ElectricPavlov

Let's first find your velocity at 7 seconds   ....we will assume there is no air resistance (you will not reach terminal velocity)

v = vo + at       vo = 0   t = 7     v then equals  9.8(7) = 68.6 m/s

now how far will you fall in the next 5 seconds (I will assume you want 5 seconds)

x = xo + vo t + 1/2 at^2

   = 0   + 68.6(5) + 1/2(9.8)(5^2) = 465.5 m  ====> distance you will fall from end of 7th sec to end of 12th second

This answer assumes you mean  (5/6) x

5/6  x  = 8/5    Multiply both sides by  6/5

6/5 (5/6) x = (6/5)8/5    Simplify

x = 48/25  = 1 23/25

Not sure EXACTLY what you are asking.....but cos^2(x)  is just the cos (x)  value   squared

example   cos (30) = .866     cos^2 (30) =  .866^2 = .75

It means the variables x and y   'vary' similarly....if x goes more positive so does y.....if x goes negative so does y

Here ya go ...each line is one step

(10 (10^(3 * -1) )^-2

(10^1(10^(-3))^-2

(10^(1-3))^-2

(10^-2)^-2

10^(-2*-2) = 10^4

0/2 = 2   is incorrect

Suppose you multiply both sides of the equation by 2

then you would have

0(2) = 2(2)

0=4                 which is FALSE

0/2 = 0  is correct

I THINK what you are needing to know is how many moles is 1.5 x 10^23 atoms...

(1.50 x 10^23 atoms) /  (6.022 x 10^23 atoms/mole ) = .249 moles

= 243^1/5  *  x^(10*1/5)  * y^(20*1/5)

=3 x^2 y^4

Here is my stab at it:

1- Qc/Qh = 1 - 288/830 = .653    or  65.3 % effeicient

65.3/100 = 230000/x

x = 352214 j    input needed       (Note: slight change in final answer calculation)

1/4 x - 3/2 x = -1/2

1/4 x - 6/4 x = -1/2

- 5/4 x = - 1/2

x = -1/2(-4/5) = 4/10