ElectricPavlov

(7G -6)(9G-5)

Yup..... that is what I 'meant'......oops !  Guess I'm not too good at this math stuff 

'x' distance       2/3 x ( 15-(-6) ) = 14        14 + -6 = 8 = x

'y' distance       2/3 x (-5-19)  =  -16        19 - 16 = 3 = y

F=   P (1=i)^n     Where F = future value  P = initial value (50000)   i= periodic interest   n = number of periods

Jose :   F = 50, 000 (1+ .04)^2  = 54,080

Patricia  i = periodic interest where the period is 3 mos (4 times a year)  .04/4 = .01   and periods = 2 yrs x 4/yr = 8

             F= 50,000 (1 +.01)^8 = 54142.84

Difference =62.84   to the nearest dollar , Patricia earned $64 more than Jose.

Slope = rise/run  =   y1-y2  /  x1-x2

-3/2 = (y-7)  / (4-(-2))

-3/2 = (y-7) / 6

6 (-3/2) = y-7

-9 = y-7

y= -2

Thanx, Melody !    

I wanted to see how Chris got the EXACT answer with a radical ....I got an equivalent numerical answer using the Law of Cosines (twice)   See handwritten diagram:

If you have the correct answer for stress (I got 1.83838 x 10^8 P )

 then, (I had to look this one up !)

Young's Modulus =  Stress / (\( \Delta \)L/ Lo)         \(\Delta \)L = change in length = .0028 m     Lo=1.5 m   (given)

FIRST find the VOLUME of the wire:

pi x r^2 x l = 5.45 x10^-7 m^3

THEN find its diameter given the NEW length of 1.5028 M

pi x r^2 x l = 5.45 x 10^-7    results in r =3.3968 x10^-4 m

then, as before   Stress = F/A      F= ma  = 6.8(9.8) = 66.64 newtons

A= area = pi x r^2    where r is the 'new ' stretched r   

     Let me know what you come up with for an answer and we will continue.....   ~EP

Yes.....because this is a VERY thin cable supporting a large force, so the stress is quite high.   Good job!