(4/(3x-6)) feet / (8/(9x-18)) seconds = (4* (9x-18) / (8 (3x-6)) = (36x -72)/(24x-48) = (3x-6)/(2x-4) f/s = 3/2 f/s
**edited....misread Q ****
.65 x = 10 for each station
x = 10/.65 = 15.385 for EACH station
15.385 x 15 = ~~ 231 seeds
Well.....from the little dot in the lower left (the 'origin) count across 7 to the right .....then 6 up.....which point is there?
Multiply their moduli and add their arguments
8*4 (cos(20+60) isin (20+60) =.........
a .5 = 3-1 + 3-2+3-3+3-4+3-5+3-6 =.111111 (in base 3 )
b 0.111111111 (see a patern here....for the next question)
c my guess is it is .1111111
you will have 3(-5) + 2(1) -4(-5) = ........
Due to the leading negative on the t^2 term, this is a dome shaped parabola
Maximum will occu at t = -b/2a b = 8 a = -1
use this value of 't' in the equation to find the maximum value ....
1/60 +1/30 = 1/x
(1+2)/60 = 1/x
60/3 = x = 20 min
There is a trig identity tan (a+b) = (tan a + tanb )/(1- tan a tan b)
so this equals tan (20 + 10) = tan30 = sin 30 / cos 30 = 1/2 / sqrt3/2 = 1/sqrt3 = sqrt3/3
since y = 2x+4 substitute that into the second equation for 'y'
3 ( 2x+4) -4x= 6
6x+12-4x = 6
2x +12 = 6
2x = -6
x = -3 then use this value in one of th equations to find 'y''
