3p!c3ntr3 Posted September 25, 2007 Posted September 25, 2007 I Hate Maths And Homework I thought that then my mum showed me how to do quadratics it was really helpful
MoogleViper Posted September 26, 2007 Posted September 26, 2007 I seriously need help with my further maths hw. "A right angled triangle has area A cm² and perimeter P cm. A side other than the hypotenuse has legnth x cm. Form a quadratic equation in x in each of the following cases: (a) A=6, P=12, (b) A=3, P=8, © A=30, P=30. Hence, for each case above, find the possible values of x whenever real solutions exist." Please help.
Supergrunch Posted September 26, 2007 Posted September 26, 2007 I seriously need help with my further maths hw. "A right angled triangle has area A cm² and perimeter P cm. A side other than the hypotenuse has legnth x cm. Form a quadratic equation in x in each of the following cases: (a) A=6, P=12, (b) A=3, P=8, © A=30, P=30. Hence, for each case above, find the possible values of x whenever real solutions exist." Please help. Well, it'll take a fair bit of algebra (which I'm having a go at), but you need to define the other sides of the triangle in terms of x (pythagoras helps here) and thus define A and P in terms of x, and work from there.
MoogleViper Posted September 26, 2007 Posted September 26, 2007 I don't know how to define them. a²+b²=c² c=sqroot(x²+somethingelse²) Am I being really stupid or something?
Supergrunch Posted September 26, 2007 Posted September 26, 2007 Try calling the sides x, y and z, with z being the hypotenuse. We know three things: x^2 + y^2 = z^2, xy/2 = A, and x + y + z = P. Rearranging the second gives y = 2A/x, and we can substitute this into the first to find z. Then substitute for y and z in the third, and simplify. This gives a quartic, I think you should be able to use the fact that x > 0 to get a quadratic. Edit: I've made an error somewhere, I'm only getting imaginary solutions. But the principle should work. Edit 2: Yep, I checked my working and you get a quadratic.
MoogleViper Posted September 26, 2007 Posted September 26, 2007 I'm doing imaginary numbers. Forget to mention that. Anyway for the first one I get x= 3+-root-6 Is that right? I've checked it and no it's not right. I know what I've done wrong, I've put x²+y²=z² as x+y=z which I know is wrong but I don't know what to do.
Supergrunch Posted September 26, 2007 Posted September 26, 2007 I'm doing imaginary numbers. Forget to mention that. Anyway for the first one I get x= 3+-root-6 Is that right? I've checked it and no it's not right. I know what I've done wrong, I've put x²+y²=z² as x+y=z which I know is wrong but I don't know what to do. Use the method I gave, using the 3 equations, then just simplify. To give you something to aim for, you should get: 2P(x^2) - (4A+(P^2))x + 4AP = 0 Then simply substitute in values of A and P and solve. I think only equation 2 has imaginary solutions, and if the question is as you said it is then you don't need to give them. (out of interest, is this FP1?) Edit: A bit more help - although x is greater than 0, that's a red herring - this question is simply straightforward algebra. Using what I said in my previous post, substituting into the third equation we get: x + 2A/x + sqrt((x^2) + ((2A/x)^2)) = P Rearrange this so the sqrt is on one side, then square both sides and simplify to get the quadratic I gave earlier.
UK Posted October 18, 2007 Posted October 18, 2007 I've tried wiki and google, but in DT, can someone show me a page on chips? Cheers. (sorry about bumping this thread)
The fish Posted October 18, 2007 Posted October 18, 2007 I've tried wiki and google, but in DT, can someone show me a page on chips? Cheers. (sorry about bumping this thread) What kind? As its DT, you could mean computer chips, wood chips, or potato chips.
thirtynine. Posted October 19, 2007 Posted October 19, 2007 Computer chips please, will ITs a pretty broad subject, but google gave me this. http://www.intel.com/education/makingchips/index.htm
Konfucius Posted October 19, 2007 Posted October 19, 2007 Everything you need to know about chipsets (read a bit into it and must say it's really good): http://www.hardwaresecrets.com/article/191
Eenuh Posted October 20, 2007 Posted October 20, 2007 Okay, weird request but I don't have time to look for this right now (but need it this weekend!). We have to make a Flash movie that shows people more about ourselves (likes, dislikes, name, age, whatever). Now we have to look up other Flash movies and be able to show the ones we like and the style we'd like to work with. I already know I want to work with photos and paper drawn images (so a more realistic, artistic style), but I don't really know where to look for examples of something like this. Plus I don't have time to look (got other important assignments to work on right now!). So, if anyone knows some sites with these kind of Flash movies, please help me out! =D
chairdriver Posted October 28, 2007 Posted October 28, 2007 Someone help me, I got a really weird answer for this which I'm pretty sure is wrong: Determine the square roots of (15-8j).
Supergrunch Posted October 28, 2007 Posted October 28, 2007 Someone help me, I got a really weird answer for this which I'm pretty sure is wrong: Determine the square roots of (15-8j). ...hmm, I knew how to do that once upon a time. Something to do with simplifying it so that you're working out complex roots of unity I think.
Mr_Odwin Posted October 28, 2007 Posted October 28, 2007 No idea if this is the most effective way of doing it, but you can say that 15-8j = (a+bj)(a+bj) and then solve for a and b => a^2 +(b^2)(j^2)+2abj = 15-8j =a^2 -b^2 +2abj => 2ab = -8 a^2 - b^2 = 15 I think there are four roots. I followed through with the maths and found that one of them is a=-4 and b=+1. i.e. (-4+1j)^2 = 15-8j
Ninty 182 Posted October 30, 2007 Posted October 30, 2007 I have a History essay which I need some help with. How significant was the contribution of women to Britain's war effort in WW2? If possible I'd appreciate it if someone could post or email me a completed essay, if someone has already done this or one similar. If not, just something to help me out
Calza Posted November 13, 2007 Posted November 13, 2007 Right hopefully somebody could help but I'm not sure how many programmers there are (even though Basic is awful, I know). The homework was to write the code for a program that will create a random number between 1 and 100. The user has to guess that number and will be given clues such as "too low" or "too high". The program also needs to count how many guesses taken. I know it is really simple but since I don't have VB here I'm struggling to remember how it's done.
Ellmeister Posted November 13, 2007 Posted November 13, 2007 The intergral I= 3 S X times square root of X+1 dx. 0 Show that by using the substitution of u=x+1 gives same answer to if you used the integration by parts rule. Now I've attempted this, and gotten 2 different answers, one is a weird answer but the method I follow seems more right whereas the other answer is a nice integer, but the method I used seemed a little risky. So Can someone help? Much appreciated, thanks.
Shino Posted November 13, 2007 Posted November 13, 2007 Right hopefully somebody could help but I'm not sure how many programmers there are (even though Basic is awful, I know). The homework was to write the code for a program that will create a random number between 1 and 100. The user has to guess that number and will be given clues such as "too low" or "too high". The program also needs to count how many guesses taken. I know it is really simple but since I don't have VB here I'm struggling to remember how it's done. Do you want exact syntax or just pseudo-code?
Calza Posted November 13, 2007 Posted November 13, 2007 Do you want exact syntax or just pseudo-code? Pseudo-code would be fine, what ever is easier for you.
Mr_Odwin Posted November 13, 2007 Posted November 13, 2007 Right hopefully somebody could help but I'm not sure how many programmers there are (even though Basic is awful, I know). The homework was to write the code for a program that will create a random number between 1 and 100. The user has to guess that number and will be given clues such as "too low" or "too high". The program also needs to count how many guesses taken. I know it is really simple but since I don't have VB here I'm struggling to remember how it's done. I got it running in a console application quite quickly. Could need some polishing. Module Module1 Sub Main() Dim randnumber As Integer = Math.Ceiling(100 * Rnd()) Dim countnumber As Integer = 0 Dim guesscorrect As Boolean = False Console.WriteLine("Guess?") Dim guess As Integer = Console.ReadLine() While guesscorrect = False If guess = randnumber Then guesscorrect = True countnumber = countnumber + 1 Console.WriteLine("Correct-a-mundo!" & vbNewLine & "It took you " & countnumber & " guesses!") ElseIf guess < randnumber Then countnumber = countnumber + 1 Console.WriteLine("Too Low! Guess Again!") guess = Console.ReadLine() Else countnumber = countnumber + 1 Console.WriteLine("Too High! Guess Again!") guess = Console.ReadLine() End If End While Console.ReadLine() End Sub End Module The intergral I= 3 S X times square root of X+1 dx. 0 Show that by using the substitution of u=x+1 gives same answer to if you used the integration by parts rule. Now I've attempted this, and gotten 2 different answers, one is a weird answer but the method I follow seems more right whereas the other answer is a nice integer, but the method I used seemed a little risky. So Can someone help? Much appreciated, thanks. Very quickly I scribbled on a bit of paper and it ended up as 116/15 ~=7.73 Could be wrong though. The programming made my brain tired.
Ellmeister Posted November 13, 2007 Posted November 13, 2007 I got -52/15 for once answer. And 3 for the other =[
Shino Posted November 13, 2007 Posted November 13, 2007 void guess() { int r= Int(Rnd * 100), number, counter; print("Insert a guess number"); scan(number); while (number!=r) { if (number>r) { print("too high"); counter++; } else if (number<r) { print("too low"); counter++; } else if (number==r) { counter++; print("congrats!"); print(counter); } } } Edit: Oh, well...
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