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1*8*2*8*3*8*4*8*5*8*6*8*7*8*8*8*9*8

= (8*8*8*8*8*8*8*8*8)*(1*2*3*4*5*6*7*8*9)

= $$8^9\cdot 9!$$

It should be pointed out that the "cross partial derivatives" are not always equal, but usually they are !

Alan, sorry

Hi Melody

I don't have an answer to this question yet, other than taking Nauseated's formula on trust, but I can tell you why your logic is wrong.

For your 5,3 setup, there will be times when one or even two of the printers have no computer connected to them at all, (at odds with the requirement).

Consider one of your possibilities, one where each of the five computers is connected to a single printer. Those connections could be to just two, or even just one of the printers. Those possibilities have to be subtracted from your total. The problem is how to count them.

Has anyone noticed odd changes -- like space-time distortions where yesterday is tomorrow, and the universe seems a little more salty?

Maybe it’s just me :)

The cross-partial derivatives are defined as:

z_xy = ∂/∂y(∂z/∂x)   and z_yx = ∂/∂x(∂z/∂y)

so, with z = x^0.3y^0.5

First do a partial derivative with respect to x: z_x = 0.3x^-0.7y^0.5

Now do a partial derivative with respect to y on the result:   z_xy = 0.15x^-0.7y^-0.5

Now try z_yx yourself.  (Ah, heureka's given the game away!)

.

 z = x^0.3 y^0.5 

I. partial derivative of z with respect to x :

$$\dfrac{\partial z}{\partial x} =0.3x^{-0.7}y^{0.5}$$

II. partial derivative of z with respect to y :

$$\dfrac{\partial z}{\partial y}=0.5x^{0.3}y^{-0.5}$$

Differentiate again to find second - order
partial derivatives:

We can also differentiate  $$\small{\frac{ \partial z } { \partial x } \end{array}$$ with respect to y, to find out how it changes when y increases.
This is written as $$\small{\frac{\partial^2 z}{\partial x\ \partial y}}\end{array}$$ and is called a cross-partial derivative:

III. cross partial differential

$$\boxed{\ \dfrac{\partial^2 z}{\partial y\ \partial x}
=\dfrac{\partial^2 z}{\partial x\ \partial y} \ }$$

$$\dfrac{\partial z}{\partial x\ \partial y}=
0.15\ \cdot x^{-0.7}y^{-0.5}$$

What is this bull s**t? Let me guess: you do everything 338 times?

A wrong answer with OCD!

arc tan is the same as inverse tan.

Just use atan

I think that you mean

$$\\sin^2\;\frac{\pi}{7}+cos^2\;\frac{\pi}{7}\\\\
$Well this is very simple because $\\\\
sin^2\theta +cos^2\theta = 1\qquad ALWAYS$$

Mr. Graham invested some of his $18,000 in bonds that made a 5% profit and the rest in bonds that made a 12% profit. If the profit on the 12% bonds was $885 more than the profit on the 5% bonds, how much did Mr. Graham invest in the 5% bonds?

Let x be the 5% portion and  (18000-x) be the 12% portion                       

5% of X +885= 12% of (18000-X)

0.05X+885 = 0.12(18000-X)

can you finish this and then check that it is correct.  ?

tan x is negative so x could be in the 2nd or 4th quadrant.  Cos x and sin x could be neg or positive.

3pi/2 < x < 2pi      This is the 4th quad    SO

Cos(x) is positive and   sin(x) is negative

tanx = opp / add     SO  

opp=5     adj = 6    hypot = sqrt(25+36) = sqrt(61)    [or a constant times these]

$$\\cos(x)=\frac{6}{\sqrt{61}}\\\\
sin(x)=-\frac{5}{\sqrt{61}}\\\\$$

You use the Pythagorean Theorem

OR

use the distance formula     (which comes from the Pythagorean Theorem )

I don't know what a cross partial differential is but i think I can answer the partial differentials 

Could another mathematician finish it and check what I have done please ?

I am not sure it I am even using the correct symbols.     

$$\\z = x^{0.3} y^{0.5}\\\\
\frac{\delta z}{\delta x}=0.3x^{-0.7}y^{0.5}\\\\
\frac{\delta z}{\delta y}=x^{0.3}*0.5y^{-0.5}=0.5x^{0.3}y^{-0.5}\\\\\\
$The rest is for my benefit, I'd like another mathematician to check it please.$\\\\
z = x^{0.3} y^{0.5}\\\\
\frac{dz}{dx}=0.3x^{-0.7}y^{0.5}+x^{0.3}*0.5*y^{-0.5}\;\frac{\delta y}{\delta x}\\\\
\frac{dz}{d y}=0.5x^{0.3}y^{-0.5}+y^{0.5}*0.3x^-0.7\;\frac{\delta x}{\delta y}\\\\\\\\\\$$

$${\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}$$

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}$$

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{12}}}{{\mathtt{2}}}}$$

$${\frac{{\mathtt{13}}}{{\mathtt{2}}}}$$

$${\mathtt{6.5}}$$

Hi Gibson,

You probably already know that I am not a physics person but I will give it a shot :)

I have not taken the swinging bit into any consideration.  I don't think that you can because you do not know what its 'radius' is.

F=ma                 m=10   pounds

s=37 inches,  t=1/3 sec,   u=0,  v=?     a=?      But I need to find a

So if I am to use physics formulas then it could have s,t,u and a   BUT not v

$${\mathtt{S}} = {\mathtt{ut}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{{\mathtt{at}}}^{{\mathtt{2}}}$$

$$\\37=0t+\frac{1}{2}a\left(\frac{1}{3}\right)^2\\\\
37=\frac{1}{2}*a*\frac{1}{9}\\\\
37*2*9=a\\\\
a=666\;inches/s^2\\\\\\
F=ma\\\\
F=10*666\\\\
F=6660\;pounds\;inches/s^2\\\\
F=6660/12\;pounds\;feet/s^2\\\\
F=555\;pounds\;feet/s^2\\\\$$

NOTE: A small error has been fixed.   Maybe it is right now.         

$${{\mathtt{0}}}^{{\mathtt{0}}}$$

$${\mathtt{ERROR}}$$

$${{\mathtt{0}}}^{{\mathtt{0}}} = \underset{{\tiny{\text{Error: indeterminante}}}}{{{\mathtt{0}}}^{{\mathtt{0}}}}$$

$${\frac{{\mathtt{2}}}{{\mathtt{9}}}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{9}}}}\right)$$

$${\frac{{\mathtt{2}}}{{\mathtt{81}}}}$$

$${\mathtt{0.024\: \!691\: \!358\: \!024\: \!691\: \!4}}$$...

$${\mathtt{y}}{\mathtt{\,\times\,}}{\mathtt{5\,412\,687\,930.023\: \!56}} = {\mathtt{2\,341\,235\,239\,278\,390\,278\,238\,373}}$$

$${\mathtt{y}} = {\frac{{\mathtt{2\,341\,235\,239\,278\,390\,278\,238\,373}}}{{\mathtt{5\,412\,687\,930.023\: \!56}}}}$$

$${\mathtt{y}} = {\mathtt{432\,545\,764\,608\,343.030\: \!784\: \!004\: \!655\: \!926}}$$...

$${\sqrt{{\mathtt{3}}}}$$

$${\sqrt{{\mathtt{3}}}}$$ or $${\mathtt{1.732\: \!050\: \!807\: \!568\: \!877\: \!3}}$$...

$${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{729}}}}$$

$${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3}}}}$$

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3}}$$

$${\mathtt{9}}$$

$${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{729}}}} = {\mathtt{9}}$$

Put x=0 into each of the equations and see which one/s give you an answer of y=-3

Let me know how you get on :)

$$\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right){\mathtt{\,\times\,}}{\mathtt{3\,500}} = {\mathtt{875}}$$

does that make sense to you ?

I will use LaTex to help with the display. :)

$$\frac{15}{10}=\frac{3*5}{2*5}=\frac{3}{2}$$

 Yes, I know, you already know that :) 

You are allowed to do the same things with letters (pronumerals) as you can with numbers - so just follow the same rules.  :)

$$\\\(1)\;\;\frac{ab}{ac}=\frac{a*b}{a*c}=\frac{b}{c}\qquad$The a's cancelled out$\\\\\\
(2)\;\;\frac{2e}{4f}=\frac{2*e}{2*2*f}=\frac{e}{2f}\qquad$The 2's cancelled out$\\\\\\
(3)\;\;\frac{4a^2b}{2ab}=\frac{\not{4}^2*\not{a}*a*\not{b}}{\not{2}*\not{a}*\not{b}}=\frac{2a}{1}=2a \qquad$The rest cancelled out$\\\\\\
(4)\;\;\frac{a}{b}*\frac{b}{c}*\frac{c}{a}=\frac{a*b*c}{b*c*a}=\frac{1}{1}=1\qquad$They all cancelled out$\\\\\\
(5)\;\;\frac{(3x+2)^2}{6x}*\frac{x^4}{(6x+4)}\\\\
=\frac{(3x+2)*(3x+2)}{6x}*\frac{x*x*x*x}{2(3x+2)}\\\\
=\frac{(3x+2)}{6x}*\frac{x*x*x*x}{2}\qquad$the (3x+2)'s cancelled out$\\\\
=\frac{(3x+2)}{6}*\frac{x*x*x}{2}\qquad$an x cancelled out$\\\\
=\frac{(3x+2)*x*x*x}{6*2}\\\\
=\frac{(3x+2)x^3}{12}$$