n}\leq\frac{1}{m}[/math] because larger denominators mean smaller numbers.So:[math]-6 \geq x[/math] and [math]x \geq 6[/math]or, proceeding from left to right -x -> +x:[math]x \leq -6[/math] and [math]6 \leq x[/math].the domain expressed would be [math][-\infty,-6]U[6,\infty][/math].

What math problem has never been solved?

Well, you can see a list here.One of my favorite problems is that of determining Ramsey numbers.  You can show in any group of six people, there must be three people all of whom know each other, or three people all of whom are complete strangers.  To see this, draw the people as six dots (say in a circular formation), join two of them by a blue line if they know each other, and by a red line if they are strangers.  There will be 15 such lines, and you'll always find a red triangle, or a blue triangle or both.  (Proving that this always happens takes a bit of work, but is not too difficult.)  You need six people; five is not sufficient: you can join up five people with red or blue lines in such a way that there are no triangles of either color.   This smallest value, 6, is an example of a Ramsey number.  Thus the Ramsey number R(3,3) = 6.What about R(4,4)?  How many people do you need to ensure that there will always be a group of four people all of whom know each other, or four people all of whom are complete strangers?  This value turns out to be 18; thus R(4,4)=18.Here's the kicker: R(n,n) is not known for any higher values of n.  R(5,5)?  It's known to be between 43 and 49, and that's all we know.  Ramsey numbers exist for unequal values: R(4,5) = R(5,4) = 25, for example.  But again, only a few are known.  R(3,10), for example, is between 40 and 42, but nobody can pin it down exactly.