Todays Maths lesson

[Acidic]

My techer had to throw something together for my class today, for our one hour lesson. See if you get the same answers:
1. Complex numbers.
z1=3+i
z2=2-i <--Edited: now correct
Find a) z1z2
b) z1/z2

2. Calculate x & y as rationals
53x*25y=1/5
7[sp]x[/sup]*492y=1

3. Find x if:
|x 6| = |2 1 3|
|x x| |1 2 3|
|3 1 1|


4. Find d2y/dx2 if y=1/qubrt(x3+2)

5. Two dice are rolled. The score is the smaller of the two numbers that appear. If the sumber is the same then the score is that number. What is the probability of getting a score of 3?

Can't be bothered to write the rest.

[fundu]

The 5th is 7/36

[Acidic]

correct.

[jemidiah]

1a. z1z2
(3 + i)(2 - 1) = (3 + i)(1) = 3 + i
If you meant (2 - i) then
(3 + i)(2 - i) = 6 - 3i + 2i + 1 = 7 - i

1b. z1/z2
(3 + i)/(2 - 1) = (3 + i)/1 = (3 + i)
If you meant (2 - i) then
(3 + i)/(2 - i) = 3/2 - 3/i + i/2 - 1 = 1/2 + 7i/2 = (7i + 1)/2

2. Those simplify to 6x + 4y = -1; x + 4y = 0; x = -1/5, y = 1/20

3. Not sure on, never seen matrices(?) like that

4. y=1/qubrt(x3+2) = (x3+ 2)-1/4; simple chain rule makes -1/4 * (x3 + 2)-5/4 * 3x2. It can be slightly simplified more, but it's nearly self-defeating unless written out.

5. Already said

I hope I haven't made any stupid little mistakes in my algebra :)

[Acidic]

Firstly, yes I did mean 2-i, sorry.

1.a) correct
b) wrong, it should be 1+i

2. Almost, look at your 6x + 4y = -1 eqn again.

3. I will try to clear that mess up :)
By using square brackets, I meant the determinant, not the matrix (but yes you use doing "Matrix Maths"). The forum messed it up though and I didn't notice untill now. It should be:

[x 6] = [2 1 3]
[x x] [1 2 3]
[3 1 1]


so the determinant on the left equals the determinant on the right.

4. by qubrt I meant quberoot, therefore y=(x3+2)-1/3
and remember that I'm not asking for:
dy/dx
but:
d2y/sx2
so you need to differentiante it again.

[jemidiah]

Told ya 'bout those stupid mistakes :)

I needed more sleep, but oh well; here goes again:

1b. It's an odd one to simplify, being imaginary and all, so I'll just conveniently skip over it

2. Of course; it should be 6x + 4y = -2, so x = -2/5, y = 10

3. Sorry. My calculator would be able to do that, but we skipped right over determinants back in Algebra II (I know because she said, 'We're just going to skip determinants and come back if there's time,' which there wasn't)

4. Told ya' I was tired :)

(2x4 + 2)/((x3 + 2)7/3). I'd show the work, but this time I did it on paper

[Acidic]

I'll do 1b

3+i/2-i = (3+1/2-i) * (2+i/2+i)
= 6+5i+i2/4-i2
= 5+5i/5
= 1+i

2. X is right, but y is supposed to be 1/10

3. Determinant of a 2by2 matrix is ad-bc
of a 3by3 is more complex, but your calc should be able to do it.

4. I'll check when I get home.

[Acidic]

4. Here's what I got:
4x4/(x3+2)7/3 - 2x/(x3+n)3/4

[fundu]

1. Solved

2. Solved

3. x = 3

4. (2x^4 - 4x) / (x^3 + 2)^(7/3)

Acidic was right just he mistyped the 4/3 as 3/4

5. Solved


Thus change the status of thread as solved.......