Find all roots for the trigonometric equation $$\sin^2 x + 0.5\sin x=0 $$.

To start with, let us rewrite the equation as $$\sin x (\sin x + 0.5) =0 \Rightarrow a) \sin x =0, \quad b) \sin x = -0.5.$$ Then, $$a) \sin x=0\Rightarrow x = \pi k, k\in Z$$ $$b) \sin x=-0.5 \Rightarrow x = \arcsin (- 0.5) + 2\pi k, k\in Z \Rightarrow x = -\frac{\pi}{6} + 2\pi k, k\in Z,$$ where $$Z$$ denotes the set of integers. Therefore, $$x \in \{ \pi k, -\frac{\pi}{6} + 2\pi k \}, k\in Z$$.

Joel received 16 candies for his birthday. Being kind, he gave 75% of those to his friends. How many candies Joel kept for himself?

Let us denote the number of candies Joel gave away with $$x$$. Knowing that the total number of candies is 16, we can write the following relation $$16:x = 100:75$$, from which we can calculate $$x = \frac{16\cdot 75}{100} = 12$$. Therefore, Joel gave away 12 candies, leaving him with 16-12 = 4 candies.

Test the convergence of the series $$\sum_{n=2}^\infty \frac{1}{n\ln n}$$.

To test the convergence of the series we will apply the Maclaurin-Cauchy test. The test states that for given $$N\geq 0$$ and continuous function $$f(n)$$ defined on the interval $$[N,\infty) $$, on which it is monotone decreasing, the series $$\sum_{n=N}^\infty f(n)$$ converges if and only if the improper integral $$\int_N^\infty f(x) dx$$ is finite. In our case, $$f(n)=\frac{1}{n\ln n}$$ and it is defined on the interval $$[2,\infty)$$. Both $$n$$ and $$\ln n$$ are monotone increasing functions on $$[2,\infty)$$, thus $$n \ln n$$ is monotone increasing as well and $$\frac{1}{n\ln n}$$ is monotone decreasing, implying that we can use the Maclaurin-Cauchy test. Recalling that $$d\ln x = \frac{1}{x} dx$$, we compute the following integral $$\int_{2}^n\frac{1}{x\ln x} dx = \int_{2}^n \frac{1}{\ln x}d\ln x = \ln(\ln x)|_2^n \rightarrow \infty, n\rightarrow \infty.$$ Therefore, the integral $$\int_2^\infty \frac{1}{x\ln x} dx$$ is infinite and, by Maclaurin-Cauchy test, the series $$\sum_{n=2}^\infty \frac{1}{n\ln n}$$ diverges.