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#1 Re: Help Me ! » more complex numbers.... » 2006-08-24 05:04:02

dont tell me its to hard :-(...... I really could do with the help

#2 Re: Help Me ! » more complex numbers.... » 2006-08-24 04:53:40

Please can anyone help me with these?

#3 Re: Help Me ! » more complex numbers.... » 2006-08-22 22:30:13

Cant anyone answer these questions??

#8 Help Me ! » more complex numbers.... » 2006-08-21 22:23:48

stacey
Replies: 5

I also to have some engineering questions....seeing as boombastictiger started some off...heres mine:

1

Consider the sytem of equations:


x - 6y + 2z = 5
2x - 9y - z = 14
4x + 12y - 3z = 19


A: By reducing the augmented matrix to row echelon form, find the solution to the equations.

B: Leaving the first two equations unchanged, the third is changed to : -3x + 6y + 14z = -31

C: Show the equations have infinitely many solutions and give a general formula for them.

D: Leaving the first two equations unchanged, the third si changed to: 3x - 24y + 16z = 9 , WHAT HAPPENDS NOW? GIVE REASONS FOR ANSWER


2

A: Show the equation x³ -3x² + 2x + 1 = 0 has the root between -1 and zero.  Use Newton's method to find it, correct to three decimal places.

B: If the complex numbers z and w are given by:

z = 3 - 5i and w = -1 + 2i

      FIND

z - w, zw,  z      , plus the modulus and argument of z
                 /w

C: Plot z on the Argand Diagram

D Solve the equation: 4z³ + 3z² + 94z - 130 = 0 ,   GIVEN one solution is: -1-5i



THANKS, PLEASE SHOW WORKING OUT ASWELL IF YOU CANup

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