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If

\(z\) is a complex number satisfying \(z + \dfrac{1}{z} =1\), calculate \(z^{10}+ \dfrac{1}{z^{10}}\). 

\(\begin{array}{|rcll|} \hline \left( z+\dfrac{1}{z} \right)^2 &=& z^2+\dfrac{1}{z^2} + \dfrac{2z}{z} \\\\ 1^2 &=& z^2+\dfrac{1}{z^2} + 2 \\\\ z^2+\dfrac{1}{z^2} &=& 1-2 \\\\ \mathbf{z^2+\dfrac{1}{z^2}} &=& \mathbf{-1} \\ \hline \end{array} \begin{array}{|rcll|} \hline \left( z^2+\dfrac{1}{z^2} \right)\left( z+\dfrac{1}{z} \right) &=& z^3+\dfrac{1}{z^3} + z+\dfrac{1}{z} \\\\ (-1)*1 &=& z^3+\dfrac{1}{z^3} + 1 \\\\ z^3+\dfrac{1}{z^3} &=&-1-1 \\\\ \mathbf{z^3+\dfrac{1}{z^3}} &=& \mathbf{-2} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \left( z^2+\dfrac{1}{z^2} \right)\left( z^3+\dfrac{1}{z^3} \right) &=& z^5+\dfrac{1}{z^5} + z+\dfrac{1}{z} \\\\ (-1)*(-2) &=& z^5+\dfrac{1}{z^5} + 1 \\\\ z^5+\dfrac{1}{z^5} &=&2-1 \\\\ \mathbf{z^5+\dfrac{1}{z^5}} &=& \mathbf{1} \\ \hline \end{array} \begin{array}{|rcll|} \hline \left( z^5+\dfrac{1}{z^5} \right)^2 &=& z^{10}+\dfrac{1}{z^{10}} +\dfrac{2z^5}{z^5} \\\\ 1^2 &=& z^{10}+\dfrac{1}{z^{10}} + 2 \\\\ z^{10}+\dfrac{1}{z^{10}} &=&1-2 \\\\ \mathbf{z^{10}+\dfrac{1}{z^{10}}} &=& \mathbf{-1} \\ \hline \end{array}\)

edited: Thank you Alan, thank you MaxWong

Suppose I throw a fair die, and then you throw the fair die. What is the probability that your number is strictly greater than mine? 

Make a grid as shown below.  These 36 groups represent every possible pair of throws. 

In each group, call the first number your throw and the second number my throw. 

Count the ones where the second number is strictly greater than the tirst. 

1,1   1,2   1,3   1,4   1,5   1,6  

2,1   2,2   2,3   2,4   2,5   2,6  

3,1   3,2   3,3   3,4   3,5   3,6  

4,1   4,2   4,3   4,4   4,5   4,6  

5,1   5,2   5,3   5,4   5,5   5,6  

6,1   6,2   6,3   6,4   6,5   6,6 

You see there are 15 such groups that satisfy the requirement.  So the probability is 15/36, or 5/12. 

_.

Thank you Guest !

Thank you all for replying! 

If tan A = 1/2 and tan B = 1/3, then find tan(2A + B).

tan(2A + B) = 1.333333333 

\(f(x) = \frac{2x-8}{x^2 -2x - 3} \qquad\text{ and }\qquad g(x) = \frac{3x+9}{2x-4}\)

I found    \(f( \frac{3x+9}{2x-4}) \)       and then I simplified it all down.

There was a lot of algebra and I have not checked my result.  Checking could be done by graphing the original substituted function

anyway I ended up with    [ -40(x^2+....)]  /  [-15(x^2+....) ]   if this is correct I would assume that the horizontal asymptote is   

40/15 = 8/3

But there could be 100 mistakes in my working and also in my logic.

101 + 97 = 198

93 + 89 = 182

Subtract 198 from 182, so 16 is the common length.

so we have 4(198 + 182 + ... + 6)

This is an arithmatic sequence so 

(6 + 198)/2 * number in sequence

We know we add 12 each time so 198-6 = 192, 192/12 = 16

+ 1 for the first term

204/2 , * 17

Then, multiply 4, and that is your answer.

Please don't post so many questions at once!

Please also fix your latex.

However I know the questions you are asking :)

for the marvin one, 12c5 to get to 5,7 and 7c3 * 5c1 passing throguh 4,3

Now you calculate.

4 choices for each ball, 4 boxes, you can solve now.

these are the groups

4-0-0-0, 3-1-0-0, 2-2-0-0, 2-1-1-0, and 1-1-1-1

The number is answer.

C)

I answered this a few minutes ago.

Go see my answer earlier.

We know we have the groups

4-0-0-0, 3-1-0-0, 2-2-0-0, 2-1-1-0, and 1-1-1-1.

There are 4 ways for 4-0-0-0 to be arranged, 

4 for 3-1-0-0, 4c2 = 6 for 2-2-0-0, and 4*3 and 1 for 2-1-1-0, and 1-1-1-1, respectively

SO we have 35

I tried what you said up until the last step, I'm probably missing ONE tiny bit but I can't seem to figure out how to do it...

RQ = sqrt(6)  

Thanks!  I will try and do that! :) :)

1 .    a and b    =    -1/14 +- sqrt141 / 14    =    0.776739   and   −0.919596

             (via Quadratic Formula)

     so (a-4)(b-4) = 15.857

Use this "closed form" formula to sum them up;

∑[4*(198 - 16n), n, 0, 12 ] = 5,304

i=0;j=0;m=0;t=0;a=(5,7, 9,11);r= (1, 2, 3, 4);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

3465 m +  1731, where m =0, 1, 2, 3, .........etc.

The smallest positive integer = 1,731

Look for a pattern in the (x+y) terms. The (x-y) terms remain the same.

Here's a hint!

(101+97)(101-97)=792

(93+89)(93-89)=728

(5+1)(5-1)=24

Notice that the end numbers are always 4, and the pattern. (How the first number decreases)

Hope this helps!

Terms mean any number with different exponents or numbers unable to be added. (5x and 5x^2 for example)

So, 10^2 is a term