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anonymous
 5 years ago
Algebra, Turingtest :D
anonymous
 5 years ago
Algebra, Turingtest :D

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TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6FOIL Inopeki :D (a+b)(c+d)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6Good, (no need to write the 1 though) (3p+t)(2p4t)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh right 6p^2+3p*4t+t*2p+5t?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6not 5t... everything else is right though

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6You don't see it? (4t)(t)=4t^2 look at the last terms in the parentheses

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6so you have what then, before simplifying?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.06p^2+3p*4t+t*2p+4t^2?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6can you simplify it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.06p^2+3p*4t+t*2p4t^2 6p^2+6p^24t^24t^2?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6no... 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) ???

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6right, so then we have...?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6yes :D and how does that simplify?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont know actually, maybe 6p^214pt+pt4t^2?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6they 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
 5 years ago
Best ResponseYou've already chosen the best response.0But dont i need to multiply them to get 10pt?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6no, just add the coefficients factoring is one way to see it... factor out 'pt' from the middle two terms: \[6p^212pt+2pt4t^2=6p^2+(122)pt4t^2=6p^210pt4t^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh wait, i get it now, it would me multiplication if it was 10pt^2

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6not 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
 5 years ago
Best ResponseYou've already chosen the best response.0Oh 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
 5 years ago
Best ResponseYou've already chosen the best response.6if you multiplied you get \[(2pt)(12pt)=24p^2t^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since you would be multiplying both p and t with itself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Whatever, were getting off track :)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6not 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)(s3p^3)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6hey I'm green! never been a mod for math before :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Congratulations! You deserve it :)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6thanks :) I can't delete things though, don't see the point oh well... (2s^2+p)(s3p^3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(2s^2+p)(s3p^3) 2s^3+2s^3(3p^3)+p(s)+3p^4

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6so far so good now try to simplify...

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6oh wait, small error

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.62s^3+2s^3(3p^3)+p(s)+3p^4 ^^ should be???

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02s^3+2s^2(3p^3)+p(s)3p^4 ^^ plus that :)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6good catch I missed the sign lol !!! yes excellent now can you simplify it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks :D 2s^3+2s^2(3p^3)+p(s)3p^4 2s^36s^2p^3+sp3p^4?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6perfect last one: (a+b)(ab)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a^2ab+abb^2? a^22abb^2?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6not quite, I don't think those terms in the middle add...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The simplification or the first one?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6a^2ab+abb^2 ^^^^^ what is x+x ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Some kind of coaching class going on here?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6And that is a very important formula @pratu Inopkie wants to learn Algebra, a good cause in my book :)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6Thanks but I think I have an idea what I want to show him, we've been working together for a while ;)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6@Inopeki We just showed that\[(a+b)(ab)=a^2b^2\]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\[a^2b^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so a^2b^2=(a+b)(ab)?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6Yes, 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\[x^24=0\]we can factor this..\[(x2)(x+2)=0\]which as I mentioned befor by the 'zero factor property' means that we need to solve\[x2=0\]and\[x+2=0\]to find the zeros of our polynomial

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But isnt (x2)(x+2) 0 because the 2s negate?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6foil it out and tell me yourself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2+2x2x+2*2 Well pellet

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6right which simplified too what?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6so 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)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6yes, sorry, bad notation on my part f(1)=3 f(0)4 etc.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But how does that work?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Functions, maybe im not ready for that subject

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6No you are, perhaps I haven't been clear enough whenever I write x^24 If I want to know where this graph hits the xaxis it is useful to call this a function f(x) here we have f(x)=x^24 Think about a graph: When the graph is toughing the xaxis it is at y=0, correct?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6I just 'called' it a function, and made it graphable...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, when x and y intersect its 0,0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(x)=y+2x3 So if we want to figure out y we need to plug in something for x

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6well, 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 ycoordinate 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
 5 years ago
Best ResponseYou've already chosen the best response.6hence it's better to think of graphs in terms of functions, represented in Cartesian coordinates by y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What are cartesian coordinates?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6the regular ones you know xaxis and yaxis

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6there 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^24 f(0)=04=4 f(1)=14=3 f(2)=44=0 so we found a zero simply by calling this a function and putting numbers into it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0One question, what do we gain from functions, what are the numbers on the RHS?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6which numbers on the RHS?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Like f(0)=04=4< ^ 

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6well that's the simplified form of f(0): f(0)=4 check it by plugging in x=0 into the function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But if we know that f(0)=4 why do we do it?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6lol, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Then teach me! This is important!

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6Really, 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.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6the 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
 5 years ago
Best ResponseYou've already chosen the best response.0So what you are saying is that if f(x)=y x=y?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What 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
 5 years ago
Best ResponseYou've already chosen the best response.6no, 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
 5 years ago
Best ResponseYou've already chosen the best response.6scratch the *for instance :/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Gotta go eat dinner, brb

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6look at a graph like a circle (I know you haven't studied them much yet this way)dw:1326043068666:dwthis 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
 5 years ago
Best ResponseYou've already chosen the best response.6drawing a line from x to its corresponding y value you should know we need a vertical linedw:1326043155512:dwsee how one xcoordinate corresponds to two different y values y1 and y2? But what about y=2x+5 ?dw:1326043286670:dwthis graph has exactly one value of y for each value of x, so it is a function.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6the 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
 5 years ago
Best ResponseYou've already chosen the best response.6when 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
 5 years ago
Best ResponseYou've already chosen the best response.0Its not. EXEPT the one that says 2  0<< cause that means that there is only one value of x ^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The others seem to have two values of x

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6actually 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?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6That's right FFM does the curriculum meet your approval? I doubt it with your standards :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x  y  2  0 =f(2)=2? im sorry, i dont really understand 3  1 5  6 3  6 1  0

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6f(2)=0 that means when x=2, y=0, yes?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6that'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
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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
 5 years ago
Best ResponseYou've already chosen the best response.6actually what is f(3) ?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6exactly, so not a function :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(3)=1+6 so its not a function!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And that means that f(3) could be graphed in a circle?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6right 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
 5 years ago
Best ResponseYou've already chosen the best response.0Y cant have to places

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6thanks FFM, praise form Caeser :) dw:1326046327322:dwnow look at the points of our possible function above. is there any place we can draw a vertical line that would intersect two points?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6lol *Praise from Ceaser

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6no way to draw a vertical line and hit more than one point you say?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6more than 3? You only need to hit 2 points to show that this is NOT a function and here it is:dw:1326046665544:dwso again we show by this that f(3) has more than one value...

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6are the following functions?dw:1326046773020:dw(I'll fill in different graph above)

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6yes f(x)=y because we have defined it that way now:dw:1326046848239:dwfunction?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6draw the vertical line that hits more than one point on the graph if that is true

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No wait, there are no points

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1326046981078:dw

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6there are infinite points on a line.... a line intersects another line at a pointdw:1326047055394:dwP is our intersect point can you draw a vertical line that hits the curve at more than one point?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, so y is a vertical line?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then this could be a function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1326047157974:dw not function

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6now you got the idea :Ddw:1326047282819:dwfunction?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1326047285288:dw

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6just one line is sufficient, but yes :)dw:1326047413725:dwfunction?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6good :) x  y  1  3 2  3 3  3 4  3 function?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But if it was 13 14 It would not

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6good x  y  0  3 0  2 0  1 0  0 I think you already proved you know this, but function?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6nice job, exactly slightly different: f(x)=x+3 function?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess that depends on what you plug in for x?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It doesnt show multiple values of y, just x+3

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6perfect observation :) every x we put in gives us only one y. y=0 is y a function of x?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Only if we know that 0=x

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6not 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
 5 years ago
Best ResponseYou've already chosen the best response.6tricky one, but now you know ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, that chart constitutes a function. But if there were no chart?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6They 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
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1326048301595:dw

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6yes but I don't want the point (0,0) y=0 is a line: it's actually the xaxis the line x=0 is the yaxis think about that for a minute

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, right. dw:1326048543616:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Didnt really think of that

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6exactly, and does the graph of y=0 pass the vertical line test?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6are you saying that you can draw a vertical line that intersects it twice? show...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OH you mean that? Never mind lol

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6so is the graph indicative of a function or not?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6perfect, 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
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1326049009726:dw

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6yes, so is it a function?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6be 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
 5 years ago
Best ResponseYou've already chosen the best response.0No wait, this becomes f(x)=x right?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6no, 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
 5 years ago
Best ResponseYou've already chosen the best response.0I think i get it, its because there is no y?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but f(x) has endless possibilities

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6no 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
 5 years ago
Best ResponseYou've already chosen the best response.0Ohhh, but if you had y corresponding to multiple values of x it would be a function.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6let me try to clear a few things up... what is the dependent and what is the independent variable here?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The dependent variable is y because if its a function or not depends on y having multiple values corresponding to one value of x.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6right, 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
 5 years ago
Best ResponseYou've already chosen the best response.0x is the dependent one i think

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6exactly, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Because we can have more of y but not more of x

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6exactly, look at the new version of our rule above (in quotes)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, that sounds right.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.6so whenever we see f(t)=r for instance which is dependent and which ins independent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We should move to another "thread" this one is lagging for me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0r is the dependent and t is independent
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