Melody

I would like to see this one answered too.    

That is good.  

It is a good way to do these but a lot of teachers teach what I think is a more complicated method.

8x^2 - 10x - 25

8*-25=-200

So I want 2 numbers that add to -10 and multiply to -200

The bigger one will be negative and the smaller one will be positive (absolute values I mean)

The fairly obvious combination is   -20 and 10

-10x is replaced with -20x+10x

= 8x^2 -20x+10x - 25

Now factorise in pairs

= 4x(2x-5) + 5(2x-5)

= (4x+5)(2x-5)

\(\frac{sec²(\frac{x}{4})-1}{cos(\frac{6}{x})-cos(\frac{2}{x})}\\~\\ =(sec²(\frac{x}{4})-1)\cdot \frac{1}{cos(\frac{6}{x})-cos(\frac{2}{x})}\\~\\ =\frac{sin^2(\frac{4}{x})}{Cos^2(\frac{4}{x})}\cdot \frac{1}{cos(\frac{4}{x}+\frac{2}{x})-cos(\frac{2}{x})}\\~\\ \)

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\(cos(\frac{4}{x}+\frac{2}{x})-cos(\frac{2}{x})\\~\\ =cos(\frac{2}{x})cos(\frac{4}{x})-sin(\frac{2}{x})sin(\frac{4}{x})-cos(\frac{2}{x})\\~\\ =cos(\frac{2}{x})[cos^2(\frac {2}{x})-sin^2(\frac{2}{x})]-sin(\frac{2}{x})[2sin(\frac{2}{x})cos(\frac{2}{x})]-cos(\frac{x}{2})\\~\\ =cos(\frac{2}{x})\{cos^2(\frac {2}{x})-sin^2(\frac{2}{x})-2sin^2(\frac{2}{x})-1\}\\~\\ =cos(\frac{2}{x})\{-sin^2(\frac {2}{x})-sin^2(\frac{2}{x})-2sin^2(\frac{2}{x})\}\\~\\ =cos(\frac{2}{x})\{-4sin^2(\frac{2}{x})\}\\~\\ =-4cos(\frac{2}{x})sin^2(\frac{2}{x})\\~\\ \)

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\(\frac{sin^2(\frac{4}{x})}{Cos^2(\frac{4}{x})}\\~\\ =\frac{(sin(\frac{4}{x}))^2}{(Cos(\frac{4}{x}))^2}\\~\\ =\frac{(2sin(\frac{2}{x})cos(\frac{2}{x}))^2}{(Cos^2(\frac{2}{x})-sin^2(\frac{2}{x}))^2}\\~\\ =\frac{4sin^2(\frac{2}{x})cos^2(\frac{2}{x})}{Cos^2(\frac{4}{x})}\\~\\ \)

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\(=\frac{4sin^2(\frac{2}{x})cos^2(\frac{2}{x})}{Cos^2(\frac{4}{x})}\cdot\frac{1}{-4cos(\frac{2}{x})sin^2(\frac{2}{x})}\\~\\ =\frac{-cos(\frac{2}{x})}{Cos^2(\frac{4}{x})}\\~\\ =-cos(\frac{2}{x})Sec^2(\frac{4}{x})\\~\\\)

\(\boxed{=-cos\left(\frac{2}{x}\right)Sec^2\left(\frac{4}{x}\right)}\)

Hi Jenny,

I am not sure what your teacher has in mind.

This is that my idea is.

The linear graph is at the bottom.  I have just done it on a number line but I am not sure if this is really what your teacher wants.

( Ginger is also answering so she might have a different idea. )

Oh, you need to write the names of the planets on there too. I'd add the sideways above, or below, the dots.

I've just drawn this with Excel and I don't know how to add the names sbut you need todo that by hand.

Label the sun too.  

See if you can make sense of that....    

I divided all the metre distances  by  1.5 x 10^11   because that is the metre distance from the Sun to the Earth.

I did this because the question tells you to devise a suitable linear scale and Astronomical units make a great scale!  

Ask more questions if you want but think about it first.  

I think that would be more helpful   

I am not actually sure what you are being asked to do.

I would assume it means  R2 is replaced by R2- 1/2 R1

in which case the answer would be

\(\begin{bmatrix} 4 && 3 && 0 \\ -6-2 && -3-1.5 && 12-0 \end{bmatrix}=> \begin{bmatrix} 4 && 3 && 0 \\ -8 && -4.5 && 12 \end{bmatrix}\)

That is really cool.   Thanks Heureka.

No matter how many marbles he takes with his first choice Ron can still lose if he is not playing strategically.

I guess we are suppose to assume that Ron is playing the ultimate game and so it Martin.

I have got nothing more.  

Coolstuff,

Is CalculatorUser supposed to be able to find your edits?