Rom

\(\text{let's use the navigational assignment of degrees to the clock, i.e. }\\ \text{12 is 0 degrees increasing as we move clockwise back to 360 degrees again at 12}\\ \phi_m = 360\dfrac{min}{60} = 360\dfrac{48}{60} = 360 \dfrac 4 5 = 72\cdot 4 = 288^\circ\\ \phi_h = 360 \dfrac{60hr + min}{720} = 360 \dfrac{60\cdot 6 + 48}{720} = \dfrac 1 2(360+48) = 204^\circ\\ |288^\circ - 204^\circ| = 84^\circ\)

\(\text{I assume this is one question and that both must be satisfied}\\ x^2 + (2a+1)x + a^2 \text{ has solutions of }\\ x = \dfrac{-(2a+1)\pm \sqrt{(2a+1)^2 -4a^2}}{2}\\ \text{in order for these to be integers}\\ (2a+1)^2 -a^2 \text{ must the be the perfect square of an odd number}\)

\((2a+1)^2 - 4a^2 = \\ 4a^2 + 4a+1-4a^2 = 4a+1 \\ \text{and this must be the perfect square of an odd number if }x \text{ is to be an integer}\)

\(\text{The values of }a \text{ that satisfy this are }\\ (2,6,12,20,30,42) \\ \text{and all of these lead to 2 unique integer solutions of the given quadratic}\)

You know how to do it but you want to see if anyone else does?

So when no one bothers with it you can bask in the illusion of your supremacy?

Here's an idea... p i s s off.

\(Carystax = (43000-11000)(0.2) +(45600-43000)(0.4) = 7440.00L\)

\(Megantax = (43000-11000)(0.2) + (150000-43000)(0.4) + (158900-150000)(0.45) = 53205.00L\)

sorry couldn't get pounds sterling sign to work

xxxxxx

\(235935623_{74} \pmod{15} = \\ 2 + 3(14) + 5+9(14) + 3+5(14)+6+2(14)+3 \pmod{15} = \\ 285 \pmod{15} = 0\\ \text{so }a=0 \)

\(1001_n = n^3+1 = (n+1)(n^2-n+1) \text{ is never prime}\)

\(\text{let (a,b,c,d) be the packs sorted from lightest to heaviest}\\ \text{the following four equations must hold}\\ a+b=34\\ a+c=37\\ b+d=43\\ c+d=46\)

\(\text{we then have a pair of equations which can be satisfied in two ways}\\ b+c=39 \text{ and } a+d=41\\ \text{or}\\ b+c=41 \text{ and }a+d = 39\)

\(\text{Solving the first set of 6 equations gets }\\ (a,b,c,d) = (16, 18, 21, 25)\\ \text{Solving the second set gets }\\ (a,b,c,d) = (15, 19, 22, 24)\\ \text{and thus 25 is the heaviest possible weight for pack d}\)

\(\text{I think I'd just use the quadratic formula and see what comes out }\\ r_{1,2} = \dfrac{19 \pm \sqrt{19^2 + 4\cdot 63\cdot 30}}{126} = \dfrac{19\pm 89}{126} \\ r_{1,2} = \dfrac 6 7,~-\dfrac{5}{9} \)

\(\text{so }63x^2-19x-30 = \\ 63 \left(x - \dfrac 6 7\right)\left(x + \dfrac 5 9\right) = \\ (7x-6)(9x+5)\)

what does 1001_n mean?