CPhill

A= 0.15 (d - 7,550) + 755

So

3455 = .15(d - 7550) + 755   subtract 755 from both sides

2700 = .15(d - 7550)     divide both sides by .15

2700 / .15   = d - 7550    add 7550 to both sides

2700 / .15   + 7550  = d = $25550

3 - (k + 1) = -2      add (k + 1) to both sides

3 = -2 + (k + 1)     add 2 to both sides

5  = k + 1              subtract 1 from both sides

4  = k

Whatever n+1 and n-1 are, they are two integers apart....

So......if we divided (n+1)!  by (n-1)!  the result  = (n+1)* n

So we have

(n+1)*n = 90

n^2 + n = 90    subtract 90 from both sides

n^2 + n - 90  = 0  factor

(n -9) (n +10)  = 0      so, either n - 9 = 0   or n + 10 = 0

And, since were assuming a positive integer, n = 9

So

(9 +1) ! / (9 - 1) !  =

10! / 8!  =

10 * 9  =

90

We can set this up as two vectors, the first for the plane and the second for the wind...so we have

u = < 500cos310, 500sin310 >      v = < 45cos225, 45sin225 >

Adding the x components, we have 500cos310 + 45cos225 = 289.57

Adding the y components, we have 500sin310 + 45sin225 = -414.84

The resultant is given by √(289.57² + (-414.84)² ) = about 505.9 km'hr

The direction is given by tan^-1(-414.84/289.57) = about (- 55.08) degrees  = about 304.91 degrees

This makes sense......the wind is helping to increase the speed of the plane but it is also blowing it toward a more "southerly" direction.....

OK.....

Everyone has their own abilities......yours are probably different from another's........constantly comparing yourself to others is a losing game.....play to your strengths, work on your weknesses.....that's all I've got......

There are infinite possibilites for this....

The equation will be in the form

x^2 / 121  - y^2 / b^2   = 1        where "a" = 11 and a^2  = 121

I'm going to use the following function to model this

T = ab^x      where T is the temperature after some "x" minutes have elapsed

So we have, when x = 2

323 = ab²     →   323/b²  = a

And when x = 5   we have

288 = ab⁵  =  (323/b²)b⁵   = 323 b³     divide both sides by 323

288/323 = b³    take the 3rd root of each side

(288/323)^(1/3)  = b = about .9624

So, a = 323/(.9624)² =  about 348.66

So, our function is

T = 348.66(.9624)^x

So, to find out when they will be at 120 degrees, we have

120 = 348.66(.9624)^x        divide both sides by 348.66

120/348.66  =  (.9624)^x    take the log of each side

log ( 120/348.66 )  = log (.9624)^x     and by a log property, we have

log ( 120/348.66 )  =  x log (.9624)    divide both sides by log(.9624)

log ( 120/348.66 ) / log (.9624)= x = about 27.83 minutes

The distance is just 4 - (-2)  = 6

(a + 4)^2 - (a - 4)(a + 4)    expanding, we have

a^2 + 8a + 16 - [a^2 - 16]  =

8a + 32