CPhill

Excellent, MathNerd(version2.0)......!!!!!

y=4 cos(2x+2pi)  write this as

y = 4 cos  [ 2 (x+ pi) ]

The  2  tells us how many periods are in 2pi rads....so each period is pi rads in length 

M = miles driven....and we have

18.95 + .17M =  40       subtract 8.95 from both sides

.17M  = 21.05         divide both sides by .17

M  = 21.05 / .17   =21.05/.17 = 123.82 ≈ 123 miles

Pretty easy.....just multiply every element by 3  =

12     -6

  3     12

You did fine, EMP.....you answered this :

[x^2-x-30] / [ x-6] = 8      [which is what I would have assumed ].....but....

Since he/she said no brackets....I assumed the literal form  of

x^2 - x -30/x - 6 = 8

Either way....they have a choice of two presentations.....LOL!!!!

Taken literally...we have

x^2-x-30/x-6 = 8

x^2 - x -30/x -14  = 0   multiply everything by x

x^3 - x^2 - 30 - 14x  = 0

x^3 - x^2 - 14x - 30  = 0

By the Rational Roots Theorem, 5 is a root....and we can use synthetic division to find the remaining polynomial / roots

5 [  1     - 1      -14       -30  ]

               5       20        30

      1       4         6          0

The remaining polynomial  is :

x^2 + 4x  + 6    ......this has no real roots......using the quad formula, the other two roots are :

-2 ± i√2

Thanks, heureka  !!!!.....that's pretty neat....I've never heard of this before and it's not taught in our schools in the US  [ at least not that I know of ]

I assume that the drawback is if the vertices of the polygon are not all grid points???

P.S. -  I look at the proof when I'm less sober than I am now......

Very well presented, EMP.....!!!!!!

A =  (1/2)s^2sin(60)  =   [ √3/4] s^2

dA/dt  =  dA/ds * ds/dt

dA /ds =  [√3/2]s

ds/dt  =  3in/s  .....so.......

dA / dt  when s = 15  = [√3/2]*(15in)* 3in/s  = (45/2)√3 in^2 / s  ≈  38.97 in^2

If the sides are increasing at 3in/s....the perimeter increases at 9in/s

Average  =   Sum of the numbers  / the number of them   =  144 / 4  =   36