b) R.T.P.
Squaring both sides,
Multiplying both sides by 16,
Can you finish it from here?
c) seems to be missing some context... what are and ?
Also this looks like it would belong in the Mathematics subforum (i.e. the 2U one).
I'm having trouble answering these questions, especially b) and c). Any help/answers would be greatly appreciated
maths.bmp
a and b.bmp
Last edited by atamiabwv; 8 Apr 2018 at 12:16 AM.
b) R.T.P.
Squaring both sides,
Multiplying both sides by 16,
Can you finish it from here?
c) seems to be missing some context... what are and ?
Also this looks like it would belong in the Mathematics subforum (i.e. the 2U one).
Last edited by fan96; 7 Apr 2018 at 11:31 PM.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
For part (b), literally follow the hint: after squaring both sides and simplifying, you should get
32 + 16√2 < 38 + 12√2,
which simplifies further to
√2 < 3/2,
which is the same inequality in part (a).
I can't help you with part (c), since I don't know what a and b are.
^{Bachelor of Science (Advanced Mathematics)/Bachelor of Arts III, USYD}
Last edited by fan96; 8 Apr 2018 at 1:36 AM.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
That's really helpful, thank you very much! I've taken it a bit further for the LHS, but I'm not sure what do to for the RHS. I've attached my working: IMG_6433.bmp
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
Subtract 2:
Multiply by -1 (which flips the inequality sign)
Which is the result in b)
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
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