Melody

Why do you need help quick?

Should your question have more importance to the answerers than the other questions?

Why would that be?

This falls out really easily if you just expand it out and add like terms.

\([(q+4)(p+3)]-[(q+2)(p+1)]=44\\ etc\)

Maybe there is a quicker way to do this but here goes.

\(a\equiv (3^{2n}+4)^{-1}\pmod{9}\\ a\cdot (3^{2n}+4)\equiv 1\pmod{9}\\ a\cdot (3^{2n}+4)=9M+1\qquad \text{where M is an integer}\\ a\cdot (9^n+4)=9M+1\\ a\cdot 9^n+4a=9M+1\\ 4a-1=9M-a\cdot 9^n\\ 4a-1=9(M-a\cdot 9^{n-1})\\ 4a-1=9N \qquad \text{Where N is an integer}\\ \text{By inspection (or euclidean equation) }a=-2\;\;works\\ -2\pmod{9}\equiv 7\pmod{9}\\ a=7+9P  \qquad \text{Where P is an integer}\\ \text{So if } a \text{ is divided by 9 the remainder will be 7}\)

You can check all the numbers from 1 to 8 but this is the only one of those that works.  

LaTex:

a\equiv (3^{2n}+4)^{-1}\pmod{9}\\
a\cdot (3^{2n}+4)\equiv 1\pmod{9}\\
a\cdot (3^{2n}+4)=9M+1\qquad \text{where M is an integer}\\
a\cdot (9^n+4)=9M+1\\
a\cdot 9^n+4a=9M+1\\
4a-1=9M-a\cdot 9^n\\
4a-1=9(M-a\cdot 9^{n-1})\\
4a-1=9N \qquad \text{Where N is an integer}\\
\text{By inspection (or euclidean equation) }a=-2\;\;works\\
-2\pmod{9}\equiv 7\pmod{9}\\

a=7+9P  \qquad \text{Where P is an integer}\\
\text{So if a is divided by 9 the remainder will be 7}

Thanks Alan,

I really do not know what I am doing here. (I have forgotten it)

However.

\(|z|=5\;\;\quad so\;\; \qquad z=5e^{\theta i}\\ |w|=2\;\;\quad so\;\; \qquad w=2e^{\alpha i}\\ |\bar w|=2\;\;\quad so\;\; \qquad \bar w=2e^{-\alpha i}\\~\\ zw=5e^{\theta i}*2e^{\alpha i}=10e^{(\theta+\alpha)i}\\ |zw|^2=100\\ \left| \frac{z}{w} \right|^2=\frac{25}{4}=6.25\\~\\ z\bar w=5e^{\theta i}*2e^{-\alpha i}=10e^{(\theta-\alpha)i}=6+8i\\ \theta-\alpha=atan(\frac{8}{6})=atan(1.\dot3)\\ z\bar w=6+8i=10e^{(atan(1.\dot 3))i} \)

 I think that is right as far as it goes but it does not answer all your questions.

LaTex:

|z|=5\;\;\quad so\;\; \qquad z=5e^{\theta i}\\
|w|=2\;\;\quad so\;\; \qquad w=2e^{\alpha i}\\
|\bar w|=2\;\;\quad so\;\; \qquad \bar w=2e^{-\alpha i}\\~\\
zw=5e^{\theta i}*2e^{\alpha i}=10e^{(\theta+\alpha)i}\\
|zw|^2=100\\
\left| \frac{z}{w} \right|^2=\frac{25}{4}=6.25\\~\\
z\bar w=5e^{\theta i}*2e^{-\alpha i}=10e^{(\theta-\alpha)i}=6+8i\\
\theta-\alpha=atan(\frac{8}{6})=atan(1.\dot3)\\
z\bar w=6+8i=10e^{(atan(1.\dot 3))i}

Continued on the original.  I will delete this thread later on.

Draw the triangle on a piece of paper.

Use Pythagoras's theorem. to find the length of the hypotenuse.

Now can you do it?  If not tell us what you have found so far.

A lecture hall contains 20 chairs, all lined in a row. What is the number of ways that five chairs can be chosen, so that no two chairs are adjacent?

There will be 4 children each with an empty chair on their right, that is 8 seats taken.   But since they are tied in pairs, it is just 4 positions gone.

The fifth child will sit to the very right of all the others but there may not be an empty chair to his right.  So now there are 9 seats but just 5 positions taken.

There are now 11 seats left that all have their own position.

So there are 16 positions in total and I want to decide where the 11 can go.

 So that is 16C11 = 16C5 = 4368 

so I think there re 4368 arrangements and that is assuming that all the chairs are identical and all the children are identical.

Well it is a line intersecting with a circle so it could be 0,1 or 2 times, but you are already told that there is only one point so the line must be a tangent.

If you want to check I would use the perpendicular distance formula fro a point to a line.  I have done that and the distance is indeed 3.

And three is the radius of the circle so the line is a tangent to the circle (only 1 point of intersection)

You could just solve the 2 simultaneously.  That is the normal way

Or you could say that the equation of the line is 

\(3x+4y=15\\ 3x=15-4y\\ x=\frac{15-4y}{3}\\ \text{So a point on the line is } (\frac{15-4y}{3},y)\)

use the normal distance formula and put the distance =3   solve for y then solve for x

I'll assume that Alonzo does not get the ball any other time

Alfonzo

first pass  4 choices

2nd   3 choices

3rd    3 choices

4th    3

5th    3

6th    3

7th    3

8th    1     Alfonzo

4*3^6 = 2916 possible sequences.