ElectricPavlov

(1/9)^3 = 1/9 x 1/9 x 1/9 =  1 / 729

THIS IS WHAT HAPPENS WHEN QUESTIONS ARE WRITTEN UNCLEARLY OR WITH MISTAKES

OK....if you want them BOTH to receive the SAME amont when they EACH turn 21

Amount available when 17 y/o turns 21  = 10,000(1.02)^16 = 13727.86

    x = amount they BOTH will get

    Money after 17y/o gets first share = 13727.86-x

Money when the 12 y/o turns 21 is the same 'x ' and this equals : 

x = (13727.86-x)(1.02)^20    Solve for 'x'

x = (13727.86-x)(1.4859473)      then x = 8205.68          THey BOTH will get this amount when they EACH turn 21

OK.....done BOTH ways ....THIS is the end of the 'saga'

~THE END   hahahahahaha

Sorry , Guest....this WAS addressed in the Question AND the naswer EP gave.

"no vision that they are to get equal mounts when they reach age 21"

and

"..and the remaining will be distributed in 36 quarters and the amounts can be different?????     Well there is a whole bunch of possibilites depending on how it is split when the first kid turns 21 !  "

IF the question said they were to receive the same amounts when they turn 21 , that would be different...BUT IT DOESN'T SAY THAT !

Of course not.

We will have to assume that they are both EXACTLY 12 and 17  today (they both had birthdays today)

then we have 10,000 invested at 2% per quarter and that 1/2 of the amount will be distributed in 16 quarters and the remaining will be distributed in 36 quarters and the amounts can be different?????     Well there is a whole bunch of possibilites depending on how it is split when the first kid turns 21 !     

  Wouldn't it be better to split the accumulated amount EQUALLY when the first kid turns 21 and then let the remainder continue to grow until final disbursement to the second child when she/he turns 21 ????     We'll need some more info to proceed.....BUT here is the way I would do it

10000(1+.02)^16 = 13727.86    when the 17 y/o turns 21......give half of the money to them  $6863.93

6863.93   will continue to grow for 20 more quarters before being disbursed to the younger child

6863.93(1.02)^20 = 10199.44   to the younger child when she/he turns 21    

This MAY seem unfair, but it is NOT....the FIRST child could leave the money deposited and get the same amount if she/he waits until the younger turns 21 to receive the money.

Egads !  I do not know that I could figure this one out without a little more info......but I will try....lets put the 250 pound force on the x-axis

Forces vertical =  200 sin x      &   150 sin y      Since the  250 pound will not exert any vertical force because it is strictly horizontal these two must be equal and opposite THEN   200 sin x = - 150 sin y    ... .(EQUATION 1)

Horizontal forces   200 cos x   &  150 cos y    and 250    I wil have to assume that the two smaller horizontal forces are balancing the larger 250 force

 so   200 cos x + 150 cos y = 250    ... . (EQUATION 2)

We now have two equations with two unknowns and should be able to solve for the angles x and y.....

I'm missing some trig rule to figure out these two equations, BUT by successive trial and elimination (i.e interprolation)  I come up with (approximately)  

250 horizontal to the right

200 @ 143.77 degrees

150 @  232     degrees     Sorry.....I got STUCK !

Basically , if you solve the equation for 'y' it will be in the slope intercept form:  y = mx + b

9x+8y= -72

8y = -9x -72

y = -9/8 x - 9        m = -9/8 = slope   -9 = b = y intercept

Force perpindicular is found by the cosin of 15 degrees

cos15   = adjacent/hypotenuse   = adj/225    Where adj is the perpindicular force (f2)

225 cos 15 = adj = 217.333 = f2

You can use pythagorean theorem to find the friction force or use the sin function....let's use the sin function to find the answer and check it with the Pythag th.

sin 15 = opposite/hypotenuse = opp/225

225 sin 15 = opposite       where opposite is the force of friction (f1) being exerted to keep the box from sliding.

f1 = 58.234 

Use Pythagorean Theorem to check:    58.234^2 + 217.333^2 = 225^2        ????      Yep !