You can check the solution here:
[url]http://www.wolframalpha.com/input/?i=y%5E2+%2B+3xy+-+10x%5E2+%2B+y+%2B+5x+%3D+0[/url]
Okay, no, I'm supposed to complete the square and I'm not sure how I do that with this. Help.
completing the square was my worst nightmare in alg 2, my teacher didn't fully understand either
[QUOTE=JohanGS;45788143]Okay, no, I'm supposed to complete the square and I'm not sure how I do that with this. Help.[/QUOTE]
Give the problem.
[QUOTE=JohanGS;45788143]Okay, no, I'm supposed to complete the square and I'm not sure how I do that with this. Help.[/QUOTE]
Just group terms by their y component and treat x as constant, then do exactly what you'd normally do.
y^2 + 3xy - 10x^2 + y + 5x = 0
y^2 + (3x+1)y + (-10x^2 + 5x) = 0
(y + (3x+1)/2)^2 - ((3x+1)/2)^2 + (-10x^2 +5x) = 0
etc.
[QUOTE=JohanGS;45788143]Okay, no, I'm supposed to complete the square and I'm not sure how I do that with this. Help.[/QUOTE]
Well yeah that's just a sort of the actual way to solve quadratics. In 90% of the cases it is skipped. I never even saw the proof with completing the square in high school. They just told me that the solutions were the crap with the square root and 4ac. Kinda tricky to do it by yourself if you've never seen it before...
Can someone show me the steps for finding the derivative for (x^2 + 1)^(x^2 + 1)? Somewhere along the way I do something I'm not supposed to do.
I don't know if I'm helping you or not, but this is the general case (an apostrophe is the same thing as a derivative here, for notation's sake), it's a clusterfuck :
[IMG]http://quicklatex.com/cache3/ql_6008d16d8e0b90c1af9190d67da3ea60_l3.png[/IMG]
I hope I didn't do anything wrong (not according to [URL="http://www.wolframalpha.com/input/?i=d%2Fdx+%28f%28x%29%5Eg%28x%29%29"]Wolfram Alpha[/URL]), I also don't know what's up with the natural log of a general function and how I should avoid taking the log of a potentially negative function (restrict the domain or some gay rigorous shit like that), although in your case they are strictly positive, which is nice. I also used h=ln(f) * g just to keep notation simple and to outline the derivative of an exponential.
I think the exercise expected you to just do the "rewrite it as an exponential"-trick on the second line (which is the key trick to figuring out exponential functions). Using the properties of the derivative of an exponential and that of a logarithm, you can do everything explicitly from there (i.e. replace f and g with (x^2 +1) and see what happens)
[QUOTE=JohanGS;45853873]Can someone show me the steps for finding the derivative for (x^2 + 1)^(x^2 + 1)? Somewhere along the way I do something I'm not supposed to do.[/QUOTE]
The base and exponent are the same, so you can apply the chain rule with x^2 + 1 and x^x.
[QUOTE=Number-41;45854536]I also don't know what's up with the natural log of a general function and how I should avoid taking the log of a potentially negative function[/QUOTE]
If f is negative and g is non-constant (so g' is nonzero) then you've already got a poorly defined function.
What is -1 to some non-integer power?
If you have a consistent (i.e. continuous) answer, that's equivalent to choosing ln(-1).
Similarly, making that choice for an arbitrary (possibly negative) f is equivalent to choosing a continuous ln(f).
I've got a question concerning logic.
A=>B
If A is true and B is true, then A=>B is true.
If A is true and B is false, then A=>B is false.
If A is false and B is true, then A=>B is true.
I don't understand that last part, imagine if A is "Eva bringing candy" and B is "Adam is happy". If all is true then Adam is happy if Eva brought candy, but he's also happy if she doesn't bring candy.
Can someone clarify?
If A is false, A => B is always true. Falsehood implies everything. The last part is a special case of this, where you happen to imply a truth.
For example, you can take any false equation (say, 8=3) and multiply both sides by zero to get the true equation 0=0.
If we suppose Adam is always happy, and we suppose Eva doesn't bring candy, then we can say that Adam would be happy if Eva brought candy (Adam is always happy, after all).
Recently in my maths class, I've started a probability subject (earlier this week) and today I had to do things with Venn diagrams. Easy enough, but I got to the last question of the exercise I had to do and I haven't the foggiest how to actually do it. I've tried but all I seemed to do was draw circles everywhere and get nothing done. The answers in the back of the book just have the answer, could anyone lend us a hand?
[img_thumb]http://i.imgur.com/YqjsSdL.jpg[/img_thumb]
I need the working as well, not just the answer
ti-89 or n-spire for calculus? (and above, since i'll be taking it to college)
This is an interesting problem. With circles, you can't actually make a Venn diagram with four sets. You need either spheres or something more irregular.
[t]http://i.imgur.com/aEGVUA8.png[/t]
Basically, you divide all the pizza into crispy or pan (no pizza is neither, no pizza is both). You can say that the set of crispy pizzas is the complement of the set of pan pizzas, and vice versa.
Draw sets for onions and mushrooms - there are nonzero probabilities that each, both, and neither are used with pan and crispy. This means that onions should intersect with crispy, pan, and mushrooms but it is neither a subset nor a superset of any of these. Similarly, mushrooms intersects crispy, pan, and onions but is neither a subset nor a superset of any of these.
[QUOTE=ZeFruitNazi;45888227]ti-89 or n-spire for calculus? (and above, since i'll be taking it to college)[/QUOTE]
Neither, do it by hand.
But if you want complete overkill go for the nspire cx cas. Though it's probably outlawed in all math courses by now.
[QUOTE=ZeFruitNazi;45888227]ti-89 or n-spire for calculus? (and above, since i'll be taking it to college)[/QUOTE]
Does your course actually require a graphing calculator?
it's very recommended. i've been doing fine with doing limits by hand and checking on a graphing app on my phone, i just want to be prepared in case i get into something deep in college.
Never needed one. They weren't allowed on our Calc exam.
Actually, one of the exam questions was to graph a function. Something dirty with an exponential, and all the exercises during the year were rational functions. Goddamn typical.
My point being that I find it weird that you'd ever need one, because being able to graph something "by hand" is really valuable and adds to your understanding of functions, especially when you get to Complex Analysis (in my experience).
All of my math courses so far have explicitly forbidden calculators. All I have ever needed it for was basic physics.
[IMG]http://latex.codecogs.com/gif.latex?%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20x-y+az%3D1%20%5C%5C%202x-y+z%3D-1%20%5C%5C%20ax+y-z%3D1%20%5Cend%7Bmatrix%7D%5Cright.[/IMG]
For each real number a, how many solutions are there?
[QUOTE=Krinkels;45888549]This is an interesting problem. With circles, you can't actually make a Venn diagram with four sets. You need either spheres or something more irregular.
[t]http://i.imgur.com/aEGVUA8.png[/t]
Basically, you divide all the pizza into crispy or pan (no pizza is neither, no pizza is both). You can say that the set of crispy pizzas is the complement of the set of pan pizzas, and vice versa.
Draw sets for onions and mushrooms - there are nonzero probabilities that each, both, and neither are used with pan and crispy. This means that onions should intersect with crispy, pan, and mushrooms but it is neither a subset nor a superset of any of these. Similarly, mushrooms intersects crispy, pan, and onions but is neither a subset nor a superset of any of these.[/QUOTE]
Are you telling me this isn't a thing of beauty?
[img]http://upload.wikimedia.org/wikipedia/commons/1/19/Venn6.svg[/img]
[editline]4th September 2014[/editline]
[QUOTE=JohanGS;45891458][IMG]http://latex.codecogs.com/gif.latex?%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20x-y+az%3D1%20%5C%5C%202x-y+z%3D-1%20%5C%5C%20ax+y-z%3D1%20%5Cend%7Bmatrix%7D%5Cright.[/IMG]
For each real number a, how many solutions are there?[/QUOTE]
Solve just like you would any other set of simultaneous equations, just be extremely careful when dividing by anything - you don't want it to be zero!
For example, to start off, I'd add the second and third equations to get:
(a+2)x = 0
So now if a =/= -2 we get x=0
To proceed, suppose that a isn't -2 and continue with x=0, then deal with the special case of a=-2 afterwards. (you'll find that the second and third equations are equivalent, and so you get an infinite number of solutions)
[sp]One value of a gives infinite solutions, one gives no solutions, and the rest give precisely one[/sp]
[QUOTE=Joey90;45891729]Are you telling me this isn't a thing of beauty?[/QUOTE]
No, I said 'irregular', not 'ugly'.
[QUOTE=ZeFruitNazi;45890512]it's very recommended. i've been doing fine with doing limits by hand and checking on a graphing app on my phone, i just want to be prepared in case i get into something deep in college.[/QUOTE]
Unless you come across a situation in the class where you'd absolutely need one (e.g. you need to do 13 iterations of Newton's Method, or they ask you how to graph something on a specific calculator), I wouldn't. Once you're in college, any situation where you would want or need a graphing calculator can be handled with a computer.
oh ok, i'll hold off on buying a calculator for a while, then
Hey does anyone wanna remake this thread. I'm not even posting in here and it'd be nice to see it on the front page for a little while, maybe attract new math students and so on.
This is a good idea, but we need a clever way to write 4.
[QUOTE=Krinkels;45901244]This is a good idea, but we need a clever way to write 4.[/QUOTE]
2^2 goddamn ima genius
How about [IMG]http://s29.postimg.org/nqyvtdcdf/MSP12132203dd0c65c0970hd0000620a72ihf135a541.gif[/IMG]
Good luck getting that in the title.
How about 1+2+3+4+5+6+...
:v:
[IMG]http://latex.codecogs.com/gif.latex?1%20-%20%5Cfrac%7B1%7D%7B2%7D%20+%20%5Cfrac%7B1%7D%7B4%7D%20-%20%5Cfrac%7B1%7D%7B8%7D+...+%5Cfrac%7B1%7D%7B2%5E%7B2n%7D%7D[/IMG]
The answer is supposed to be
[IMG]http://latex.codecogs.com/gif.latex?%5Cfrac%7B2%20+%202%5E%7B-2n%7D%20%7D%7B3%7D[/IMG]
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