anonymous
  • anonymous
Algebra, Turingtest :D
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
FOIL Inopeki :D (a+b)(c+d)
anonymous
  • anonymous
ac+ad+bc+bd
TuringTest
  • TuringTest
nice (x+3)(x-4)

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More answers

anonymous
  • anonymous
x^2-4x+3x-12
TuringTest
  • TuringTest
good, simplify
anonymous
  • anonymous
x^2-1x-12
TuringTest
  • TuringTest
Good, (no need to write the 1 though) (3p+t)(2p-4t)
anonymous
  • anonymous
Oh right 6p^2+3p*-4t+t*2p+5t?
TuringTest
  • TuringTest
not 5t... everything else is right though
anonymous
  • anonymous
-5t
TuringTest
  • TuringTest
no... look closely
anonymous
  • anonymous
-5t^2?
TuringTest
  • TuringTest
-4t^2 :P
anonymous
  • anonymous
Huh :O
TuringTest
  • TuringTest
You don't see it? (-4t)(t)=-4t^2 look at the last terms in the parentheses
anonymous
  • anonymous
Ohhh, now i see it
TuringTest
  • TuringTest
so you have what then, before simplifying?
anonymous
  • anonymous
6p^2+3p*-4t+t*2p+4t^2?
TuringTest
  • TuringTest
-4t^2 :P
anonymous
  • anonymous
Oh right xd
TuringTest
  • TuringTest
can you simplify it?
anonymous
  • anonymous
6p^2+3p*-4t+t*2p-4t^2 6p^2+6p^2-4t^2-4t^2?
TuringTest
  • TuringTest
no... how did you the p^2 in the middle? we had 6p^2+3p(-4t)+t(2p)-4t^2 parentheses look cleaner btw, you should use them here what is 3p(-4t) and t(2p) ???
anonymous
  • anonymous
-12pt? 2pt?
TuringTest
  • TuringTest
right, so then we have...?
anonymous
  • anonymous
6p^2-12pt+2pt-4t^2
TuringTest
  • TuringTest
yes :D and how does that simplify?
anonymous
  • anonymous
I dont know actually, maybe 6p^2-14pt+pt-4t^2?
anonymous
  • anonymous
No
TuringTest
  • TuringTest
they are like terms in the middle -12pt+2pt=-10pt this goes for any combination of variables a^2bc+3a^2bc=4a^2bc wr^4dt^3+3wr^2dt+6wr^4dt^3=7wr^4dt^3+3wr^2dt notice in this last example we couldn't combine all the terms because one had different exponents. We can only combine when they are the exact same in exponents and number of variables.
anonymous
  • anonymous
But dont i need to multiply them to get -10pt?
TuringTest
  • TuringTest
no, just add the coefficients factoring is one way to see it... factor out 'pt' from the middle two terms: \[6p^2-12pt+2pt-4t^2=6p^2+(12-2)pt-4t^2=6p^2-10pt-4t^2\]
anonymous
  • anonymous
Oh wait, i get it now, it would me multiplication if it was -10pt^2
TuringTest
  • TuringTest
not necessarily. You can't multiply things unless the equation tells you too, or you do a trick like multiply both sides if you have -10pt^2+7pt you cannot simplify that...
anonymous
  • anonymous
Oh ok. But what i meant was that if it were to tell us to multiply it would look like this -10pt^2 and not this -10pt
TuringTest
  • TuringTest
if you multiplied you get \[(2pt)(-12pt)=-24p^2t^2\]
anonymous
  • anonymous
Oh right
anonymous
  • anonymous
Since you would be multiplying both p and t with itself
TuringTest
  • TuringTest
right
anonymous
  • anonymous
Whatever, were getting off track :)
TuringTest
  • TuringTest
not really, you need to get this down too. It's all related, and it's good to know where you're at. (2s^2+p)(s-3p^3)
TuringTest
  • TuringTest
hey I'm green! never been a mod for math before :)
anonymous
  • anonymous
Congratulations! You deserve it :)
TuringTest
  • TuringTest
thanks :) I can't delete things though, don't see the point oh well... (2s^2+p)(s-3p^3)
anonymous
  • anonymous
(2s^2+p)(s-3p^3) 2s^3+2s^3(-3p^3)+p(s)+3p^4
TuringTest
  • TuringTest
so far so good now try to simplify...
TuringTest
  • TuringTest
oh wait, small error
anonymous
  • anonymous
OH the -
TuringTest
  • TuringTest
2s^3+2s^3(-3p^3)+p(s)+3p^4 ^^ should be???
anonymous
  • anonymous
2s^3+2s^2(-3p^3)+p(s)-3p^4 ^^ plus that :)
TuringTest
  • TuringTest
good catch I missed the sign lol !!! yes excellent now can you simplify it
anonymous
  • anonymous
Thanks :D 2s^3+2s^2(-3p^3)+p(s)-3p^4 2s^3-6s^2p^3+sp-3p^4?
TuringTest
  • TuringTest
perfect last one: (a+b)(a-b)
anonymous
  • anonymous
a^2-ab+ab-b^2? a^2-2ab-b^2?
TuringTest
  • TuringTest
not quite, I don't think those terms in the middle add...
anonymous
  • anonymous
The simplification or the first one?
TuringTest
  • TuringTest
a^2-ab+ab-b^2 ^^^^^ what is -x+x ?
anonymous
  • anonymous
Oh right, 0
TuringTest
  • TuringTest
so (a+b)(a-b) is?
anonymous
  • anonymous
a^2-b^2?
anonymous
  • anonymous
Some kind of coaching class going on here?
TuringTest
  • TuringTest
And that is a very important formula @pratu Inopkie wants to learn Algebra, a good cause in my book :)
anonymous
  • anonymous
ok...
anonymous
  • anonymous
Shall I help?
TuringTest
  • TuringTest
Thanks but I think I have an idea what I want to show him, we've been working together for a while ;)
TuringTest
  • TuringTest
@Inopeki We just showed that\[(a+b)(a-b)=a^2-b^2\]That relation is called the 'Difference of Squares' formula. It is used all the time, as you will see, as a quick way to factor anything of the form\[a^2-b^2\]
anonymous
  • anonymous
so a^2-b^2=(a+b)(a-b)?
anonymous
  • anonymous
Its an identity.
TuringTest
  • TuringTest
Yes, it's always true for that form. This is handy as you will see because if you remember how factoring helped us solve other equations, it will do the same for us here: For instance\[x^2-4=0\]we can factor this..\[(x-2)(x+2)=0\]which as I mentioned befor by the 'zero factor property' means that we need to solve\[x-2=0\]and\[x+2=0\]to find the zeros of our polynomial
anonymous
  • anonymous
But isnt (x-2)(x+2) 0 because the 2s negate?
TuringTest
  • TuringTest
foil it out and tell me yourself
anonymous
  • anonymous
x^2+2x-2x+-2*2 Well pellet
TuringTest
  • TuringTest
right which simplified too what?
anonymous
  • anonymous
x^2-4
TuringTest
  • TuringTest
so that's not always zero, is it? if x=1 then f(x)=-3, not zero what the expression f(x) is equal to here depends on what we put for x. There are in fact only two values for x that will make f(x)=0 Which should make sense by what I said yesterday: The order of the equation is 2, so the total multiplicity of its roots is 2 (fundamental theorem of algebra)
anonymous
  • anonymous
isnt that f(1)=
TuringTest
  • TuringTest
yes, sorry, bad notation on my part f(1)=-3 f(0)-4 etc.
anonymous
  • anonymous
But how does that work?
TuringTest
  • TuringTest
which part?
anonymous
  • anonymous
Functions, maybe im not ready for that subject
TuringTest
  • TuringTest
No you are, perhaps I haven't been clear enough whenever I write x^2-4 If I want to know where this graph hits the x-axis it is useful to call this a function f(x) here we have f(x)=x^2-4 Think about a graph: When the graph is toughing the x-axis it is at y=0, correct?
TuringTest
  • TuringTest
I just 'called' it a function, and made it graph-able...
anonymous
  • anonymous
Yeah, when x and y intersect its 0,0
anonymous
  • anonymous
oooh
anonymous
  • anonymous
f(x)=y+2x-3 So if we want to figure out y we need to plug in something for x
TuringTest
  • TuringTest
well, wait now, if you have a y in the function then it is not f(x) here's one thing that I bet is confusing you: f(x)=y when we write a line equation y=mx+b we are saying something about the graph: that the y coordinate is proportional to the x coordinate in this way. However this is rather simplistic. It is really more that we have a function f(x)=mx+b and we have chosen to graph it by letting the y-coordinate of our graph be f(x) they are effectively the same here, but a function can be graphed in many different types of coordinates, so y would not be used unless we are in Cartesian coordinates as we are now.
TuringTest
  • TuringTest
hence it's better to think of graphs in terms of functions, represented in Cartesian coordinates by y
anonymous
  • anonymous
What are cartesian coordinates?
TuringTest
  • TuringTest
the regular ones you know x-axis and y-axis
anonymous
  • anonymous
Yeah
TuringTest
  • TuringTest
there are other coordinate systems like polar, spherical, and cylindrical coordinates If we tried to graph our function in polar coordinates, y would have no meaning. however the zeros of the function are independent of how we choose to graph it. The math will be the same. Hence when we want to know the zeros of some polynomial we call it f(x), and that allows us to put numbers into it and see what happens f(x)=x^2-4 f(0)=0-4=-4 f(1)=1-4=-3 f(2)=4-4=0 so we found a zero simply by calling this a function and putting numbers into it.
anonymous
  • anonymous
One question, what do we gain from functions, what are the numbers on the RHS?
TuringTest
  • TuringTest
which numbers on the RHS?
anonymous
  • anonymous
Like f(0)=0-4=-4<--- ^ |
TuringTest
  • TuringTest
well that's the simplified form of f(0): f(0)=-4 check it by plugging in x=0 into the function
anonymous
  • anonymous
But if we know that f(0)=-4 why do we do it?
TuringTest
  • TuringTest
lol, why /don't/ we want to do that?! Almost all things in physics can be represented by function. If you want to know when the water pressure of a system will become too high, what trajectory to shoot your spaceship, or where a quantum particle will be at a certain time, you must analyse functions in this way.
anonymous
  • anonymous
Really?
anonymous
  • anonymous
Then teach me! This is important!
TuringTest
  • TuringTest
Really, though of course actual physics problems are usually (though not always) a bit trickier than just plugging in a number but yes, functions are instrumental to calculus, and therefor physics As I said earlier calculus is about analyzing functions... I guess we should start to talk about functions in a very detailed way.
anonymous
  • anonymous
Yes please :D
TuringTest
  • TuringTest
the definition of a function for you will be: if you have something like y=f(x) (where f(x) is some arbitrary polynomial with x's in it) and for every x you put in you get exactly one number for y back, then y=f(x) is a function, and y is a function of x try to soak that in, I'll tell you more about it momentarily
anonymous
  • anonymous
So what you are saying is that if f(x)=y x=y?
anonymous
  • anonymous
Well not really
anonymous
  • anonymous
What you are saying is that if x is one and y is 2 and you add 1 to x you add one to y aswell?
TuringTest
  • TuringTest
no, just that we get only one value of y for each x for instance f(x)=y I defined that, whether or not it's a function however if for every x we put in we get only one y back, then f(x) is a function, and the variable y is a function of the variable x
TuringTest
  • TuringTest
scratch the *for instance :/
anonymous
  • anonymous
Gotta go eat dinner, brb
TuringTest
  • TuringTest
look at a graph like a circle (I know you haven't studied them much yet this way)|dw:1326043068666:dw|this is not a function because there is more than one y for each value of x... see you in a bit, I'm gonna have breakfast.
TuringTest
  • TuringTest
drawing a line from x to its corresponding y value you should know we need a vertical line|dw:1326043155512:dw|see how one x-coordinate corresponds to two different y values y1 and y2? But what about y=2x+5 ?|dw:1326043286670:dw|this graph has exactly one value of y for each value of x, so it is a function.
TuringTest
  • TuringTest
the idea above is what's called the 'vertical line test': if you can draw a vertical line at any point on the graph and intersect it twice, then it is not a function.
TuringTest
  • TuringTest
when you return look at this say we have a table of values for some y=f(x) and we want to know if f(x) is a function.... look at the table x |y=f(x) --------- -1 | 2 3 | 7 2 | 0 5 | 2 Is this a function? why or why nmot?
anonymous
  • anonymous
Its not. EXEPT the one that says 2 | 0<---------------< cause that means that there is only one value of x ^
anonymous
  • anonymous
The others seem to have two values of x
TuringTest
  • TuringTest
actually look at the x's each x has one y, so it is a function we have that f(5)=f(-1)=5 so two different x's give the same y, but that is okay the only thing against the rules would be to have something like this x | y ------ 1 | 3 2 | 5 1 | 4 because this means that f(1) has two values: f(1)=3 and 4 so this is not a function... look at the x to see which gives more than one f(x)... (I am using the terms f(x) and y here interchangeably) x | y ----- 2 | 0 3 | 1 5 | 6 3 | 6 1 | 0 function or not ant why?
anonymous
  • anonymous
Yeah
TuringTest
  • TuringTest
That's right FFM does the curriculum meet your approval? I doubt it with your standards :P
anonymous
  • anonymous
x | y ----- 2 | 0 =f(2)=2? im sorry, i dont really understand 3 | 1 5 | 6 3 | 6 1 | 0
TuringTest
  • TuringTest
f(2)=0 that means when x=2, y=0, yes?
anonymous
  • anonymous
Ooooh
TuringTest
  • TuringTest
that's all there is to that so is there any f(x) on the list that gives more than one number for a single x?
anonymous
  • anonymous
So, x | y ----- 2 | 0 = f(2)=0 3 | 1 = f(3)=1 5 | 6 = f(5)=6 3 | 6 = f(3)=6 1 | 0 = f(1)=0 So, no.
TuringTest
  • TuringTest
actually what is f(3) ?
anonymous
  • anonymous
Oh! 1 and 6!
TuringTest
  • TuringTest
exactly, so not a function :)
anonymous
  • anonymous
f(3)=1+6 so its not a function!
anonymous
  • anonymous
YEah :D
anonymous
  • anonymous
And that means that f(3) could be graphed in a circle?
TuringTest
  • TuringTest
right not 1+6=7 of course, it equal the set\[f(3)=\left\{ 1,6 \right\}\] It doesn't mean it would be a circle but it would fail the vertical line test, watch...
anonymous
  • anonymous
Of course.
anonymous
  • anonymous
Y cant have to places
anonymous
  • anonymous
two*
anonymous
  • anonymous
:)
TuringTest
  • TuringTest
thanks FFM, praise form Caeser :) |dw:1326046327322:dw|now look at the points of our possible function above. is there any place we can draw a vertical line that would intersect two points?
anonymous
  • anonymous
No way!
TuringTest
  • TuringTest
lol *Praise from Ceaser
TuringTest
  • TuringTest
no way to draw a vertical line and hit more than one point you say?
anonymous
  • anonymous
Not more than 3.
anonymous
  • anonymous
And there are 5 so..
TuringTest
  • TuringTest
more than 3? You only need to hit 2 points to show that this is NOT a function and here it is:|dw:1326046665544:dw|so again we show by this that f(3) has more than one value...
anonymous
  • anonymous
Yeah
TuringTest
  • TuringTest
are the following functions?|dw:1326046773020:dw|(I'll fill in different graph above)
anonymous
  • anonymous
f(x)=y?
TuringTest
  • TuringTest
yes f(x)=y because we have defined it that way now:|dw:1326046848239:dw|function?
anonymous
  • anonymous
NO
TuringTest
  • TuringTest
draw the vertical line that hits more than one point on the graph if that is true
anonymous
  • anonymous
No wait, there are no points
anonymous
  • anonymous
|dw:1326046981078:dw|
TuringTest
  • TuringTest
there are infinite points on a line.... a line intersects another line at a point|dw:1326047055394:dw|P is our intersect point can you draw a vertical line that hits the curve at more than one point?
anonymous
  • anonymous
Oh, so y is a vertical line?
anonymous
  • anonymous
Then this could be a function
anonymous
  • anonymous
|dw:1326047157974:dw| not function
anonymous
  • anonymous
But yours is
TuringTest
  • TuringTest
now you got the idea :D|dw:1326047282819:dw|function?
anonymous
  • anonymous
No.
TuringTest
  • TuringTest
show
anonymous
  • anonymous
|dw:1326047285288:dw|
TuringTest
  • TuringTest
just one line is sufficient, but yes :)||dw:1326047413725:dw|function?
anonymous
  • anonymous
Yes
TuringTest
  • TuringTest
good :) x | y ----- 1 | 3 2 | 3 3 | 3 4 | 3 function?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
For all of them
anonymous
  • anonymous
But if it was 1|3 1|4 It would not
TuringTest
  • TuringTest
good x | y ----- 0 | 3 0 | 2 0 | 1 0 | 0 I think you already proved you know this, but function?
anonymous
  • anonymous
No
anonymous
  • anonymous
0 has 4 values of y
TuringTest
  • TuringTest
nice job, exactly slightly different: f(x)=x+3 function?
anonymous
  • anonymous
I guess that depends on what you plug in for x?
anonymous
  • anonymous
Actually, no
anonymous
  • anonymous
It is a function
anonymous
  • anonymous
It doesnt show multiple values of y, just x+3
TuringTest
  • TuringTest
Excellent!!!!!!!!
anonymous
  • anonymous
Yes! :D
TuringTest
  • TuringTest
perfect observation :) every x we put in gives us only one y. y=0 is y a function of x?
anonymous
  • anonymous
Only if we know that 0=x
TuringTest
  • TuringTest
not so, plug in various x into the function f(x)=0, we get the table x | y ---- 0 | 0 1 | 0 2 | 0 3 | 0 and so on... and I think you would agree that this chart constitutes a function.
TuringTest
  • TuringTest
tricky one, but now you know ;)
anonymous
  • anonymous
Yes, that chart constitutes a function. But if there were no chart?
TuringTest
  • TuringTest
They are all in agreement, the chart, the graph, everything. we don't need x=0 to get one value of y back; we always get one value of y. The same value of y, namely zero. do you know what the graph of y=0 looks like?
anonymous
  • anonymous
|dw:1326048301595:dw|
anonymous
  • anonymous
Thats 0,0
TuringTest
  • TuringTest
yes but I don't want the point (0,0) y=0 is a line: it's actually the x-axis the line x=0 is the y-axis think about that for a minute
anonymous
  • anonymous
Oh, right. |dw:1326048543616:dw|
anonymous
  • anonymous
Didnt really think of that
TuringTest
  • TuringTest
exactly, and does the graph of y=0 pass the vertical line test?
anonymous
  • anonymous
No
TuringTest
  • TuringTest
are you saying that you can draw a vertical line that intersects it twice? show...
anonymous
  • anonymous
OH you mean that? Never mind lol
TuringTest
  • TuringTest
so is the graph indicative of a function or not?
anonymous
  • anonymous
Yes, it is
TuringTest
  • TuringTest
perfect, what about the line x=0? (standard cartesian notation here , don't get hung up on the names of variables, we could easily have them to t and p or whatever)
anonymous
  • anonymous
|dw:1326049009726:dw|
TuringTest
  • TuringTest
yes, so is it a function?
anonymous
  • anonymous
Yes.
TuringTest
  • TuringTest
be careful can we draw a vertical line that touches the graph at more than one point? what would the table of values look like?
anonymous
  • anonymous
No wait, this becomes f(x)=x right?
TuringTest
  • TuringTest
no, this would be f(x)|x=0 (read f of x such that x is zero) here f(x) takes on a range of values from negative to positive infinity: x | f(x) ------ 0 | 0 0 | 1 0 | 2 0 | -1 0 | -2 etc. not a function, right?
anonymous
  • anonymous
I think i get it, its because there is no y?
anonymous
  • anonymous
but f(x) has endless possibilities
TuringTest
  • TuringTest
no this chart is the same as x | y ------ 0 | 0 0 | 1 0 | 2 0 | -1 0 | -2 I told you that y and f(x) are interchangeable y is the name of the coordinate which we are using to represent f(x) so here they are the same the chart itself is of the same type as you saw before, we have one value of x corresponding to multiple (in this case infinite) values of y, so it is not a function.
anonymous
  • anonymous
Ohhh, but if you had y corresponding to multiple values of x it would be a function.
TuringTest
  • TuringTest
right :)
anonymous
  • anonymous
Ah, now i see!
TuringTest
  • TuringTest
let me try to clear a few things up... what is the dependent and what is the independent variable here?
anonymous
  • anonymous
The dependent variable is y because if its a function or not depends on y having multiple values corresponding to one value of x.
TuringTest
  • TuringTest
right, but I'm not sure about your reasoning, so what about f(y)=x=y^2+3y which is dependent and which is independent?
anonymous
  • anonymous
x is the dependent one i think
TuringTest
  • TuringTest
exactly, so here x is a function of y we could have therefor more than one y value for an x value in this case, but not more than one x value for each y, if it is to be a function. so really the rule is better stated this way: "a relation is a function if each value of the independent variable corresponds to exactly one value of the dependent variable" because we can change the names or swap positions of the variables
anonymous
  • anonymous
Because we can have more of y but not more of x
anonymous
  • anonymous
exactly
TuringTest
  • TuringTest
exactly, look at the new version of our rule above (in quotes)
anonymous
  • anonymous
Yeah, that sounds right.
TuringTest
  • TuringTest
so whenever we see f(t)=r for instance which is dependent and which ins independent?
anonymous
  • anonymous
We should move to another "thread" this one is lagging for me
TuringTest
  • TuringTest
I agree go ahead
anonymous
  • anonymous
r is the dependent and t is independent
TuringTest
  • TuringTest
exactly :)

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