TuringTest, repetition!

- anonymous

TuringTest, repetition!

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- TuringTest

repetition?

- anonymous

You know, practising something you have learned

- anonymous

?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

Oh maybe thats not a word xd

- anonymous

revision

- TuringTest

factor\[x^2y^4-x^4y^2\]

- anonymous

Ohhh, im thinking of the swedish word..

- TuringTest

repetition is a word, just not the right one

- anonymous

Again??

- anonymous

GCF of x^2 and x^4 is x^2
GCF of y^2 and y^4 is y^2
x^2*y^2(x^2*y^4/x^2*y^2)-x^2*y^2(x^4*y^2/x^2*y^2)?

- TuringTest

yes that should work :)
and yes, /again/ GT lol
continue Inopeki.

- anonymous

x^2*y^2(x^2*y^4/x^2*y^2)-x^2*y^2(x^4*y^2/x^2*y^2)=x^2*y^2(y^2-x^2)?
Damn thats a long line of numbers!

- TuringTest

lol yeah, but you got the right answer :D
but I think there's one more thing you can do, look closely.

- anonymous

x^2*y^2(x^2*y^4/x^2*y^2)-x^2*y^2(x^4*y^2/x^2*y^2)=-x^4*y^4?

- anonymous

x^2*y^2(x^2*y^4/x^2*y^2)-x^2*y^2(x^4*y^2/x^2*y^2)=(-x^4)*y^4?

- TuringTest

who who you got much closer with
x^2*y^2(y^2-x^2)
but is there something familiar here?

- anonymous

x^2*y^2(y-x)(y+x)?

- TuringTest

yes :)
good job, tricky one!

- anonymous

Thanks :DDD

- anonymous

one step closer to quantum mechanics

- anonymous

:D baby steps are still steps, right?

- TuringTest

true that :)
hey you're going plenty fast you've got time...
hmmm
do you know how to find the slope of a line give two points?

- anonymous

Yes, if i remember right.

- TuringTest

ok
(1,4) (6,12)
what is the slope between the points?

- anonymous

Well
12+4
----- Doing alright?
6+1

- anonymous

- actually

- TuringTest

gotta subtract, yes
the reason for this is important to understand you should think about it if you can

- anonymous

No wait

- anonymous

12-4
----- =8/5
6-1

- TuringTest

right
rise over run
remember that rise and run are about changes, and to find the change in x or y we must subtract, so it makes sense.
what about the equation of the line? any ideas?

- anonymous

\[\Delta y \div \Delta x\]

- TuringTest

perfect, even better way to think of it :)
now the equation of the line that passes through
(1,4) (6,12)
do you know how to find that?

- anonymous

|dw:1326067972678:dw|

- anonymous

I need to get x first then substitute to get y?

- TuringTest

no, you use the 'Point-Slope' form of the line:\[y-y_1=m(x-x_1)\]where m is the slope (which you have already found) and (x_1,y_1) are the coordinates of one of your points (it doesn't matter which) can you get the equation of the line now?

- anonymous

Wait,
12-4=8/5(6-1)?

- TuringTest

no, just use one of the points, the other x and y without subscripts (the little number 1) are just left as x and y...

- anonymous

Oh, y-12=8/5(x-6)?

- TuringTest

right, that is the equation in point-slope form
there is another form called slope-intercept form
that looks like\[y=mx+b\]where b is the y-intercept.
Our equation will be in this form if we solve for y, so do that.

- anonymous

y=8/5(x-6)-12?

- TuringTest

it would be +12 right?
but distribute 8/5 to the parentheses as well to get slope-intercept form.

- anonymous

y=12+(8/5)*(x-6)?

- TuringTest

yes now distribute the 8/5...

- TuringTest

in order to get rid of the parentheses...

- anonymous

y=12+(8/5)x-6?

- TuringTest

you forgot to distribute to the 6...

- anonymous

y=12+(8/5)x-(6)?

- TuringTest

(8/5)(x-6)
distribute the 8/5 to each term, that is how distribution always works.

- TuringTest

a(b+c)=ab+ac

- anonymous

But i did that before!

- TuringTest

and what did you get?

- anonymous

Aw man, i need some sleep. school starts tomorrow and its 1:30am XD Im screwed!

- TuringTest

lol
Thought so...
goodnight, good work :)

- anonymous

Goognight! Thanks again for teaching me all this :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.