Rom

\(\cos^2(\theta) = 1\\ \text{there are only two values }x \text{ such that }x^2=1\text{, these are}\\ x=1,~x=-1\\ \cos(\theta)=1 \Rightarrow \theta = -360^\circ,~\theta=0^\circ\\ \cos(\theta)=-1 \Rightarrow \theta = -180^\circ,~\theta = 180^\circ\)

\(a^{-1}\equiv a \pmod{24} \Longleftrightarrow a^2 \equiv 1 \pmod{24}\\ a^2-1 = 24k,~k \in \mathbb{Z}\)

\(|\{1,5,7,11,13,17,19,23\}|=8\)

\(\text{We notice that these are primes such that }p^2 > 24\\ \text{There must be an algebraic reason for this and I'll see if I can dig it up}\)

\(3^2 = 9<10\\ 18^2 = 324 > 300\\ \{x : x^2 \in [10,300]\} = \{4,5,\dots, 17\}\\ | \{4,5,\dots, 17\}| = 17-4+1 = 14\)

\(\text{What's that? Lagrange multipliers?}\\ \text{The method of Lagrange multipliers involves solving}\\ \nabla (f(s,h) - \lambda (c(s,h)-C))\\ \text{In this problem }\\ f(s,h) = vol(s,h) = s^2 h\\ c(s,h)=6sh+6s^2 (\text{see the previous answer})\\ C=15\)

\(\text{our Lagrange equation becomes}\\ \nabla \left(s^2 h - \lambda (6sh-6s^2 - 15)\right)= (0,0,0)\\ \text{this leads to the system of equations }\\ 2sh - \lambda(6h - 12s)=0\\ s^2 - 6s\lambda = 0\\ 6sh-6s^2-15 = 0\\\)

\(\text{and these are solved for }s,~h,~\lambda\\ \text{This seems like a lot more work than the previous answer. Why do it?}\\ \text{In most problems you can't cleanly solve for one in terms of the other and substitute like we did}\)

\(\text{the wording is kind of odd. Is the bottom open? Sort of a weird way to orient the problem if so}\\ \text{I'm going to assume both the top and bottom are closed}\\ \text{we've got the two different types of materials so let's compute the two areas separately}\\ \text{assume the box has square side length }s \text{ and height }h\\ area_{side}=4sh\\ area_{tb} = 2s^2\\ vol = s^2h\)

\(\text{we can convert areas into cost using the materials prices}\\ C = 4sh\cdot (1.5) + 2s^2 \cdot (3) = 6sh + 6s^2\)

\(\text{I"m going to assume you haven't been taught Lagrange multipliers yet}\\ \text{so we have to use the cost function to write the volume in a single variable}\\ 15 = 6sh + 6s^2\\ h = \dfrac{15-6s^2}{6s}\\ vol = s^2 h = s^2\dfrac{15-6s^2}{6s} = \dfrac{15s-6s^3}{6}\)

\(\text{Now to find the extrema of the volume we set it's derivative equal to 0}\\ \dfrac{dvol}{ds} = \dfrac{15-18s^2}{6} =\dfrac{ 5-6s^2}{2} = 0\\ 5 = 6s^2\\ s = \sqrt{\dfrac 5 6}~ft \text{ (we can ignore the negative solution)}\)

\(vol = \dfrac 1 6 \left(15\sqrt{\dfrac 5 6} -6 \left(\dfrac 5 6\right)^{3/2}\right) = \dfrac 5 3 \sqrt{\dfrac 5 6}~ft^3\)

\(\dfrac{120}{20} = \dfrac{93}{x}\\ 120x = 1860\\ x = 15.5~lbs\)

\(\text{a line perpendicular to a line with slope }m \text{ has slope }-\dfrac 1 m\\ y=-\dfrac 2 3 x + 5 \text{ has slope }m = -\dfrac 2 3\\ -\dfrac 1 m = \dfrac 3 2\\ \text{now we use point slope form of a line}\\ (y-1) = \dfrac 3 2 (x-8)\\ y = \dfrac 3 2 x - 11\)

\(2000 \pmod{24} = 8\\ 2am+8 = 10am\\ \)

I told you it was in the other thread

well done