VCE Stuff > VCE Specialist Mathematics
VCE Specialist 3/4 Question Thread!
dc302:
--- Quote from: Bhootnike on December 05, 2011, 08:04:51 pm ---yeahh thats right haha,
hence 'im stuck' :p
I know i shouldnt have even gone there..., so do you think it was something to do with not converting tanx to sin/cos ?
or just some arithmetic?
--- End quote ---
Cancelling is a big nono ;)
Bhootnike:
--- Quote from: b^3 on December 05, 2011, 08:13:11 pm ---move everytihng to the one side then used null factor law.
4tan(x)-4tan3(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan2(x))=0
tan(x)=0 or -8-4tan2(x)=0
x=0 or tan2(x)=2
x=0, or or no solution
So x=0, ,
EDIT: forgot sols
--- End quote ---
legend! thanks bro!
just ah... wouldnt:
-8-4tan2(x)=0
-4tan2(x)=0 +8
tan2(x)= 8 /-4
tan2(x)=-2 ?
so yeah no solution because u cant take the sq.root of a negative number anyways!
And thanks alot guys, will remember this for next time! :D
+1
b^3:
--- Quote from: Bhootnike on December 05, 2011, 08:21:29 pm ---
--- Quote from: b^3 on December 05, 2011, 08:13:11 pm ---move everytihng to the one side then used null factor law.
4tan(x)-4tan3(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan2(x))=0
tan(x)=0 or -8-4tan2(x)=0
x=0 or tan2(x)=2
x=0, or or no solution
So x=0, ,
EDIT: forgot sols
--- End quote ---
legend! thanks bro!
just ah... wouldnt:
-8-4tan2(x)=0
-4tan2(x)=0 +8
tan2(x)= 8 /-4
tan2(x)=-2 ?
so yeah no solution because u cant take the sq.root of a negative number anyways!
And thanks alot guys, will remember this for next time! :D
+1
--- End quote ---
YES! I did it in my head, and didn't type it, I've lost it already :(. Stupid holidays :|
Bhootnike:
ok sweet, atleast im getting something right ahaha... yaya.
thanks again guyss
Bhootnike:
Was wondering what other people's approach to this question is :
Use vector methods to prove that the midpoint of the hypotenuse of a right-angled triangle
is equidistant from all vertices.
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