Melody

Do you mean ..?

$(e^{5\pi/6} + e^{4\pi/3})^9 (e^{5\pi/6} - e^{4\pi/3})^9\\ =[(e^{5\pi/6} + e^{4\pi/3}) (e^{5\pi/6} - e^{4\pi/3})]^9\\ =[ (e^{2*5\pi/6} - e^{2*4\pi/3}]^9\\ =[ (e^{5\pi/3} - e^{8\pi/3}]^9\\ =[ e^{5\pi/3}(1 - e^{3\pi/3}]^9\\ =[ e^{5*9\pi/3}][(1 - e^{\pi})]^9\\ =e^{15\pi}[1+9 (- e^{\pi})+\binom{9}{2}(- e^{\pi})^2+\binom{9}{3}(- e^{\pi})^3+\binom{9}{4}(- e^{\pi})^4+\binom{9}{5}(- e^{\pi})^5+\binom{9}{6}(- e^{\pi})^6+\binom{9}{7}(- e^{\pi})^7+\binom{9}{8}(- e^{\pi})^8+\binom{9}{9}(- e^{\pi})^9]\\ =e^{15\pi}[1-9 (e^{\pi})+\binom{9}{2}(e^{\pi})^2-\binom{9}{3}(e^{\pi})^3+\binom{9}{4}(e^{\pi})^4-\binom{9}{5}( e^{\pi})^5+\binom{9}{6}(e^{\pi})^6-\binom{9}{7}(e^{\pi})^7+\binom{9}{8}(e^{\pi})^8-\binom{9}{9}(e^{\pi})^9]\\ =e^{15\pi}[1-9 e^{\pi}+\binom{9}{2}(e^{2\pi})-\binom{9}{3}e^{3\pi}+\binom{9}{4}(e^{4\pi})-\binom{9}{5}5e^{\pi}+\binom{9}{6}e^{6\pi}-\binom{9}{7}e^{7\pi}+9e^{8\pi}-e^{9\pi}]\\ $

LaTex

(e^{5\pi/6} + e^{4\pi/3})^9  (e^{5\pi/6} - e^{4\pi/3})^9\\
=[(e^{5\pi/6} + e^{4\pi/3})  (e^{5\pi/6} - e^{4\pi/3})]^9\\
=[ (e^{2*5\pi/6} - e^{2*4\pi/3}]^9\\
=[ (e^{5\pi/3} - e^{8\pi/3}]^9\\
=[ e^{5\pi/3}(1 - e^{3\pi/3}]^9\\
=[ e^{5*9\pi/3}][(1 - e^{\pi})]^9\\
=e^{15\pi}[1+9 (- e^{\pi})+\binom{9}{2}(- e^{\pi})^2+\binom{9}{3}(- e^{\pi})^3+\binom{9}{4}(- e^{\pi})^4+\binom{9}{5}(- e^{\pi})^5+\binom{9}{6}(- e^{\pi})^6+\binom{9}{7}(- e^{\pi})^7+\binom{9}{8}(- e^{\pi})^8+\binom{9}{9}(- e^{\pi})^9]\\
=e^{15\pi}[1-9 (e^{\pi})+\binom{9}{2}(e^{\pi})^2-\binom{9}{3}(e^{\pi})^3+\binom{9}{4}(e^{\pi})^4-\binom{9}{5}( e^{\pi})^5+\binom{9}{6}(e^{\pi})^6-\binom{9}{7}(e^{\pi})^7+\binom{9}{8}(e^{\pi})^8-\binom{9}{9}(e^{\pi})^9]\\
=e^{15\pi}[1-9 e^{\pi}+\binom{9}{2}(e^{2\pi})-\binom{9}{3}e^{3\pi}+\binom{9}{4}(e^{4\pi})-\binom{9}{5}5e^{\pi}+\binom{9}{6}e^{6\pi}-\binom{9}{7}e^{7\pi}+9e^{8\pi}-e^{9\pi}]\\
 

let the number be x

(Five times a number)is divided by (7 more than that number minus 4)    =  18

5x                                is divided by (      7+x-4              )                           =18

                        5x         is divided by           x+3                                          =18

$\frac{5x}{x+3}=18\\ 5x=18x+54\\ -13x=54$ 

you can finish it.

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It is possible that the question intended

$\frac{5x}{x+7}-4=18$

But I think the first interpretaion is more reasonable.

Say person 1 choses 2 cards then person 2 then person 3

In order for this to happen they all have to get 1 red and 1 blue card

lets say I tie a red and a blue together.  This can be RB or BR    2*2*2 = 8 ways this can happen. (with the 3 pairs)

altogether there are   6!/(3!3!) ways to order the cards  =  20 ways

So I think the likelyhood that they will each have a red card is    8/20  =  2/5

What is the average of all positive integers that have four digits when written in base 3, but two digits when written in base 4? Write your answer in base 10.

Base 3     3^3 base10 =1000base3 = 27 base 10  smallest

                 3^4-1 base10 = 2222base3 = 80 base10  biggest

Base 4     4^1 = 10 base 4 = 4 base10    smallest

                 4^2-1=33 base4 = 15 base 10   biggest

There is no overlap so there is no answer.    0/0 is undefined

Unless I have made a mistake with my drawing I do not think there is a set answer to this question.

I have two different sketches here that both match the data given and angle ABC has 2 diferent values. 

Namely 50 degrees and  48 degrees.  I suppose the answers are close. Maybe it is rounding error in the construction.

Angle ABC will always be a half of angle AOC though.   

(The angle at the centre is twice the angle at the circumference when botha standing on the same arc )

F1,   F2,  F3      c1, c2, c3, c4, c5, c6

How many ways can   the flavours 1,1,2,2,3,3    be soted in a row

6!/(2!2!2!) = 6!/ (2*4) = 3*5*6 = 90

Attn  Keihaku

I am waiting for a response from you.

Opposite angles in a cyclic quad are suplementar

Givew those equalities I can see that

The smallest would add with the biggest to get 180 degrees

The largest angle could be

$\frac{\angle q}{4}=\frac{\angle r}{5}\\ \angle q=0.8*\angle r\\~\\ \angle q + \angle r =180^ \circ\\ 0.8*\angle r + \angle r =180^ \circ\\ 1.8*\angle r =180^ \circ\\ \angle r =100^ \circ\\ \angle q = 80 ^\circ\\~\\ \frac{\angle p}{3}=\frac{\angle r}{5}\\ \frac{\angle p}{3}=\frac{100}{5}\\ \angle p = 60^\circ\\ so\\ \angle s = 120 ^\circ$

LaTex:

\frac{\angle q}{4}=\frac{\angle r}{5}\\
\angle q=0.8*\angle r\\~\\
\angle q + \angle r =180^ \circ\\
0.8*\angle r + \angle r =180^ \circ\\
1.8*\angle r =180^ \circ\\
\angle r =100^ \circ\\
\angle q = 80 ^\circ\\~\\
\frac{\angle p}{3}=\frac{\angle r}{5}\\
\frac{\angle p}{3}=\frac{100}{5}\\
\angle p = 60^\circ\\
so\\
\angle s = 120 ^\circ

I answered on the original which was ONLY 5 hours ealier.