ElectricPavlov

Upright:

What is your question?      Maybe.....

Perhaps this?  (sorry...sideways)

Use Quadratic Formula

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a =1   b = -5  c = -8    to find   x =   (5-sqrt57)/2      x = (5+sqrt57)/2

3    collect like terms and simplify

1    given

4    simplify

5    factor ----> equality

2    expand

Take LOG of both sides to obtain:  (I  assume this is  :   2^(x-1) = 4^(x/8)      )

(x-1) LOG 2  = x/8 (LOG 4)    and solve for x

(x-1)(8/x) = LOG4/LOG2

8-8/x =2

-8/x = -6

x= 8/6

Terry gets a 5 minute head start and covers 5 min x 60 m/min = 300 m

  then Tao starts at 1405       They have  2000 - 300 = 1700 meters to cover before they meet

     the distance is closing at (40+60)= 100 m/min

         1700 m  / (100 m/min) = 17 minutes after Tao starts   14:05 + 17 = 14:22    (02:22 PM)

                                                                                      (corrected...THANX!)

Hour hand moves 360 degrees in 12 hours  ....   30 degrees per hour or 15 degrees per 1/2 hour

   so 2:30  would be 5    1/2 hour segments   5 x 15 = 75 degrees= angle of hour hand.    The MINUTE hand will be at 180 degrees (pointing directly downwawrd)      180 degrees - 75 degrees = 105 degrees.

hmmmm

1 2 3 4 5 6 9 10 12 15 18 20 30 36 45 60 90   ?

v=0 when car stops

v=vo + at       vo= 20 m/s    a=-5 m/s^2

0=20 +(-5)t

t= 4 seconds

x=xo + vot + 1/2 at^2

x = 0 + 20(4) + 1/2 (-5)(4)^2

    =  80 + (-40)= 40 m       (as Chris found)