Melody

I answered this question - or one almost identical a couple of days ago.

Admittedly I was not sure of my answer but Alan gave me points for it so I guess it was ok :/

Thanks Alan :)

I think of three numbers. The mean is equal to the mode and this is equal to 4. Find ALL the possible sets of 3 numbers.

4,4,4     I think that is it!     (assuming that there is only one mode)

Yes, it is doing the same for me.

I will let sdmin know :)

Mmm

the equation c^2 = p can have zero, one, or two solutions. What can you say about the number of solutions of the equation c^3 = p?

\(c^2=p\\so\\c=\pm\sqrt p\\ \text{no real solutions if }p<0\\ \text{one real solutions if }p=0\\ \text{2 real solutions if }p>0\\\\~\\ If\\ c^3=p \;\;then\;\; c=\sqrt[3]{p}\\ \text{If p=0 then c=0}\\ \text{If p<0 then }c=-\sqrt[3]{|p|}\\ \text{If p>0 then }c=+\sqrt[3]{|p|}\\~\\ \text{asinus is correct, no matter what real value p has, there is exactly on solution for c} \)

Thanks asinus :)

I see Alan beat me to it :))

Integrate the following:  ∫x^2 sin^3 (x) dx. Also, please show steps of solution.

I thank you for any help.

\(\int\;x^2sin^3x\;dx\\ =\int\;x^2sinx\;sin^2x\;dx\\ =\int\;x^2sinx\;sin^2x\;dx\\ \qquad \qquad cos2x=cos^2x-sin^2x\\ \qquad \qquad cos2x=1-2sin^2x\\ \qquad \qquad 2sin^2x=1-cos2x\\ \qquad \qquad sin^2x=\frac{1-cos2x}{2}\\ =\int\;x^2sinx\;\frac{1-cos2x}{2}\;dx\\ =\int\;\;\frac{x^2sin(x)-x^2sinxcos(2x)}{2}\;dx\\ =\frac{-1}{2}\int\;[x^2sin(x)cos(2x)-x^2sin(x)]\;dx\\ \)

\(\qquad \qquad sin(\alpha-\beta)+sin(\alpha+\beta)\\ \qquad \qquad =sin(\alpha)cos(\beta)-sin(\beta)cos(\alpha)+sin(\alpha)cos(\beta)+sin(\beta)cos(\alpha)\\ \qquad \qquad =2sin(\alpha)cos(\beta)\\ \qquad \qquad so\\ \qquad \qquad 2sin(x)cos(2x)=sin(x-2x)+sin(x+2x)\\ \qquad \qquad 2sin(x)cos(2x)=sin(-x)+sin(3x)\\ \qquad \qquad 2sin(x)cos(2x)=sin(3x)-sin(x)\\ \qquad \qquad sin(x)cos(2x)=\frac{1}{2}[sin(3x)-sin(x)]\\ =\frac{-1}{2}\int\;[x^2sin(x)cos(2x)-x^2sin(x)]\;dx\\ =\frac{-1}{2}\int\;\frac{x^2}{2}[sin(3x)-sin(x)]-x^2sin(x)]\;dx\\ =\frac{-1}{2}\int\;[\frac{x^2}{2}sin(3x)-\frac{x^2}{2}sin(x)-x^2sin(x)]\;dx\\ =\frac{-1}{2}\int\;[\frac{x^2}{2}sin(3x)-\frac{3x^2}{2}sin(x)]\;dx\\ =\frac{-1}{4}\left(\int\;[x^2sin(3x)]dx\;-\;\int[3x^2sin(x)]\;dx\right)\\ \text{Now work out each of these two integral seperately using integration by parts.}\)

Maybe I will finish it later  (it is 1:36am)  anyway, that is a good start for you :)

Hi all,

I know everyone here has their own ideas on what the proper use of this forum is.

Some people want it to be a serious and dedicated help site for young people who really want to learn mathematics.

I want that too of course but I want the forum to have a lighter side as well. 

I have always wanted it to be like a after school drop in centre for students who enjoy mathematics and who want to be with others who are like minded.  As such it would (and is) a social centre for many people, including myself. I have never minded people socializing here so long as it does not monopolize the forum or diminish the forum's vaue as a place to learn, teach and enjoy mathematics. 

I am pleased that ClownPrinceofChaos and HighSchoolCalculus have shared their music with the rest of us.

Thanks guys :)

I really like this Matthew, I didn't know you wrote music !! 

You do not use gravity at all.

You may use the radius of the earth depending on the question.

Hi Sara :)

Lets see :)

\(\sqrt3\;x^2-2\sqrt2\;x-2\sqrt3=0\\~\\ \triangle=(-2\sqrt2)^2-4*\sqrt3* -2\sqrt3 \\ \triangle=(8+24) \\ \triangle=32\qquad \text{Positive so this means it has 2 real roots} \\~\\ \)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-b \pm \sqrt{\triangle} \over 2a}\\ x = {2\sqrt2 \pm \sqrt{32} \over 2\sqrt3}\\ x = {2\sqrt2 \pm 4\sqrt{2} \over 2\sqrt3}\\ \\~\\ x = {2\sqrt2 \pm 4\sqrt{2} \over 2\sqrt3}\times \frac{\sqrt3}{\sqrt3}\\ x = {2\sqrt6 \pm 4\sqrt{6} \over 6}\\ x = {\sqrt6 \pm 2\sqrt{6} \over 3}\\ x = {\sqrt6 (1\pm2) \over 3}\\ x=\frac{3\sqrt6}{3}\qquad or \qquad x=\frac{-\sqrt6}{3}\\ x=\sqrt6\qquad or \qquad x=\frac{-\sqrt6}{3}\\ \)

I checked this answer by graphing it and it is correct.