HSC 2013 Maths Marathon (archive) (1 Viewer)

[ocatal]

Re: HSC 2013 2U Marathon

Domain: -4 < x < 4
Range:All real y
??

It is a semicircle with a radius of 3.

 

[Menomaths]

Re: HSC 2013 2U Marathon

It is a semicircle with a radius of 3.

Oh yeah I didn't see that isn't -4 < x < 4 the same as -3<= x <= 3 or am I going full retard?

edit: probs would've saw that if it was in latex lol

 

[ocatal]

Re: HSC 2013 2U Marathon

Oh yeah I didn't see that isn't -4 < x < 4 the same as -3<= x <= 3 or am I going full retard?

edit: probs would've saw that if it was in latex lol

Haha no problem. Hmm that boundary would also include the numbers in between 3 and 4 (i.e. 3.1, 3.2 etc.) so I don't think its right.

 

[Menomaths]

Re: HSC 2013 2U Marathon

Haha no problem. Hmm that boundary would also include the numbers in between 3 and 4 (i.e. 3.1, 3.2 etc.) so I don't think its right.

*Full retard mode activated*

 

[Sy123]

Re: HSC 2013 2U Marathon

 

[Badsiii]

Re: HSC 2013 2U Marathon

Dom 2 < x < 3 including

Range: 0 < y < 1 including

??

 

[Badsiii]

Re: HSC 2013 2U Marathon

how did you assume this was zero?

I didn't assume.. The last function value at f(last)= 0

 

[Sy123]

Re: HSC 2013 2U Marathon

Dom 2 < x < 3 including

Range: 0 < y < 1 including

??

Yea well done

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[Menomaths]

Re: HSC 2013 2U Marathon

Yea well done

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(5(+-)sqrt5)/10?

 

[Badsiii]

Re: HSC 2013 2U Marathon

yeah but where did you get zero?

Look at the graph/ garden. Then look at the last 5th function value it's on ground point with f(last) so it's =0

 

[Shazer2]

Re: HSC 2013 2U Marathon

Yea well done

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[Sy123]

Re: HSC 2013 2U Marathon

Yea correct

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(this one is different I assure you)

 

[Badsiii]

Re: HSC 2013 2U Marathon

Yea correct

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(this one is different I assure you)

1-cosx= sin^2(x)/cos(x)
Cos(x) - cos^2(x)= sin^2(x)
Sin^2(x) + cos^2(x)=1
Hence, cos(x)=1
X=pi/2

 

[Menomaths]

Re: HSC 2013 2U Marathon

Are you allowed to put 0 in the RHS?

 

[Sy123]

Re: HSC 2013 2U Marathon

1-cosx= sin^2(x)/cos(x)
Cos(x) - cos^2(x)= sin^2(x)
Sin^2(x) + cos^2(x)=1
Hence, cos(x)=1
X=0

Not quite

 

[Badsiii]

Re: HSC 2013 2U Marathon

Are you allowed to put 0 in the RHS?

X=pi/2 my bad

 

[Badsiii]

Re: HSC 2013 2U Marathon

Not quite

X=pi/2 silly mistake

 

[ColdLipstick]

Re: HSC 2013 2U Marathon

Another Q I wrote ^.^

 

[RealiseNothing]

Re: HSC 2013 2U Marathon

1-cosx= sin^2(x)/cos(x)
Cos(x) - cos^2(x)= sin^2(x)
Sin^2(x) + cos^2(x)=1
Hence, cos(x)=1
X=pi/2

Your first line isn't right.

 

[Sy123]

Re: HSC 2013 2U Marathon

X=pi/2 silly mistake

Still not there

This is a rather deceptive question lol