technically it rounds to $2.38 but yeah
see edited version
\(\int 4 \cos(\pi x) ~dx = 4 \cdot \dfrac 1 \pi \sin(\pi x) = \dfrac{4}{\pi}\sin(\pi x)\)
\((2-i)z + (2+i)\bar{z} = 20\\ 2(z+\bar{z})-i(z-\bar{z}) = 20\\ 4Re(z) - 2iIm(z)= 20\\ \text{On the real axis }Im(z)=0\\ 4Re(z) = 20\\ Re(z) = 5\\ z=Re(z)+i Im(z)= 5 + i0 = 5\)
\(\dfrac{1+\sin(\theta)}{1-\sin(\theta)} \cdot \dfrac{1+\sin(\theta)}{1+\sin(\theta)} = \\ \dfrac{(1+\sin(\theta))^2}{1-\sin^2(\theta)} = \\ \left(\dfrac{1+\sin(\theta)}{\cos(\theta)}\right)^2 = \\ (\sec(\theta)+\tan(\theta))^2\)
\(\displaystyle \int \limits_{-\frac 1 2}^{\frac 1 2}4\cos(\pi x)-(12x^2-3) ~dx = \\ \left . \dfrac 4 \pi \sin(\pi x)-(4x^3-3x) \right|_{-\frac 1 2}^{\frac 1 2}\)
I leave you to complete it
\(\text{all the roots are integers so we have}\\ f(x) = (x-i_1)(x-i_2)(x-i_3)(x-i_4)\\ \text{The constant term is }c_0 = i_1 i_2 i_3 i_4 = 6\\ \text{3 can be a root as }3\cdot 2 = 6\\ \text{On the other hand 3 cannot be a double root as }3\cdot 3 = 9\\ \text{and there is no combination of integer factors that will multiply 9 to obtain 6}\)
\(\text{I'm showing}\\ \begin{pmatrix}1 &2 &3\\4 &5 &6\end{pmatrix}\\ \begin{pmatrix}1 &2 &4\\3 &5 &6\end{pmatrix}\\ \begin{pmatrix}1 &2 &5\\3 &4 &6\end{pmatrix}\\ \begin{pmatrix}1 &3 &4\\2 &5 &6\end{pmatrix}\\ \begin{pmatrix}1 &3 &5\\2 &4 &6\end{pmatrix}\\\)
C just has to not have a flower in common w/B (and D, but we enforce that on D not C)
My bad, let's start again
A can take four flowers, B can take 3
Then there are 3 choices for C.
If C=A there are 3 choices for D
if C!=A there are 2 choices for D
So we have
4 x 3 x (3 + 2x2) = 4 x 3 x 7 = 84
+6252 