Rom

Since the constant term is -1 you know your factors are going to be

\(6x^2 -x -1 = (a x + 1)(b x -1) = (ab)x^2+(b-a)x-1\)

\(a b =6, (b-a)=-1 \\ b=a-1 \\ a(a-1) = 6 \\ a=3, b=2\)

\(6x^2-x-1 = (3x+1)(2x-1)\)

\(-\ln(x+\sqrt{x^2+1}) = \ln\left(\dfrac{1}{x+\sqrt{x^2+1}}\right) =\)

\(\ln\left(\dfrac{1}{x+\sqrt{x^2+1}} \dfrac{x-\sqrt{x^2+1}}{x-\sqrt{x^2+1}}\right) = \ln\left( \dfrac{x-\sqrt{x^2+1}}{x^2-(x^2+1)}\right)=\)

\(\ln\left(-(x-\sqrt{x^2+1})\right) =\ln\left(-x + \sqrt{x^2+1}\right)\)

\(\mbox{so }f(-x)=-f(x) \mbox{ and f is odd.}\)

First thing you need to do is sort the data.  Doing this we get

7,17,18,20,26,26,26,32,33,34,43,43,45,48,52

By eye we can pick the minimum and maximum as 7 and 52.  These are the 1st and 5th numbers.

Also by eye we can pick the median as 32, this is the 3rd number.

Then we take the sequence less than the median which is 7,17,18,20,26,26,26 and find it's median which is 20.

This is the 2nd number.

Finally we take the sequence greater than the median 33, 34, 43, 43, 45, 48, 52, and find the median of this

which is 43.  This is our 4th number.

So our 5 number summary is

7, 20, 32, 43, 52

\(\mbox{You want a score }G \mbox{ such that }\Phi\left(\dfrac{G-544}{72}\right)=0.2 \\ \mbox{where }\Phi(x) \mbox{ is the CDF of the standard normal.}\\ \mbox{Using the trusty table we find that }\\ \Phi^{-1}(0.2)=-0.841621 \mbox{ so}\\ \dfrac {G-544}{72}=-0.841621 \\ G = 483.4\)

\(\mbox{We want to maximize }x y \mbox{ subject to }x+y=22\\ y=22-x \\ x y = (22-x)x = 22x - x^2 = -(x^2-22x) = \\ -((x-11)^2-121) = 121-(x-11)^2 \\ \mbox{This has a maximum value of }121 \mbox{ at }x=y=11\)

\(5^9 \times 5^3 = 5^{9+3}=5^{12}\)

\(\mbox{A rectangular area has perimeter }2(L+W) \mbox{ and area }L W\\ \mbox{so }2(L+W)=2000\\ L+W=1000\\ L=1000-W\)

\(A=LW = (1000-W)W = 1000W-W^2\)

\(A=-(W^2-1000W) = -((W-500)^2-250000)=250000-(W-500)^2\)

\(\mbox{It should be clear that the max area is thus }250000 sq. yds\\ \mbox{and this occurs at }W=500 yds, L=500yds\)

\(\cot(\theta)=\dfrac 1 4 \\ \dfrac {\cos(\theta)}{\sin(\theta)}=\dfrac 1 4 \\ \dfrac{\sqrt{1-\sin^2(\theta)}}{\sin(\theta)}=\dfrac 1 4\\ \dfrac{1-\sin^2(\theta)}{\sin^2(\theta)}=\dfrac 1 {16} \\ 16 -16\sin^2(\theta)=\sin^2(\theta)\)

\(16 = 17\sin^2(\theta) \\ \sin^2(\theta)=\dfrac {16}{17} \\ \sin(\theta)=\sqrt{\dfrac{16}{17}}=\dfrac{4}{\sqrt{17}}\)

\(\cos(\theta)=\sqrt{1-\sin^2(\theta)} = \sqrt{1-\left(\dfrac{4}{\sqrt{17}}\right)^2}=\sqrt{\dfrac {1}{17}}\\ \sec(\theta)=\dfrac 1 {\cos(\theta)} = \sqrt{17}\)

\(\mbox{The current flow through the internal resistance causes the voltage drop.}\\ dV=i R\\ (1.45-1.38)V = 0.35 R~~ A \Omega \\ R \Omega = \dfrac{0.07 V}{0.35 A} =0.2 \Omega = 200 m\Omega\)

\(\mbox{I'm going to assume it's linear and then check.}\\ (p-32.50)=m(N-250)\\ m=\dfrac{31-32.5}{400-250} = \dfrac{-1.5}{150}=-\dfrac 1 {100} \\ \mbox{so }(p-32.50)=-\dfrac 1 {100}(N-250)\\ p=-\dfrac{N}{100}+35\)

\(\mbox{Checking the two inside points.}\\ 32=-3+35 \Rightarrow True\\ 31.50=-3.5+35 \Rightarrow True\\ \mbox{so all the points check out and the formula is correct.}\)

(b) \(R(N)=N p(N) = N\left(35 - \dfrac N {100}\right)\)

(c) \(Profit(N) = R(N) - C(N) = N\left(35-\dfrac N{100}\right)-(30N+700)\)

\(Profit(N)=5N - \dfrac{N^2}{100}-700\)