I'm not sure of your background with modular arithmetic, so I'll just start small...
We consider all numbers with the same remainder after dividing by n to be EQUIVALENT.
For example, 2 = 5 = 8 = 11 = 300000000000002 = -1 mod 3. (Here, the "triple" bar sign would be better than the double bar...
sec(3t) = 2
invert (flip) both sides
cos(3t) = 1/2
Get a unit circle. Each point is of the form (cosine(angle), sine(angle))
We want the angles where cosine is 1/2.
These angles are:
π/3 and 5π/3
But since we have 3t, we include all solutions less than 6π. See the above reply for more details...
Have you ever seen P = e^{rt} before?
Usually, I would recommend exponential growth/decay for bacteria situations.
Furthermore,
"Doubling" and "halving" are very closely related!
Since this was the first hit when I Googled my homework, I'll resurrect this thread with my thoughts.
This holds trivially for prime numbers.
Claim: For n "square-free", \mathbb{Z}_n is (up to isomorphism) the unique abelian group of order n.
We extend the fact that \mathbb{Z}_{ab} \cong...
I guess so! There was another "seemingly active thread" about closed form expressions that caught my eye today, as did the fact that some first-poster had resurrected it after years of inactivity...
Cancel a factor of h from top and bottom. Then let h -> 0 and you'll get
\frac{-2}{x^2+2x+1} = \frac{-2}{(x + 1)^2}
Later, when you learn "shortcuts" (i.e rules of differentiation that will be proven), this second form will look a lot more familiar.
TsAmE, I might add that Calculus *methods* are not appropriate for this problem (even though it might appear in a calc book/course).
This is more about, as Mark44 has mentioned, coterminal angles and the domain of inverse trig functions.
Some people have an easier time working with degrees at...
WOW. I'll just... be on my way... at the local community college!
But seriously - that's awesome. What an opportunity (I speculate).
As for me, I'm probably (just) taking Intro to Abstract II and Advanced Calc I.... with 2 part-time jobs and part-time parenting, there's only room left for...
The integral isn't correct.
Your left-most integral bounds shouldn't have variables in it.
Your right-most shouldn't have constants.
You have used "z=" in two of the integrals, and "y=" in none.
In real numbers, the problem is immediate: no linear combination of powers of e will ever equal -1.
In complex numbers, L = W would make things a lot simpler, but that's trivial...
Part a. x^n * 1/x^n = 1.
To see this, notice that the left hand member is the same as
x^n/x^n, and any nonzero number divided by itself is equal to one.
Part b. x^n * x^-n = 1.
To see this, use the rule of exponents that says "when multiplying like bases, add the exponents". So you will have...
Gave you WHAT answer? That article doesn't explain where the error is...
Yes, you are having problems with the chain rule. If v = e^u, then dv/du = e^u, and dv = (e^u)du, which doesn't help solve the integral.
If you wanted to take dv/dx, then that would equal (dv/du)*(du/dx), but that's not...
This is the kind of problem that you shouls do once, for whatever reason (!), and then use a computer for afterward! Good luck. wolframalpha can verify your results, in case you weren't aware of that resource.
The difficulty in this prompts many to advise against exponentiating until exactly one log expression is on each side of the equality. But yeah, you can...