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I believe cosmology is an advanced course available to students who accelerate. This can be (and usually is) done at a local university which offers it.

I think this one is more so: -______-"""

= 1/[4cosxsinx + cos^2 x] = 1/[2sin(2x) + cos^2 x] not equal the question!. You made a small mistakeeee

pi/4 / | 1 |----------------dx |2sin(2x) + cosx / 0

I understand why the acidic catalyst is required, however; why does dehydration require a concentrated acidic catalyst, and why does hydration require a dilute acidic catalyst.

Is this the answer:

From my class test. Prove: [file attached]

Hereeeeee :D edit: oh damn i got beat to it : P

books wrong. If the only information given is the diagram and; If pythagoras doesn't work, then obviously the angle is not 90 degrees. or theres like extra info in invisible ink

Heres onee: Graph |x+y| = 2 labeling significant points.

CBF typing, its obvious. (sec^2 is the derivative of tan)

It's not "hard". It's just a broken question, with that little information, the only way is to measure it physically. It's like asking; prove x = 1 when y = 2, without any given relationship between x and y. edit: You could measure the sides, and prove that pythagoras works (hence it...

1000 words is about 6-8 exam pages.

As stupid as this sounds. Get a potracter and measure it. In sc they allow measurement to find angles. Solving by any other means is impossible without more information.

3 months is outrageous. 3 weeks is plenty

int tannx dx = int tann-2x(sec2x -1) dx = int tann-2xsec2x dx - In-2 = tann-1x / (n-1) - In-2

| | 1 / (1 + sinx) dx | | | 1-sinx / (1 - sin^2x) dx | | | 1-sinx / cos^2 x dx | | | sec^2 x - integral sinx / cos^2 x dx | = tanx + c1 + integral cos^-2 d (cosx) = tanx - sec x + c2

This is a slightly modified question from my half yearlyy.

Finding patterns in some sequence isn't really learning how the general solutions formulas are constructed.

In the exam, they won't penalise you if you have more then one general solution formula. You can have 1 or 2 or 3 overlapping eachother, as long as they are all viable.