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This is for you GingerAle...

First of all, don't say hypercube because a cube is not a tesseract as glorified as you want to make said cubes. Nothing you do to a cube will make it a tesseract so don't go there with your hypercubes. (unless of course, this discussion included a large and inconvenient list of shapes across a wide span of dimensions)

Second, of all, you don't read minds. Stop saying you do. Stop making a fool of yourself. Everyone knows it so please stop embarrassing yourself and just stop. Makes you sound childish af.

Third, it is the art of math NOT math of art... know what you are talking about before speaking of it. Okay?

Fourth, great job really... claps* more claps* sarcastic and slow claps* long silence*

Oh and lastly, any hyper-cuboid as you would probably put it has just one diagonal. If you actually know what a diagonal is then it would be silly to say that any shape of any dimension would have more than one diagonal.

This is for you Melody...

I respect what you said. If Gingerale wants to continue then he/she will respond and I will happily continue with Gingerale in private. (btw I am learning... it just is not math from this experience)

As for anyone else...

post HERE in PUBLIC if you would like to attempt my second problem.

Let g(x) be a horizontal shrink by a factor of 2/3, followed by a translation 4 units left of the graph of f(x)=√(6x). Write a rule for g(x) described by the transformations of the graph of f. I thought it was g(x)=√(4x+4) but maybe not, what is the correct answer and why?

$f(x)=\sqrt{6x}\\ \text{Now shrink this by a factor of 2/3}\\ u(x)=\frac{2}{3}\sqrt{6x}\\ \text{Now move the graph 4 units in the negative direction}\\ g(x)=\frac{2}{3}\sqrt{6(x+4)}$

Here are the graphs

They go     green, orange, red   (just like trapffic lights)

My bad, didn't picture it this way in my mind.

I already know this, don't know why I didn't think of it that way on this one...

Thanks!

Awesome!

Could you just explain to me how you go from:

sqrt (k + x ) = 7  - sqrt(x)   square both sides

to

k + x  =     49   -  14 sqrt(x) +  x  

+++++++++++++++++++++++++++++++++++++++++++++++++++++++

OK.....squaring both sides, we have

(√( k + x) )^2  = ( 7  - √x)^2      [squaring the left side just causes the radical to disappear]

k + x  =  ( 7  - √x) ( 7  - √x)    expand on the right

k + x  =  49  - 7√x  - 7√x  + x

k + x  =  49  - 14√x    + x   

Does that help ??

Awesome!

Could you just explain to me how you go from:

sqrt (k + x ) = 7  - sqrt(x)   square both sides

to

k + x  =     49   -  14 sqrt(x) +  x  

I don't get how  ( - sqrt (x)  )^2  =  -14sqrt(x) + x

OK...I see....let me get you there

sqrt (k + x ) + sqrt(x) - 7 = 0     rearrange

sqrt (k + x ) = 7  - sqrt(x)   square both sides

k + x  =     49   -  14 sqrt(x) +  x        subtract x from both sides

k = 49  -  14 sqrt(x)   rearrange

14 sqrt(x)  =  49  - k  square both sides

196x  =  2401  - 98k + k^2     divide both sides by 196

x  = [2401  - 98k + k^2] / 96  =  [ k^2 - 98k + 2401] / 96 

Voila....!!!!!

Here is the exact problem given in my class:

"Resolve in X":

sqrt (k + x) + sqrt (x) - 7 = 0

Now here is the answer provided by the book:

( k^2 - 98k + 2401 ) / 196    where k ∈ R

infinity  + 1   =  infinity

sqrt (k + x ) + sqrt(x) - 7 = 0

Note, Tony.......there are  infinite things that might work, here.....because we have more variables than equations......for example...

k = 49    and x  = 0

sqrt (49 + 0 ) + sqrt(0) - 7 = 0

sqrt (49 ) + 0  - 7 = 0

7 + 0 - 7   = 0

or

k = 7  and x = 9

sqrt (7 + 9 ) + sqrt(9) - 7 = 0

sqrt (16) + sqrt(9) - 7 = 0

4   +  3    - 7 =  0

7  - 7 = 0

etc  .......

This willl always happen whenever we have more variables than equations.......

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Thanks that's what I needed to know. I said "negative" but really just refered to i.

(5 - sqrt ( -59) ) / 6

(5+ sqrt (-59) ) / 6

The concept of "positive" and/or "negative" − in real number terms −  has no meaning here, because both of these will be complex solutions......

Ahh   I told you CPhill was a better guesser than me  :))

Silly me, I answered the wrong question.  Thanks Chris :)

(k-2) x^2 -5x +7 = 0

Knowing that P = 7  (P = c/a)

c /a    =   7 / (k − 2)  =  7 / 1

This implies that .....

k − 2   = 1

k = 3

What havew P,c and a got to do with anything?

(k-2) x^2 -5x +7 = 0

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {5 \pm \sqrt{25-4*7*(k-2)} \over 2(k-2)}\\ x = {5 \pm \sqrt{25-28(k-2)} \over 2(k-2)}\\ x = {5 \pm \sqrt{25-28k+56} \over 2(k-2)}\\ x = {5 \pm \sqrt{81-28k} \over 2(k-2)}\\$

You had better explain better what you are on about.   k can be anything except 2

and if x is real then 

$81-28k\ge0\\-28k\ge-81\\ k\le\frac{81}{28}\\$

If x is rational then 

$81-28k \text{ is a perfect square}$

I see CPhill is answering - he may be a better guesser than me.

$(k-2)x^2-5x+7=0\\ (k-2)x^2-5x=-7\\ (k-2)x^2=5x-7\\ k-2=(5x-7)/x^2\\ k=(5x-7)/x^2+2$

Call the amount that the first cask can hold  = A  .... and call the amount that the second cask can hold  = 2A

So the amount of 3 year old wine in the first cask = A/2....   and the amount of 3 year old wine in the second cask  = (1/3)2A  = (2/3)A

So....if he contents are mixed....the part that is 3 year old wine  =

[ A/2   +  (2/3) A ]  /  [ A + 2A]   = 

[ (3/6)A  +  (4/6)A ] / 3A =

[ (7/6)A]  / 3A  =

[ (7/6)] / 3  =

7 / 18

LOL   I meant 'cue'.

I do not think that there was any queue for this performance :))

The slope can be expressed as rise/run (Always remember: the slope is rise over run. That will help you in the future.)

Since the slope is 2/3, the rise is 2 for every 3 units of run

There is 6 units of run, which is 3 times 2

Therefore, there must be 4 units of rise, because 2 times 2 is 4.

The verticle rise between the two points is 4.

how do you find the vertical rise between two points on a line if their horizontal run is 6 and the slope of the line is 2/3

$slope=\frac{rise}{run}=\frac{2}{3}=\frac{rise}{6}\\ \qquad \frac{2}{3}=\frac{rise}{6}\\ \frac{2}{3}*\frac{6}{1}=\frac{rise}{6}*\frac{6}{1}\\ \frac{2}{\not{3}}*\frac{\not{6}^2}{1}=\frac{rise}{\not{6}}*\frac{\not{6}}{1}\\ \qquad rise=4$

Arr Ginger, you are indeed a well trained chimp :))

Performing on queue. 

Your trainer will be proud of you :)

Thanks Chris :))

Good job, Melody!!!........

I did it  a little differently.......I'll show my mehod in a day or two.....!!!!