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$$\\(-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{-2*-2}=\frac{1}{4}\\\\
but\\\\
-2^{-2}=-1*2^{-2} = \frac{-1}{(2)^2}=\frac{-1}{2*2}=\frac{-1}{4}\\\\$$

To plot a 3-d function, say z = f(x,y), you need a 3-d plotting tool.  In general there will be more than one slope at any point, depending on which direction you are interested in.  If you want the slope in the x-direction, for example, you would find the partial derivative of z with respect to x. 

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@@ End of Day Wrap    Thurs 13/11/14      Sydney,   Australia       Time 10:50 pm             ♬

Hi all,

Lots of great answers today from Alan, Voodox, CPhill, happy7, Hc2027, Anonymous4338, taufique, geno3141, TayJay, Jillybean and Heureka.  Thank you all  

[thanks for my thumbs up yesterday :) ] 

We have a "Rat Pack" back.     It is not the same pack as before, we do miss DragonSlayer554 and Zegroes and Rose98.  But still it is great to have a rat pack back again.  From my observation the new 'rat pack' includes Rosala, Tiddle, TayJay, happy7, Jillybean and Anonymous4338     

Do not take offense to the term 'Rat Pack' it is a term coined by CPhill and it is most definitely a term of endearment :)))   CPhill and I both love having an active Rat Pack on the forum  

Fri 14/11/14

1) The children have been very acive today :)  A toungue twister from Jillybean 

okay Happy - I will give you that one :)

5(19) = 800 + 450 ln (3(19)+e)  ?

e = 2.71828182845904523536028747135266249775724709369995... // Euler number

$$S(19) = 800 + 450* \ln{(3*19+ 2.71828182846 )} \\
S(19) = 800 + 450* \ln{(57+2.71828182846 ) } \\
S(19) = 800 + 450* \ln{ (59.7182818285) } \\
S(19) = 800 + 450* 4.08963820180 \\
S(19) = 800 + 1840.33719081 \\
S(19) = 2640.33719081 \\
S(19) \approx 2640$$

Hi Melody,

it is not reasonable to set  $$1^x=e^{x\times ln(1)}$$,

but in general it is easier to integrate $$e^x$$

Hi Heurka,

I don't get what you did all that for?

Why is it reasonable to set 

$$1^x=e^{x\times ln(1)}$$

I mean I know that

$$1^x=1$$

that is where i started but I do not understand why you put those lines of working in front of it.  

okay sorry!

baby dont hurt me ?

He is clever. I have tried that, and it works well for nausea caused by hyper-stimulated proton pumps, producing excess acid indigestion, from high fat foods.

Treating post CDD nausea requires something that modifies the spatial perception processing areas of the brain. Many treatments only work for a short while.

Most scientists do not understand the mechanism of the contagion of CDD, or why it often produces nausea after it is cured. Some think it is a quantum state effect known as the “spooky dumbness at a distance” theory.

This is the first thing i saw as the site opened so i'll answer this!Love,it has many forms , like the love between friends, and many more!well in general love is the care you do for someone!

integration of 1^x

$$\int{1^x}\ dx \ ?$$

  $$\begin{array}{llr}
set & \quad 1^x=e^ { x \times \ln{(1)} } & \qquad \ln{(1)} = 0\\
so & \quad 1^x=e^ { x \times 0} & \qquad x\times0= 0\\
so & \quad 1^x=e^0 & \qquad e^0=1\\
so & \quad 1^x=1
\end{array} \\
\int{1^x}\ dx = \int{1}\ dx = x + c$$

Beats me - maybe you should submit an actual question and Alan or someone else might be able to help you. :)

We all make mistakes, you certainly do not make any more than any one else.  :)))

Yes, what about it.

The point (4,3) is on that line.

because

4*4-5*3=1

Hi Melody,

Because the message started with the words "Hi Alan,..." I took it as a friendly pointer to a possible error (let's face it, I make enough mistakes!).

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Hi Alan,

Even if it is a different anon they can still acknowledge the work that you have done.

Might be a different Anonymous Melody.  That's the problem with Anonymous - he/she is a multi-headed beast!

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You know anon - picking up possible errors is fine but it is important to throw in a thank you as well.

Alan did a lot of work on your behalf!

No, I don't think so:

Or did you mean in the general expression with the a, b, c and d?  Here you need to look closely at the top right and bottom left terms to see the minus signs. At a glance they can look like a continuation of the division line!  This is a facet of the software I used to generate the expression - I've missed them myself in the past! 

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Divide numerator and denominator by 3

I assume this is:

$$T_y\times 2L^3-315L^2-180L=0$$

where the 315 and the 180 are Newtons, and L is dimensionless, so T_y is in Newtons.

Rearranging, we get:

$$T_y=\frac{315}{2L}+\frac{90}{L^2}$$

However, you need to know the value of L before you can evaluate this (L would have to be about 0.605 to get T_y as 506N)

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