All Answers 359404 Answers

ok thanks 

Hi Melody,

no

$$\\\left(a^{-12}\right)^{1/4}\\\\
=a^{-12*1/4}\\\\
=a^{-3}\\\\
=\frac{1}{a^3}$$

It might be worthwhile for you to look throught this thread.

Thanks Heureka,

Is your way the standard 'formula' presentation?

That is good :)

Well if you are going to do that you will have to do well! Otherwise it would be embarassing!

And, it would be an insult to Camelot.

Good one Chris :)

The absolute value of x is the one Aziz gave us as a counter example. (from the site he sent us to)

My query/consideration was a tad different.

cos(sin^-1(5/x))

$$\cos{(\sin^{-1}{( \frac{5}{x} )})} = \pm \sqrt{ 1 - \left(
\frac{5}{x} \right)^2 }$$

or

$$\cos{(\sin^{-1}{( \dfrac{z}{1} )})} = \pm \sqrt{ 1 - \left(z
\right)^2 }$$

Good point Alan,

I had not heard of PEDMAS till I was on this forum.

We call it BODMAS   Brackets, powerOf, etc

To me parentheses are only curly 'brackets' and these are rarely used.  

Parentheses is not a word students here would often hear.  

Yes, it is reasonable so long as I don't have to keep up for any length of time.  I am a dawdler.  

Eh, Melody....that rate is about 3.2 mi/hr......i.e., about the rate of a "medium" pace......definitely not "slow," but definitely not "fast," either.......Aziz's observation seems reasonable.......

Thanks Alan,

Alan has interpreted the question differently from me.

He thinks that you want the solution to    $$(x+9)^2 = x^2$$

If   $${{\mathtt{a}}}^{{\mathtt{2}}} = {{\mathtt{b}}}^{{\mathtt{2}}}$$    then  $$a=\pm b$$

So either x+9=x which has no solution.

or

x+9=-x and the solution is x=-4.5

I don't know Aziz

that is 1.47*60*60 = 5292m/hour = 5.292km/hour

I think the people of California must walk faster than the ones in Sydney.  

Using an exponent law, we have

(a^-2)^-1 = a^{(-2 * -1)}  = a²

I always do these with a triangle to aid me.

if x is positive then the angle is in the 1st or 2nd quads and cos of the angle could be pos or neg.

If xis neg then the angle is in the 3rd or 4th quads and cos  of the angle could be pos or neg.

now I'll look at the first quad.

$$cos\left(sin^{-1}\left(\frac{5}{x}\right)\right)$$

$$I let $\theta =sin^{-1}\left(\frac{5}{x}\right)$\\
I then found the third side using Pythagoras' Theorum.\\
Hence\\
$cos\left(sin^{-1}\left(\frac{5}{x}\right)\right)=\pm \frac{\sqrt{x^2-25}}{x}$$$

$$ok, let me see if I can make this look like Heureka's answer\\\\
$\pm \frac{\sqrt{x^2-25}}{x}$\\\\
=$\pm \sqrt{\frac{{x^2-25}}{x^2}$\\\\
=$\pm \sqrt{1-\frac{{25}}{x^2}$\\\\
=$\pm \sqrt{1-\frac{{25}}{x^2}$\\\\
=$\pm \sqrt{1-\left(\frac{{5}}{x}\right)^2$\\\\$$

There you go 

(x + 9)² = x² if x = -4.5

Ginger your post is very colourful and arty, a little French in origin perhaps?

Your grand total of 2 points looks a little pathetic so I will more than double your points for you.

It is not often that one of my thumbs can have such a dramatic effect.

But, this is the magical land of Camelot, and I am the Lady of the Land.  

Apparently, the "pm" for Aziz's time zone differs from that of heureka's.....remember to adjust your watches accordingly........

And, in Melody's time zone, "pm" is always afternoon.....even at 11 at night !!!!........so I see why she needed to steal some of these  ...........

if pn=29cm and mn=13cm, then pm=?

0--------------------0--------------------0

p                          n                         m

<-- pn = 29 cm --><-- mn = 13 cm -->

<------------- pm = 42 cm ------------->

pm = 42 cm

As one of my calculus teachers would have said.....differentiability implies continuity, but continuity does not imply differentiability.....

To see why, look at the absolute value function at x = 0. Note that, it is continuous at that point but the derivative is undefined. (What is the "slope" of the function at that point .....1  or -1 ??)

We can see this in one more way......the absolute value of a number "x" is defined as √(x²).....and taking the derivative of this gives us  .......   (1/2) (x²)^-1/2 *(2x)  =  x / √(x²) =  x / lxl     ........and note that, this derivative is undefined at x = 0  !!!!

I am impressed too !!!!

Looks like you have quite a fan club David.    

(I found these Chris, you just leave them lying around everywhere.  Finders keepers! )

Thank you CPhill. How many imperial minims in a minimum daily requirement of CN?

Ah, trying to slip some cyanide into the forum, are you DavidQD???   Well, you won't get rid of us that easily.........we're too crafty to fall for such a ruse....!!!!

{ BTW.....an impressive dispaly of LaTex !!!! }

This is a cool work of art. So is the humour.