Melody

200*0,967/0,07 = 2762.8571428571428571

Hi Juriemagic,

If x is negative then f(x) would have to be positive (so the product of them is negative)

likewise

If x is positive then f(x) will have to be negative.

Does that help?

Do you have a specific question that is troubling you?

You can work out the diagonal in terms of s  

and as a separate issue you can work it out in terms of r.

Then you can put them equal to each other to get r in terms of s   OR s in terms of r

That is true but it can be written in terms of just s or just r.

Since it is a ratio question you can use a convenient ellipse.

Let X be the centre and M be the midpoint of AC

triangle BXF₂    and  triangle BMA   are similar tiranges and they are both 30, 60 90 degree triangles.

So their sides are in the ration 1:2:sqrt3

I want the focal length (c) to be 1 and

I want B to be   \((0,\sqrt3)\)

\(c^2=a^2-b^2\\ 1=a^2-3\\ 4=a^2\\ a=2\)

So the ellipse that I will use is   

 \(\frac{x^2}{4}+\frac{y^2}{3}=1\\ \frac{y^2}{3}=1-\frac{x^2}{4}\)

Equation of line AB

\(y=\sqrt3\;x+\sqrt3\\ y=\sqrt3(x+1)\\ y^2=3(x+1)^2\\ \frac{y^2}{3}=(x+1)^2\\ \)

Solve simultaneously

\(1-\frac{x^2}{4}=(x+1)^2\\ -\frac{x^2}{4}=x^2+2x\\ -x^2=4x^2+8x\\ 5x^2+8x=0\\ x(5x+8)=0\\ x=0\quad or \quad x=-\frac{8}{5} \)

So the x value for A is   -8/3

\(AC=AB=BC=\frac{16}{5}\\ F_2F_1=2\\ \frac{AB}{F_1F_2}=\frac{16}{10}=\frac{8}{5} \)

what do you mean by  F_1*F_2    

they are points, not distances so it doesn't make sense to me.......

I'd just do it like a co-ordinate geometry question

(3,2),  (6,1)   

\(distance=\sqrt{9+1}=\sqrt{10}\)

height from midpoint =\(\sqrt{10+2.5}=\sqrt{12.5}\)

le

gradient = 1/-3      gradient of perpendicular = 3

midpoint = (4.5, 1.5)

Equ of perpendicular

Now I would find the equation of the perpendicular bisector

And I would find the equation of the circle centre (4.5,1,5) radius sqrt12.5

and solve them simultaneously.

This is not a very 'profession' method since they are complex numbers, not co-ordinate geometry, but the method will work.

\(\)

I have heard that English has more words than any other language. 

I think this is partly because alternate multiple meaning are less common.

I am learning German and one word often has dozens of meanings.

In German, context and grammar much more heavily to convey meaning.

I did this using Matrices:

\(a+2b+3c=0\mod7\\ 2a+3b+c=4\mod7\\ 3a+b+2c=4\mod7\\ \)

\(\begin{bmatrix} 1 && 2&& 3&&|&& 0\;(mod7)\\ 2&& 3 && 1&&|&& 4\;(mod7)\\ 3 && 1&& 2 &&|&& 4\;(mod7)\\ \end{bmatrix} \\~\\~\\ \begin{bmatrix} 2 && 4&& 6&&|&& 0\;(mod7)\\ 2&& 3 && 1&&|&& 4\;(mod7)\\ 3 && 1&& 2 &&|&& 4\;(mod7)\\ \end{bmatrix} \\~\\~\\ \begin{bmatrix} 3 && 6&& 9&&|&& 0\;(mod7)\\ 0&& -1 && -5&&|&& 4\;(mod7)\\ 3 && 1&& 2 &&|&& 4\;(mod7)\\ \end{bmatrix} \\~\\~\\ \begin{bmatrix} 1 && 2&& 3&&|&& 0\;(mod7)\\ 0&& -1 && -5&&|&& 4\;(mod7)\\ 0 && -5&&-7 &&|&& 4\;(mod7)\\ \end{bmatrix} \\~\\~\\ \begin{bmatrix} 1 && 2&& 3&&|&& 0\;(mod7)\\ 0&& 1 && 5&&|&& -4\;(mod7)\\ 0 && 0&& 18 &&|&& -16\;(mod7)\\ \end{bmatrix} \\~\\~\\ \begin{bmatrix} 1 && 2&& 3&&|&& 0\;(mod7)\\ 0&& 1 && 5&&|&& -4\;(mod7)\\ 0 && 0&& 4 &&|&& -2\;(mod7)\\ \end{bmatrix}\)

\(4c=-2\mod7\\ 4c=5\;\;\mod7\\ 4c=5+7\mod7\\ 4c=12\;\;\mod7\\ c=3\\~\\ b+5c=-4\mod7\\ b+15=-4\mod7\\ b=-19\mod7\\ b=-21+2\mod7\\ b=2\\~\\ a+2b+3c=0\mod7\\ a+4+9=0\mod7\\ a=-13\mod7\\ a=-14+1\mod7\\ a=1 \)

a=1,  b=2,  c=3

Under here is not supposed to display, (except as LaTex coding.)

For some reason, it keeps rendering.......

LaTex:

a+2b+3c=0\mod7\\
2a+3b+c=4\mod7\\
3a+b+2c=4\mod7\\

\begin{bmatrix}
1 && 2&& 3&&|&& 0\;(mod7)\\
2&& 3 &&  1&&|&& 4\;(mod7)\\
3 && 1&& 2 &&|&& 4\;(mod7)\\
\end{bmatrix}
\\~\\~\\
\begin{bmatrix}
2 && 4&& 6&&|&& 0\;(mod7)\\
2&& 3 &&  1&&|&& 4\;(mod7)\\
3 && 1&& 2 &&|&& 4\;(mod7)\\
\end{bmatrix}

\\~\\~\\
\begin{bmatrix}
3 && 6&& 9&&|&& 0\;(mod7)\\
0&& -1 &&  -5&&|&& 4\;(mod7)\\
3 && 1&& 2 &&|&& 4\;(mod7)\\
\end{bmatrix}

\\~\\~\\
\begin{bmatrix}
1 && 2&& 3&&|&& 0\;(mod7)\\
0&& -1 &&  -5&&|&& 4\;(mod7)\\
0 && -5&&-7 &&|&& 4\;(mod7)\\
\end{bmatrix}

\\~\\~\\
\begin{bmatrix}
1 && 2&& 3&&|&& 0\;(mod7)\\
0&& 1 &&  5&&|&& -4\;(mod7)\\
0 && 0&& 18 &&|&& -16\;(mod7)\\
\end{bmatrix}

\\~\\~\\
\begin{bmatrix}
1 && 2&& 3&&|&& 0\;(mod7)\\
0&& 1 &&  5&&|&& -4\;(mod7)\\
0 && 0&& 4 &&|&& -2\;(mod7)\\
\end{bmatrix}

4c=-2\mod7\\
4c=5\;\;\mod7\\
4c=5+7\mod7\\
4c=12\;\;\mod7\\
c=3\\~\\
b+5c=-4\mod7\\
b+15=-4\mod7\\
b=-19\mod7\\
b=-21+2\mod7\\
b=2\\~\\
a+2b+3c=0\mod7\\
a+4+9=0\mod7\\
a=-13\mod7\\
a=-14+1\mod7\\
a=1

Hi ilorty,

I put it into microsoft translator and got:

Unit 90 Multiply 6 Employees 4

I would interpret this as each of 4 employees got 6 units, where each unit consists of 90 items.  

It sometimes pays to use more than one interpreter.  My interpretation may not be right of course. 

But Max interpreted it the same way.