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I'll try and show you how to graph it by hand.  I will be back in a little while :)

I just need a point at the midline and a point at the maximum or minimum closest to the first point. Thanks again.

Yes that looks really good to me.    Nice work!   

And thanks for your gratitude, I appreciate it  

Did they want you to graph it like this or did they want you to do it with points?  Do you know?

I modified your graph just a little bit. I used the spanner tool in the top right corner to sset the units to 1,2,3 etc instead of  pi

And on the second graph y=1 I made it so it could not be moved by saying it had to be from 1 to 1. 

If you look at my graph you will see what I mean.

My goodness you are a lazy guest. I hope you have to sit for an exam because you will almost certainly fail.

Sorry Rom but I answered this question ealier today and I requested that they did some of their own thinking and interact with me.

You did not know that, I am certainly not upset with you.    I just get so annoyed/upset with all the askers here who have no desire to learn anything. 

https://www.desmos.com/calculator/exp2ewjoza

Am I doing this right? Also, I appreciate you. 

\(P[1] = \dbinom{2}{1}(0.48)(0.52) = 2 (0.48)(0.52) = 0.4992\\ \text{The table at the link got this wrong by a factor of 2}\\ \text{because they only accounted for one of the 2 paths to get to 1 girl/1 boy}\)

\(\text{The expression seems to vary from }\dfrac 3 2\\ \text{which corresponds to an equilateral triangle to }2\\ \text{which corresponds to an isosceles triangle with one leg having zero length}\\ \dfrac 3 2 \leq \dfrac{a}{b+c}+ \dfrac{b}{a+c}+\dfrac{c}{a+b} < 2\)

the parent sin funtion is 

\(y=sin \theta\)

this has a amplitude of 1, a period of 2pi (radians) , the wave centre is the line y=0  (the theta axis)   and no phase shift.

\(y=sin (n\theta) \)

This has a period of     2pi/n    nothing else has changed

\(y=a\cdot sin \theta\)

This has an amplitude of a           Everything else is the same as the parent function.

\(y=sin \theta+L\)

Lifts the parent funtion by L units.

Now what would be a good starting point do you think for this graph. It may need tweaking afterwards?

Now graph your function with desmos.  Use x instead of theta, I do not think theta will work.

Ive started you off with the parent funtions. You add what you think is write underneath and see if it works.

Why haven't you done it, or shown an attempt at it?

You were logged on for quite a while.

Is it that you just want someone to do it for you in entirety so you can go play without getting into trouble with your teacher?

c)
Using parts (a) and (b), find a formula for

\(\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}^n\)

\(\begin{array}{|rcll|} \hline \mathbf{\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}^n} &=& PD^nP^{-1} \\ &=& \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 1^n & 0 \\ 0 & 2^n \end{pmatrix}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ &=& \begin{pmatrix} \mathbf{6-5\cdot 2^n} & \mathbf{15(1-2^n)} \\ \mathbf{2(2^n-1)} & \mathbf{6\cdot 2^n-5} \end{pmatrix} \\ \hline \end{array}\)

(b)

Find a formula for  \(\mathbf{D^n}\) where D is the matrix you found in part (a).
(You don't need to prove your answer, but explain how you found it.)

\(\begin{array}{|rcll|} \hline D &=& \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \\\\ D^2 &=& \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \\ &=& \begin{pmatrix} 1^2 & 0 \\ 0 & 2^2 \end{pmatrix} \\\\ D^3 &=& \begin{pmatrix} 1^2 & 0 \\ 0 & 2^2 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \\ D^3 &=& \begin{pmatrix} 1^3 & 0 \\ 0 & 2^3 \end{pmatrix} \\ \ldots \\ \mathbf{D^n} & \mathbf{=} & \mathbf{\begin{pmatrix} 1^n & 0 \\ 0 & 2^n \end{pmatrix}} \\ \hline \end{array} \)

We're going to consider the matrix

\(\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\)

a)
Let \({P} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\).
Find the 2x2 matrix D such that

\({P} {D} {P}^{-1} = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\)

\(\text{Let $M = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix} $ }\)

Formula:

\(\begin{array}{|rcll|} \hline P^{-1} &=& \dfrac{1}{\begin{vmatrix} 3 & -5 \\ -1 & 2 \end{vmatrix}}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ &=& \dfrac{1}{6-5}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ &=& \dfrac{1}{1}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ \mathbf{P^{-1}} & \mathbf{=} & \mathbf{\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline PDP^{-1} &=& M \quad | \quad \cdot P \\ PDP^{-1}P &=& MP \quad | \quad P^{-1}P = I(\text{Identity matrix}) \\ PDI &=& MP \quad | \quad \cdot P^{-1} \\ P^{-1}PDI &=& P^{-1}MP \quad | \quad P^{-1}P = I(\text{Identity matrix}) \\ IDI &=& P^{-1}MP \\ \mathbf{D} & \mathbf{=} & \mathbf{P^{-1}MP} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{D} & \mathbf{=} & \mathbf{P^{-1}MP} \\ &=& \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \\ &=& \begin{pmatrix} 2 & 5 \\ 2 & 6 \end{pmatrix}\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \\ \mathbf{D} & \mathbf{=} & \mathbf{\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}} \quad | \quad (\text{Diagonal matrix}) \\ \hline \end{array}\)

Good. Can you give this a go now yourself?

You have asked enough questions for now. Learn what you can from the first two.

Hi Rudram,

this is how to do it 

1) What is the length of the bottom of the triangle?

2)what is Pythagoras's theory ?

3) You now have the hypotenuse of the triangle and also one of the short sides.

    You have to use Pythagoras's theorum to find the third side.

4)  The area of a triangle is   half x base x height    So find the area.

5) Find the area of the square   (Area = side squared)

6) Add the two areas together.

If you have problems show what you have done yourself and state where your problem arises. 

Please  no one answer over me. Giving a full answer will not help rudram learn :)

I definately will. Thanks

Just draw the diagram and solve for distance using Pythagoras's theorum.

Please no one answer over me

Coul you please type your worded questions in. It is easier for answerers.

Thanks!

There are three different possibilities for our first decision, each corresponding to which container we choose. So, if we choose container A, with 1/3 probability, we have a 6/10=3/5 probability of drawing green, which means we have a (1/3)(3/5)=1/5 of picking Container A and then picking a green ball. Similarly for container B the probability is (1/3)(4/10)=2/15, and the same for container C. So, the total probability is (1/5)+(2/15)+(2/15) = 7/15.

Hmmmm....I think the greatest value will be  '1'       1^k   will always be 1 when k>0      as you approach 0 the value will get mucho smaller for larger values of k  ...approaching 0 

so range would look something like    (0,1]                  (I think! )

Missing some info....or a diagram....sorry!

Nice, guest   !!!!

Income x thousands        tax = x * x%

Take home is     x -   x (x/100) =   -x^2/100 + x         Max will be at   -b/2a =  - 1/(2* (-1/100))    =    50  = x    = $50 000