Do my Penguin for me

[MadDog]

I Hate Maths And Homework

[3p!c3ntr3]

I Hate Maths And Homework

 

I thought that then my mum showed me how to do quadratics it was really helpful

[MoogleViper]

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]

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]

I don't know how to define them. a²+b²=c²

 

c=sqroot(x²+somethingelse²)

 

Am I being really stupid or something?

[Supergrunch]

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]

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]

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]

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]

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.

[UK]

Computer chips please, will

[thirtynine.]

Computer chips please, will

ITs a pretty broad subject, but google gave me this.

http://www.intel.com/education/makingchips/index.htm

[Konfucius]

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]

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]

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]

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]

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]

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]

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]

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]

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]

Do you want exact syntax or just pseudo-code?

 

Pseudo-code would be fine, what ever is easier for you.

[Mr_Odwin]

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]

I got -52/15 for once answer. And 3 for the other =[

[Shino]

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...