VCE Methods Question Thread!

[eforlano]

I understood part a, but part b is a bit confusing >< any help would be appreciated ><

[MB_]

thanks so much!!! makes sense now

Can someone please explain how to integrate 1/(2x-1)^3 ? thanks in advance ^^

Working attached

[MandhreeE]

You need to find m in terms of a by finding the intersection between the two graphs and then use that as your upper value

That makes more sense thank you

[MandhreeE]

could someone please explain why when we find an antiderivative of 1/(3x-4) the answer is 1/3In(3x-4) and not In(3x-4) i seem to be getting really confused with these laws

[HasibA]

could someone please explain why when we find an antiderivative of 1/(3x-4) the answer is 1/3In(3x-4) and not In(3x-4) i seem to be getting really confused with these laws

theres a formula on the exams 1 and 2 sheets , that relates to 1/(ax+b)^N , where N is not 1 or someshit
use that? youll get it supplied tmr !!

[MandhreeE]

theres a formula on the exams 1 and 2 sheets , that relates to 1/(ax+b)^N , where N is not 1 or someshit
use that? youll get it supplied tmr !!

Hahah thanks i know its on the formula sheet i just dont know how that was the answer even though i used the formual sheet to get it. haha

[HasibA]

Hahah thanks i know its on the formula sheet i just dont know how that was the answer even though i used the formual sheet to get it. haha

ohhh jokes.  i just mrbed, when u use that fomrula, N cannot EQUAL -1, so you have to use substituiton (idk what its called ahhaha)
like let U=(3x-4) and treat it as a 1/U, integrate that etc.
sorry, im horrid at methods ah!

[MB_]

could someone please explain why when we find an antiderivative of 1/(3x-4) the answer is 1/3In(3x-4) and not In(3x-4) i seem to be getting really confused with these laws

Its because you have to divide by the coefficient of x

[MandhreeE]

Its because you have to divide by the coefficient of x

oh! is that always when we intergrate? thank you!!! that makes a lot more sense. I was doing it for all other functions but log. thank you

[Zoe211]

Can anyone help me with VCAA 2015 Exam 1 question 10? I don't understand triangles at all and part a) and b) are completely beyond me

[Spode24363]

Simplify (log2(p^2-1)/ log2(p+1),p>-1, expressing your answer in the form k+logm(n)
thanks in advance                     

[MB_]

Can anyone help me with VCAA 2015 Exam 1 question 10? I don't understand triangles at all and part a) and b) are completely beyond me

You setup a triangle like in the image attached. Cos(x) is equal to your x and sin(x) is your y however as the radius of the circle is 2, using trig you find that your y is 2sin(x) and your x is 2+2cos(x) because the centre of the circle is at (2,0). Hopefully that helps

(Image removed from quote.)

I guess MB_ beat me to it  .

In addition to what MB_ asserted, it would be worth having a look at polar coordinates (rcosx, rsinx) for the exam. 

Cartesian coordinates (x,y) = polar coordinates (rcosx, rsinx)

(Image removed from quote.)

[bedigursimran]

Hey guys!

For exam 1, when we are dealing with a 95% Confidence interval do we simply write 1.96 or do we round to 2? Thanks.

And also, what other confidence interval can they ask us?

[larissaaa_]

GUYS HELP.

What are people writing in for the two exams? Pen or pencil? I'm so conflicted

[halo]

Hey guys!

For exam 1, when we are dealing with a 95% Confidence interval do we simply write 1.96 or do we round to 2? Thanks.

And also, what other confidence interval can they ask us?

I'm not 100% sure but for exam 1 they would most likely give us the z value they want us to use for a particular confidence interval (including whether to use 1.96 or 2 for a 95% confidence interval). But to answer your question, we generally use 1.96?