VCE Stuff > VCE Specialist Mathematics

Specialist 1/2 Question Thread!

<< < (6/84) > >>

tysh:

--- Quote from: qwertyqwerty on March 21, 2016, 02:32:21 pm ---Solve for x:
|x-4| - |x+2| = 6

--- End quote ---

One possible approach is this:

Form magnitude values:
|x-4|=x-4 if x>4, |x-4|=4-x if x<4.
|x+2|=x+2 if x>-2, |x+2|=-x-2 if x<-2.

If x>-2 and x>4, so x-4-(x+2)=6, but -6=6 is false.
If x<-2 and x<4, so 4-x-(-x-2)=6, we end up with an equation with infinite solutions (x=x), thus x<-2 for these solutions.
If x>-2 and x<4, so 4-x-(x+2)=6, 2x=-4, x=-2 is a solution.
We have x=-2 and x<-2, so x≤-2.

bk.fuse:
Hey guys
I just had my second spesh test and I had done all the school text book questions and additional questions from other textbooks and i can do all the questions off by heart, even with the practise tests my teacher gave us, i could do really easily. Even though i could do all of that, the test questions were a lot harder and even though i knew how to do all the questions, i ran out of time. Do any of you guys know any resources that i could use to practise with? Such as websites or any text books that could help so i can just practise a lot more.
Thanks!  :)

Adequace:
http://imgur.com/a/hN7pz

I need some help for part a. My working was a random method but I'm not sure why this doesn't work?
Another way which seems more likely is finding AB then using the cosine rule to find the magnitude of the angle, my problem is how do I find the length AB?

Thanks

Syndicate:

--- Quote from: Adequace on April 16, 2016, 09:42:37 pm ---http://imgur.com/a/hN7pz

I need some help for part a. My working was a random method but I'm not sure why this doesn't work?
Another way which seems more likely is finding AB then using the cosine rule to find the magnitude of the angle, my problem is how do I find the length AB?

Thanks

--- End quote ---

I believe you meant to ask: how would you find the length of O'B (because only then you could use the cosine formula to workout the angle between AO'B).

As the triangles share the chord lengh equally, it must mean that they are isosceles triangles. Hence, O'B = O'A.

So, O'A = O'B = 6cm, and AB = 8 cm


cosO' = (6^2 + 6^2 - 8^2)/ (2 x 6 x 6)

plug it in your calculator, and you will end up with 83.62 degrees.

Adequace:
Thanks, I just realised as well!

Navigation

[0] Message Index

[#] Next page

[*] Previous page