Rom

another method

\(z = r e^{i\theta}\\ z^8 = r^8 e^{i8\theta}\\ z^8 + \bar{z}^8 = r^8(e^{i8\theta}+e^{-i8\theta}) = 2r^8\cos(8\theta)\\ z = \dfrac{3+i\sqrt{3}}{2}\\ r^2 = \left(\dfrac 3 2\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 = 3\\ r^8 = \left(r^2\right)^4 = 81\\ \theta = \arctan\left(\dfrac{\frac{\sqrt{3}}{2}}{\frac 3 2}\right) = \\ \arctan\left(\dfrac{\sqrt{3}}{3}\right) = \dfrac \pi 6\\ 8\theta = \dfrac{4\pi}{3}\\ \cos(8\theta) = -\dfrac 1 2\\ 2r^8 \cos(8\theta) = 2 \cdot 81 \cdot -\dfrac 1 2= -81\)

\(6 sed + 8 (120-sed) = 856\\ -2 sed = -104\\ sed = 52\\ suv=120-52=68\)

\(\text{probably easiest to just use the cheater formula for the vertex and evaluate the expression there}\\ \text{as it's a downward facing parabola that will be the maximum}\\ \text{the cheater formula for the vertex of $ax^2+bx+c$ is $x_v = -\dfrac b {2a}$}\\ r_v = -\dfrac{40}{2\cdot (-5)} = 4\\ f_v = -5(4^2)+40(4) - 12 = -80 +160-12 = 68\)

\(\dfrac 3 x + \dfrac x 3 = b\\ \dfrac{9+x^2}{3x}=b\\ 9+x^2 = 3bx\\ x^2 - 3bx+9 = 0\\ \text{we could use the quadratic equation and do some further solving but we can see that if $b=2$}\\ \text{then we have $x^2-3bx+9 = (x-3)^2$, which provides exactly one solution, i.e. $x=3$}\)

Ok you didn't like the last answer.

(Incidentally why be so rude?  If you feel the answer is incorrect respond back with why.  Don't just silently downvote me)

So let's look at another way to count these.

There are 3 partitions of 6 using 3 numbers 1-4

(4,1,1), (3,2,1), (2,2,2)

These can be assorted among the lanes so that 

(4,1,1) can be done 3 ways

(3,2,1) can be done 6 ways

(2,2,2) can be done 1 way

Now cars must be in order in the lanes.

(4,1,1) will have 6C4 * 2C1 = 30 arrangements.  You pick 4 cars from 6, sort them, put them in their lane.  Then select 1 car from the 2 remaining 

and put it into a lane.

(3,2,1) will have 6C3 * 3C2 = 60 arrangements

(2,2,2) will have 6C2 * 4C2 = 90 arrangements

so there will be 3*30 + 6*60 + 1*90 = 90+360+90 = 540 arrangements

6 different from the last way I solved it.  I'll have to figure out what's the disparity.

\(AB=B-A\\ AC=C-A\\ n = AB \times AC\\ \hat{n} = \dfrac{n}{\|n\|}\\ \text{The scaled version of the normal vector the problem is after is $-7\hat{n}$}\)

I leave plugging and chugging to you.

\(\text{We must have the sum the lengths of any two sides greater than the length of the third side}\\ x+13> 37\\ 13 + 37 > x\\ 24 < x < 50\\ x \in \{25, 26, 27, \dots 49\}\\ \text{There are 25 integer values of $x$}\)

\(\text{The remainder theorem says that the remainder}\\ \text{when dividing $f(x)$ by $(x-c)=f(c)$} \)

\(f(-3) = 11\)