My Methods 3/4 questions thread

[bobbyz0r]

Just a quick question, how to use CAS to solve tan(x)=0.83 for the angle?

You don't need CAS. Just get the tan^-1(0.83)=x + n

[smile+energy]

You don't need CAS. Just get the tan^-1(0.83)=x + n

Yeah, but the question requires the particular angle. Do you know how?

[lzxnl]

type in tan^-1(0.83) for the basic angle

[smile+energy]

Thanks heaps

[smile+energy]

Could anyone help me to solve the question 2 part b?
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009mmCAS1-w.pdf
I don't know what the '4' and '1' mean

[nhmn0301]

Could anyone help me to solve the question 2 part b?
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009mmCAS1-w.pdf
I don't know what the '4' and '1' mean

That means definite integral, meaning to only integrate from x=1 to x=4.
For example: if you have [ 2x - 3 ] is the value that you have after apply normal anti-diff (obviously not in this question). Then what definite integral does is : [2x - 3] ⁴₁ = (2 x 4 -3 ) - (2 x 1 - 3) = 5 - (-1) = 6.
I don't know what textbook you use but if it's Essential, then read 12.5 on page 456, if it's Math Quest, then read page 441.

[smile+energy]

That means definite integral, meaning to only integrate from x=1 to x=4.
For example: if you have [ 2x - 3 ] is the value that you have after apply normal anti-diff (obviously not in this question). Then what definite integral does is : [2x - 3] ⁴₁ = (2 x 4 -3 ) - (2 x 1 - 3) = 5 - (-1) = 6.
I don't know what textbook you use but if it's Essential, then read 12.5 on page 456, if it's Math Quest, then read page 441.

Thanks so much.
I got it and I found the page.