9th June strip

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BlasTech
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9th June strip

Post by BlasTech »

Heh, those brownies definately dont like squidge :lol: panel 3 just cracked me up XD

Must compliment quent on his new outfit, and the headgear :D



Preliminary translation of the brownie language

Squidge: Hello gnomes, i am ally?
Gnomes: Ha! Bug Eye!! Kill the Scum!
Squidge: YAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA!

Dont know if there's anything in the next line, looks just like random warcries :P

And when he holds up the spear
"Eh, North Tribe?"
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Post by Astral »

Aww poor squidge XD
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Post by Fusion »

POOR SQUIDGE?!? :o
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Post by CasVeg »

If the white stag keeps licking Quentyn, he just might end up with a big white mohawk.

Okay, maybe not; but, I still can't get the image out of my head.
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Post by SolidusRaccoon »

CasVeg wrote:If the white stag keeps licking Quentyn, he just might end up with a big white mohawk.

Okay, maybe not; but, I still can't get the image out of my head.
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Post by The JAM »

[...unWARP!!!]

Good evening.


What does the LRR mean? It can't be a street name...a sector name, perhaps?


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DracoDei
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Post by DracoDei »

Branches to take on the streets from the center of town (hexagonal streetmaps remember? Although that might only be Sanctuary...)
Left
Right
Right
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Post by UncleMonty »

I'm 'way too lazy to work out what the brownies are saying, but at least Quentyn's got this tribe's attention. Hmmm... I hope the tribes aren't warring with each other!

It's getting harder to find a thread for this forum that isn't dedicated to fanfic, too. I'm glad THIS thread exists even if I don't have a clue as to what's going on in the comic!
:roll:
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Post by The JAM »

[...unWARP!!!]

Good evening.


Squidge: "Hello, Gnomes! I am ally?"

Brownie: "HAI!! Bogey! Kill the scum!"

Squidge: "EEEEYAAAAGH!!!!"

Brownies: "Ungawa! Tuk tuk tuk arrgh!"

[after Quentyn calms everyone down]

Brownie: "Eh? North Tribe?"


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Post by RedSquirrel456 »

Branches to take on the streets from the center of town (hexagonal streetmaps remember? Although that might only be Sanctuary...)
Left
Right
Right
Two wrongs don't make a right, but two rights make a one hundred eighty degree turn...
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Post by StrangeWulf13 »

Only if those rights equal 90 degrees. Which, in a circular city, they don't.

Divide 360 by 3 to get the angle...

120 degrees, going by the outside. So, to find the inside angle, we subtract 120 from 180.

The left and right turns are 60 degrees each, if you face the middle of the fork and turn until you're facing down one of the streets.

=P Little bit of math never hurt anyone. Calculus on the other hand...

...well, the United Nations is currently sending inspectors to look into suspected war crimes commited by this hideous beast in classrooms worldwide. Math teachers are advised to comply with their requests under threat of invasion by the United States without UN approval.

Ya might wanna duck and cover when you see the marines knocking on your classroom door... :P
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Post by BlasTech »

ahhh calculus ... a true weapon of maths instruction
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Post by Astral »

StrangeWulf13 wrote:Only if those rights equal 90 degrees. Which, in a circular city, they don't.

Divide 360 by 3 to get the angle...

120 degrees, going by the outside. So, to find the inside angle, we subtract 120 from 180.

The left and right turns are 60 degrees each, if you face the middle of the fork and turn until you're facing down one of the streets.
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Post by SolidusRaccoon »

BlasTech wrote:ahhh calculus ... a true weapon of maths instruction
Grrr, you are really pushing your luck today. :D
Yes, sir. I agree completely. It takes a well-balanced individual... such as yourself to rule the world. No, sir. No one knows that you were the third one... Solidus. ...What should I do about the woman? Yes sir. I'll keep her under surveillance. Yes. Thank you. Good-bye...... Mr. President.
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Post by Shyal_malkes »

((9x+2)/(3x^2-2x-8))+((7)/3x^2+x-4)

now simplify and solve as best you can!

you have never lived until you take a class that you fear will be challenging and on the fourth day you have progressed beyond any skill you have otherwise developed yet and you STILL understand every word the teacher says!
I still say the doctor did it....
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Post by Trump »

shyal_malkes wrote:((9x+2)/(3x^2-2x-8))+((7)/3x^2+x-4)

now simplify and solve as best you can!

you have never lived until you take a class that you fear will be challenging and on the fourth day you have progressed beyond any skill you have otherwise developed yet and you STILL understand every word the teacher says!
Did you mean to put an extra set of parenthesis around the 3x^2+x-4 part? Also should I assume the smiley to be 8 with an extra parenthesis?
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Aurrin
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Post by Aurrin »

I'm fairly sure he meant it like this:

Code: Select all

    9x + 2               7
---------------  + --------------
 3x^2 - 2x - 8      3x^2 + x - 4
Conquering the Universe, one class at a time...
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Post by Shyal_malkes »

somehow i must have hit a code thingy and my 8 parentheses was interperated as a smiley

great my first smiley and it was a total accident!

8) lets see if it does it again...
I still say the doctor did it....
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Post by Shyal_malkes »

yes it did and yes aurrin you are correct that is how it looks like in my notes (this was prob 64 that our teacher was reviewing with us, i forget what page number or really what book it was, not that it matters.)
I still say the doctor did it....
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CasVeg
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Post by CasVeg »

Code: Select all

          9x + 2            7
[1.1]  ------------- + ------------
       3x^2 - 2x - 8   3x^2 + x - 4
The quadratic expressions can be factored.

Code: Select all

           9x + 2               7
[1.2]  --------------- + ---------------
       (3x + 4)(x - 2)   (3x + 4)(x - 1)
Partial fraction decomposition may now be used on each of the rational expressions.

The first expression:

Code: Select all

         A        B         9x + 2
[1.3]  ------ + ----- = ---------------
       3x + 4   x - 2   (3x + 4)(x - 2)
Multiplying by the least common denominator,

Code: Select all

[1.4]  Ax - 2A + 3Bx + 4B = 9x + 2
This yields the following system of equations:

Code: Select all

[1.5]    A + 3B = 9
[1.6]  -2A + 4B = 2
or
[1.7]   -A + 2B = 1
[1.5] and [1.7] may be added to eliminate A.

Code: Select all

       5B = 10
[1.8]   B = 2
By substitution,

Code: Select all

[1.9]  A = 3
So,

Code: Select all

            9x + 2          3        2  
[1.10]  --------------- = ------ + -----
        (3x + 4)(x - 1)   3x + 4   x - 2
The second expression:

Code: Select all

          A        B            7
[1.11]  ------ + ----- = ---------------
        3x + 4   x - 2   (3x + 4)(x - 1)
Multiplying by the least common denominator,

Code: Select all

[1.12]  Ax - A + 3Bx + 4B = 7
This yields the following system of equations:

Code: Select all

[1.13]   A + 3B = 0
[1.14]  -A + 4B = 7
[1.13] and [1.14] may be added to eliminate A.

Code: Select all

        7B = 7
[1.15]   B = 1
By substitution,

Code: Select all

[1.16]  A = 3
So,

Code: Select all

              7             -3       1  
[1.17]  --------------- = ------ + -----
        (3x + 4)(x - 1)   3x + 4   x - 1
Substituting back into [1.2],

Code: Select all

          3        2       3        1
        ------ + ----- - ------ + -----
        3x + 4   x - 2   3x + 4   x - 1

          2       1
[1.18]  ----- + -----
        x - 2   x - 1
What? I like taking the scenic route. Besides, I couldn't pass up a chance to use partial fractions!

Okay, okay. . . Math is the art of doing as little work as possible. . . .

Code: Select all

          9x + 2            7
[2.1]  ------------- + ------------
       3x^2 - 2x - 8   3x^2 + x - 4
The quadratic expressions can be factored.

Code: Select all

           9x + 2               7
[2.2]  --------------- + ---------------
       (3x + 4)(x - 2)   (3x + 4)(x - 1)
The rational expressions may be added together.

Code: Select all

       (9x + 2)(x - 1) + 7(x - 2)
       --------------------------
         (3x + 4)(x - 2)(x - 1)

             9x^2 - 16
[2.3]  ----------------------
       (3x + 4)(x - 2)(x - 1)
The numerator in [2.3] is a difference of squares.

Code: Select all

          (3x + 4)(3x - 4)
[2.4]  ----------------------
       (3x + 4)(x - 2)(x - 1)
[2.4] is simplified by cancelling 3x + 4 in the numerator and denominator.

Code: Select all

           3x - 4
[2.5]  --------------
       (x - 2)(x - 1)
As this is an expression, it can't actually be "solved". However, it is worth noting that the expression is equal to zero when x = 4/3 and undefined when x = 1, x = 2, or x = -4/3. The latter is true because 3x + 4 was in the denominators in the original expression.


P.S.: When you make a post, there is an option to disable smilies. I have mine disabled by default. 8)
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