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Suppose \(\dfrac{1}{x} = \dfrac{2}{y + z} = \dfrac{3}{ z + x} = \dfrac{x^2 - y - z}{x + y + z}\).  

What is \(\dfrac{z - y}{x}\) ?

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{x}} &=& \mathbf{\dfrac{3}{ z + x}} \\\\ z + x &=& 3x \\\\ \mathbf{z} &=& \mathbf{2x} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{x}} &=& \mathbf{\dfrac{2}{y + z}} \\\\ y+z &=& 2x \quad | \quad \mathbf{z=2x} \\\\ y+2x &=& 2x \\\\ \mathbf{y} &=& \mathbf{0} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{z - y}{x}} \quad | \quad \mathbf{y=0} \\\\ &=& \dfrac{z}{x} \quad | \quad \mathbf{z=2x} \\\\ &=& \dfrac{2x}{x} \\\\ &=& \mathbf{2} \\ \hline \end{array}\)

How many rectangles are there in a chessboard ?

Hints:

\(f(4)+2f( a )= 3x^2\\ find \;\;a\\ then\\ f(a)+2f(b)=3x^2\\ find\;\;b\\ \)

Now you can solve for  f(4) using simultaneous equations. 

Please no one provide a full answer. 

4^x+6^x=9^x

\(4^x+6^x=9^x\\ 2^x(2^x+3^2)=3^x3^x\\ 2^x+3^x=3^x\cdot (\frac{3}{2})^x\\ (\frac{2}{3})^x+1= (\frac{3}{2})^x\\ let\;\;y=2/3\\ y^x+1=(\frac{1}{y})^x\\ y^{2x}+y^x=1\\ (y^{x})^2+y^x-1=0\\ y^x=\frac{-1\pm\sqrt{1+4}}{2}\\ y^x \;\;must\;\;be \;\;pos\;\;so\;\;\\ y^x=\frac{-1+\sqrt{5}}{2}\\ log(y^x)=log(\frac{-1+\sqrt{5}}{2})\\ xlog(y)=log(\frac{-1+\sqrt{5}}{2})\\ xlog(\frac{2}{3})=log(\frac{-1+\sqrt{5}}{2})\\ x=log(\frac{-1+\sqrt{5}}{2})\div log(\frac{2}{3})\\ x\approx 1.1868\)

Latex coding

4^x+6^x=9^x\\
2^x(2^x+3^2)=3^x3^x\\
2^x+3^x=3^x\cdot (\frac{3}{2})^x\\
(\frac{2}{3})^x+1= (\frac{3}{2})^x\\
let\;\;y=2/3\\
y^x+1=(\frac{1}{y})^x\\
y^{2x}+y^x=1\\
(y^{x})^2+y^x-1=0\\
y^x=\frac{-1\pm\sqrt{1+4}}{2}\\
y^x \;\;must\;\;be \;\;pos\;\;so\;\;\\
y^x=\frac{-1+\sqrt{5}}{2}\\
log(y^x)=log(\frac{-1+\sqrt{5}}{2})\\
xlog(y)=log(\frac{-1+\sqrt{5}}{2})\\
xlog(\frac{2}{3})=log(\frac{-1+\sqrt{5}}{2})\\
x=log(\frac{-1+\sqrt{5}}{2})\div log(\frac{2}{3})\\
x\approx 1.1868

64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204. 64 in total but it depends on the actual chess board, the sequence on the left shows the different possible rectangles a chess board can have. 

yeah correct, showing the steps is challenging tho. 

We will not be helping you to cheat on your homework problems.

We will not be helping you to cheat on your homework problems.

We will not be helping you to cheat on your homework problems.

x≈1.18681

Nice Photo.

25 =400 x 1/2^(y/3175), solve for y

0.0625 = 1/2^(y/3175)

y =12,700 Years - when 400 grams decays to 25 grams.

a=1; b=2007;c= b% a; if( c==5, goto4, goto5);printa," ",; a++;if(a<1000, goto2, 0)

Solutions =7  11  13  14  22  26  77  91  143  154  182  286  

All above solutions leave a remainder of 5, but only one is between 200 and 300.

Therefore, the answer =286 Boxes.

By similar triangles, BC = 24*9/12 = 18.

Thank you so much.

wait why was my post removed and why is the rest invisible....i did not say anything bad. 
I just told him to ignore the mean guest who thinks they are better than everybody else. :(

x^2 - 13x  + 1  = 0

The sum of the roots  =13

The product of the roots  = 1

But....if the product of the  roots is  1.... we can call one root x  and the other   1/x

And the sum of the roots  =  x  + 1/x  =  13

So

(x)^2 - 13(x) + 1  = 0

(1/x)^2 - 13 (1/x) + 1  = 0        add these

(x)^2 + (1/x)^2 - 13 ( x + 1/x)  + 2  = 0

(x)^2  + (1/x)^2 -13 (13)  +  2  = 0

(x)^2 + (1/x)^2  - 169 + 2  =  0

(x)^2  + (1/x)^2  -167  = 0

(x)^2 + (1/x)^2  =   167      square both sides

(x)^4  + 2 (x)^2 * (1/x)^2  +  (1/x)^2  =  27889      simplify

x^4 + 2 (x^2/x^2) +  1/x^4  = 27889

(x)^4 +  2  +  (1/x)^4   +  2  =  27889       subtract 2 from both sides

x^2 +  1/x^4   = 27887

Let the starting number of watermelons=W
[1/2W - 1/2] - Melons left after the first hour.
[1/4W - 1/4 - 1/2] - After the second hour.
[1/8W - 1/8 - 1/4 - 1/2] - After the third hour.
[1/8W - 0.875] = 6, solve for W
W = 55

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