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Giving me negative rep does nothing. Just a red instead of a green. But thanks for thinking about me.

Okay. Thanks. We have done vectors at an introductory level, but are now doing alot more work on dot product and cross product. vector AB = 2i + j -4k So you did B-A vector CD = -2i -j + 4k = -(2i + j -4k) And here you did D-C What is the name of this rule and where can I learn more about...

Show that the following are vertices of a rectangle: i+j+k (3 dimensions) A(2,1,3) B(4,2,-1) C(5,4,0) D(3,3,4) A, B, C and D are position vectors Apparently since these are just points I have to find a resultant vector(s) and then do some other calculations? Please help. Just started this...

Well...you could say that. But, the basics are covered, and for those that like maths (I do, but I'm a rare find in QLD.) you can appreciate other material that is not covered. We have three types of maths Maths A - You can get an A very easily Maths B - This is your mid-level difficulty...

True. But more appropriately perhaps, the New World Order is already here. Now, more than ever are political leaders calling for a 'one world government', 'new world order'. Also, the way you said that: 'The New World Order is coming' may come across to those unfamiliar with reality as a...

Yes. In fact, ig you turn up without one, they'll have like 3-4 spares ready to give you. I'm in grade 12 now, and you'd be spewing if you saw our curriculum. For example, binomial probability. They don't teach permutations/combinations/combinatorics at all. You use binombdf or binomcdf on the...

This cubic is one that I made up. I'm just tired of seeing these sorts of equations and using a graphics calculator. I'm curious as to how you'd solve them. I'm in Queensland, and unfortunately a graphics calculator is used for everything. So yeah.

Ahhh.. You are a legend!. You're right, a rational root would have to be a/b, where integers. So, I have ruled out the use of rational roots theorem. I didn't realize that answer could also be irrational. Okay, next strategy is...sum products?. What is that?.

I plugged the equation into an online solver getting one rational root, and two complex roots. the rational root is: -1.35729514140789 But this is not one of the fractions that I have?.

Okay. So, the factors of the constant term will be: 1, 6 , 4 , 2 , 12 , 24 , 3 , 8 and of the leading co-efficient: 1, 7 So the factors will 'have' to be these, if there are any rational roots: +/- 1, 6 , 4 , 2 , 12 , 24 , 3 , 8 ......._____________________ 1, 7 So...

Okay,so from my equation: 7x^3 − 5x^2 − 2x+24 = 0 Possible candidates for a rational roots will be: +/- 1, 7 , 6 , 4 , 2 , 12 , 24 , 3 , 8 And then I would have to examine the list of these possibilities as fractions?

Thankyou :)

Suppose we wish to solve x^3 − 5x^2 − 2x+24 = 0 given that x = −2 is a solution. We know that (x+2) will be a facor, and we can then assume that the cubic can be factored as follows: (x+2)(x^2+ax+b) Then by synthetic division: .....(X+2)(x-3)(x-4)=0, x = -2, 3, 4 BUT, say the equation given...

Wow. That was a very logical argument. You back up what you have to say really well.

It's a map based on Indigenous populations.

'A hard time?'. I think we deserve an apology day from the Aborigines.

I didn't realize that you were king of the forums. But I have seen the light, and can now understand why telling the truth is the wrong thing to do. After all, your incredible online reputation stands testament to just how great you are. And thanks for calling me a truther. You just prove my...

That's a ridiculous proposal. But you are obviously joking, so I assume you are a sane individual and that was posted in jest.

LOL:haha:

Whats wrong with ynet?