http://www.wolframalpha.com/
This math engine can solve a lot of mathematical equations including differential equations.
Aly's ~ Help Me With Math Thread ; ;
|
Aly's ~ Help me with math thread ; ;
Hsieh said: http://www.wolframalpha.com/ This math engine can solve a lot of mathematical equations including differential equations. yea i dont care for sites, makes it more real asking ppl here like if it was face to face. but thx for sharing that..
I agree wolfram isn't much for learning. It's for when you're trying to compute a fourier transform by direct integration and don't feel like integrating.
Oh good, a math thread. Anyone care to explain what an 'effective interest rate' is? I have to find whatever that thing is with a 13% annual interest rate compounded monthly. I swear the past two semesters have been nothing but financial word problems.
"you can compound daily my *** mom"
Carbuncle.Sterling said: » Oh good, a math thread. Anyone care to explain what an 'effective interest rate' is? I have to find whatever that thing is with a 13% annual interest rate compounded monthly. I swear the past two semesters have been nothing but financial word problems. or: (1+.13)=(1+x)^12, 1.0102=1+x, x=1.02% interest rate. I'm not 100% sure on that as almost every problem I've ever gotten has given me a monthly or weekly compounding interest rate, and I've had to figure out the effective annual (thus i'm not entirely sure I'm solving the right problem). If it helps at all... my only notes given on this rate are the following:
When annual interest rate of 20% is compounded quarterly, $100 becomes $121.55 after 1 year. 20% is the nominal rate and 21.55% is called the effective annual rate. My homework question literally says, simply: "What is the effective interest rate of an investment with 13% annual interest rate compounded monthly?". It is the "other way around Sawtelle" and used to compare differing types of rates usually. The effective interest rate is calculated as if compounded annually.
The effective rate is calculated in the following way, where r is the effective annual rate, i the nominal rate 13%, and n the number of compounding periods per year (12 for monthly compounding) In General r = (1+i/n)^n - 1 So r = (1+.13/12)^12 - 1 You can finish the work. It will be over 13% of course so you will know you have the correct answer.
[+]
Oh thank god, an equation. That's what I needed, thank you!
Edit: Solved by someone better at this than me
I use mathematica to check my homework, for algebra it won't break a sweat, and only uses a very small fraction of its capabilities. It's nice for factoring, if you don't see how to factor something. When it was calculating limits there was a noticeable delay, and I was seeing how to calculate derivatives with it. There's a free Linux version that is similar but I can't remember the name. Just remember it is only good as a tool, never completely rely on it, if it gives me different answer I go back to my paper and figure out what I did wrong. Plus there's the learning curve with the syntax, but it is well documented.
Speaking of these math checkers, the problem I continuously run into is that you can very rarely use natural log when working things out in wolfram alpha. It seems to constantly change my ln to a log base 10. Makes it pretty much useless for checking my work on natural log problems and dang it if I have to go through and remember what ln cancels to.
Carbuncle.Sterling said: » Speaking of these math checkers, the problem I continuously run into is that you can very rarely use natural log when working things out in wolfram alpha. It seems to constantly change my ln to a log base 10. Makes it pretty much useless for checking my work on natural log problems and dang it if I have to go through and remember what ln cancels to. If you're using logs to make an equation tractable like, for example, with maximum likelihood estimators and the like the base of the log won't matter as it won't change the result. There really aren't a whole lot of instances where the base of the log would matter so long as you're consistent. Siren.Kalilla said: » volkom said: » should just post your problems~ they probably could be all solved in a few minutes Well, people have been explaining their reasoning etc. Linear inequalities are fairly simple. Treat them like an equation (can add/subtract/divide/multiply on both sides etc); just keep the inequality sign consistent. If you divide or multiply by a negative number, flip the sign; eg: x+5 > 8 Therefore x > 3 (note it is explicitly greater than here, NOT equals) x+5 >= (greater than or equal to) 8 Therefore x >= 3 (again, be consistent. in an inequality, a variable is not simply one discrete value in most cases, which is what the "equals" sign implies). 3 > 2 [* -1] -3 < -2 Note the sign flip, just to show you why it is necessary. Etc... Quote: The quantification of business modeling. It is the ultimate in your "creative thinking" application to real world situations. Like fitting the appropriate curve to empirical data so you can extrapolate within the standard deviations with a particular amount of uncertainty defined by our clients.[\quote] That sounds amazing. It's kind of like physics (physical modelling) but within a business context... kind of thing I could really enjoy! Need more girls like you that can appreciate maths! Jazzy said: The quantification of business modeling. It is the ultimate in your "creative thinking" application to real world situations. Like fitting the appropriate curve to empirical data so you can extrapolate within the standard deviations with a particular amount of uncertainty defined by our clients.[\quote] Bahamut.Raenryong said: » That sounds amazing. It's kind of like physics (physical modelling) but within a business context... kind of thing I could really enjoy! Need more girls like you that can appreciate maths! Well it IS a rush. I think a lot of girls appreciate maths (I know mine does *giggles) but I think folks in general get scared by it. I don't think they teach estimating or "sanity checking" anymore in school. I always have some basic idea of what the answer should be before I get started in most cases. Luckily math is rather simple in that it has rules and you can build almost anything just like legos! Don't let math get complicated, just because it might get complex! Lego
BTW, if there is anyone here that is particularly knowledgeable with Chemistry maths, please message me. I have a handful of people who I know can help me with my math maths, but absolutely no one yet who might be able to help with Chem- and I actually sort of love Chem maths once I get the concepts down.
Aly Saxena said: » Ok that equation was already done :/ But here is a new one I'm confused on... Find the perimeter and area for each figure (use pi = 3.14)  I have the answers for both but I only need to understand how to get the perimeter. I have the details for the area answer, so dont need that. Answers: A = 48cm^2 P= 28cm When I did the perimeter...I got P=4s...= 4(6) = 24cm but it doesnt seem right if I just add 4 from the middle of the triangle. So how do you get 28cm? I'm so confused ; ; I only just skimmed the thread, but noticed alot of the answers to this were needlessly complicated. Sorry if this was already cleared up/simplified. The perimeter is the sum of the outside edges. 5+5+6+6+6 (28cm) for the shape as a whole, 5+5+6 (16cm) for the triangle alone and 6+6+6+6 (24cm) for the square alone. The area of the whole shape is the area of the square (6 * 6)cm^2 + the area of the triangle. To find the area of a triangle, I find it best to turn it into a square, then half it. So multiply the height by the length then divide by 2: (4 * 6) / 2 = 12 Triangle area = 12cm^2 Square area = 36cm^2 Whole area = sum of the 2 parts (48cm^2)
[+]
Ok for compound inequalities
-10 ≤ -3/2x + 2 ≤ -1 I tried this on my own.... (-2)(-10) ≤ (-3/2x)(-2) + 2(-2) ≤ -1(-2) -20 ≤ -3x -4 ≤ 2 This doesn't look right :/... Original:
-10 ≤ -3/2x + 2 ≤ -1 Subtract 2 from the middle and both sides: -12 ≤ -3/2x ≤ -3 Multiply by -2 (flip the signs because multiplying by a negative number): 24 ≥ 3x ≥ 6 Divide by 3: 8 ≥ x ≥ 2 You could save a step by multiplying by -(2/3) instead of multiplying by -2 and then dividing by 3, but I wanted to keep it simple. I think. It's been a long time since I had to do that. <_<
[+]
Ooooooh I see ok, yes simple is better in my case lol Thank you!
The more important question is... Why can't we copy and paste that equation into google and have it come up with the value of X?!
/fume Science has failed us!!!!!
[+]
Ok so what if it had 2 fractions in the equation :/
like this: -2/3z + 9 ≥ 2/3z + 21 Is it same steps as what you gave? Subtract 9 from all of them? Oh wait... hmmm subtract 9 from 21, combine like terms? You should multiply both sides by a common denominator to get rid of the fraction and then solve for z. This step isn't necessary but working with fractions should be avoided at all costs unless you absolutely have to since it's easier to make a mistake that way.
It would turn into something like: 3*(-2/3z + 9) ≥ (2/3z + 21)*3 -2z + 27 ≥ 2z +63 Then solve for z. Aly Saxena said: » -2/3z + 9 ≥ 2/3z + 21 Subtract 9 from both sides, subtract Multiply by -3 for both sides Divide by 4 for both sides or... Multiply both sides by the common denominator, 3 Simplify... Subtract 27 from both sides, Subtract 2z from both sides Divide by -4 for both sides Leviathan.Celille said: » You should multiply both sides by a common denominator to get rid of the fraction and then solve for z. This step isn't necessary but working with fractions should be avoided at all costs unless you absolutely have to since it's easier to make a mistake that way. It would turn into something like: 3*(-2/3z + 9) ≥ (2/3z + 21)*3 -2z + 27 ≥ 2z +63 Then solve for z. Just watch domain restrictions. Z can't be 0. I also end up getting: -4z ≥ 36 z <= -9 but according to wolfram alpha the correct answer is: edit: Oh, I'm doing 2 different equasions there. Is it -2/(3z) + 9 ≥ 2/(3z) + 21 or (-2/3)z + 9 ≥ (2/3)z + 21 if it's the former then there are domain restrictions, if it's the latter then x <= -9 would be correct. Yeah kalilla, if you multiply (or divide) a negative across an inequality it reverses the direction of the inequality.
simple example: 3 < 5 -1(3) < -1(5) -3 < -5 is incorrect, you need to reverse the inequality to -3 > -5 Odin.Registry said: » but according to wolfram alpha the correct answer is: It should be not |
All FFXIV and FFXI content and images © 2002-2025 SQUARE ENIX CO., LTD.
FINAL FANTASY is a registered trademark of Square Enix Co., Ltd.