ElectricPavlov

The FIRST equation simpifies to:  4xy + 4y^2=0

solve the second for x= 6-y/2    and substitute into the FIRST

4y(6-y/2) + 4 y^2 =0

24y -2^y^2 +4y^2=0

2y^2 +24y = 0

y (2y +24)= 0      so y = 0   or  -12      substitute these values into the second equation to get x = 6   or  12    so x,y  =  (6,0)    or (12,-12)

COT (15) =   adjacent/ opposite   =   x(the horizontal distance) / 14500

14500 cot (15) = x = 54114.74 ft

ANYTHING (except '0')    raised to the 0 power is = 1    so the WHOLE NUMERATOR = 1

=

1/y^2

2 sqrt12  + 4 sqrt 75

=

2 sqrt(4*3)  + 4 sqrt (25*3)

=

2*2 sqrt3    + 4*5 sqrt3

4 sqrt3  +20 sqrt3

24 sqrt3

Take the derivative twice then substitute t=3 into the equation

x'  = -12sin(5t) + 4cos(5t)

x" = -12cos(5t) - 4sin(5t)

    = -12cos(15) -4sin(15)

    = -12.62  units/time^2       I think.

Guest is right....Thanx for the correction !

x' = - 60 sin (5t) + 20 cos(5t)

x" = - 300 cos (5t) - 100sin (5t)

when t = 3     then x" = -315.66

That's as reduced as it can go as a FRACTION.

Take the derivative twice then substitute t=3 into the equation

x'  = -12sin(5t) + 4cos(5t)

x" = -12cos(5t) - 4sin(5t)

    = -12cos(15) -4sin(15)

    = -12.62  units/time^2       I think.

10000000000000000000000000000000000000000000000000+my d**k = 10000000000000000000000000000000000000000000000000

Stupid.

Simplify it first       73.0 (9.8)  /(2*(38^2)) = sin (theta)

.247715 = sin(theta)

Arcsin (.247715) = theta = 14.34 degrees

 x = at^2-bt^3  

The first derivative will be the velocity  v=  2at - 3bt^2 

The second derivative will be the acceleration= 2a - 6bt    

   SInce t= 2 in this question,  Acceleration(@t=2) = 2a-12b