Rom

\(4\cos^2(2\theta-\pi)=3\\ \cos(2\theta-\pi) = \pm \dfrac{\sqrt{3}}{2}\\ \text{We need the 3 smallest positive values of $\theta$ so we have}\\ 2\theta - \pi = -\dfrac {5\pi}{6}, -\dfrac{\pi}{6}, \dfrac{\pi}{6}\\ 2\theta = \dfrac \pi 6,~\dfrac{5\pi}{6},~\dfrac{7\pi}{6}\\ \theta = \dfrac{\pi}{12},~\dfrac{5\pi}{12},~\dfrac{7\pi}{12}\)

I should have been more careful.

My statement is true if for all x such that f(x) is defined, then f(-x) is defined as well.

\(\text{Your first differential equation!}\\ \text{I don't know how much you've been taught. The usual way to solve this is}\\ f'(x) = 5f(x)\\ \dfrac{f'(x)}{f(x)}=5\\ \text{Integrate both sides}\\ \ln(f(x)) = 5x+c\\ f(x) = e^{5x+c} = Ce^{5x}\)

\(i = e^{i \pi/2+2\pi k},~i \in \mathbb{Z}\\ z^2 = e^{i (\pi/2+2\pi k)}\\ z = \exp\left(i\dfrac{\frac \pi 2 + 2\pi k}{2}\right) =\\ \exp\left(i\left(\frac \pi 4 + \pi k\right)\right),~k = 0,1 = \\ e^{i\pi/4},~e^{i5\pi/4}=\\ \pm \dfrac{1}{\sqrt{2}}(1+i)\)

can you multiply two matrices together?

The top question wants you to multiply A times B, drag that to the box, then multiply that by C, and drag that product to the box.

The bottom question wants you to multiply B times C first, drag that into the box, then multiply A times that and put it in the box.

Well presumably the velocity vs. time graph is actually a group of graphs, one for each directional component of the velocity vector.

Simply look up the value for each component on the graph at the time you want the speed,

then take the square root of the sum of the square of each directional component.

That will be the instantaneous speed.

\(\text{any function $f(x)$ can be decomposed into its even and odd parts}\\ f(x) = e(x)+o(x)=\dfrac{6}{x+2}+x^2+2^x\\ e(x) = \dfrac 1 2 \left(f(x)+f(-x)\right)\\ o(x) = \dfrac 1 2 \left(f(x) - f(-x)\right)\\~\\ o(1) = \dfrac 1 2(f(1) -f(-1))\\~\\ \text{I leave you to plug and chug}\)

\(2x - y = a\\ -6x+3y=b\\ \text{Add 3 times equation 1 to equation 2}\\ 0=3a+b\\ b = -3a\\ \dfrac a b = \dfrac{a}{-3a} = -\dfrac 1 3\)

\(\text{From the vertex we have}\\ (y-3) = a(x+3)^2\\ \text{Now from the point}\\ (0-3) = a(-6+3)^2\\ -3 = 9a\\ a=-\dfrac 1 3\\ y = -\dfrac 1 3 (x+3)^2 + 3\)