CPhill

Thanks, heureka.....that WAS a tough one   !!!!

2z  + i   =  iz +  3

2 (a + bi)  + i   =  i ( a + bi) + 3

2a + 2bi + 1i  =  ai - b + 3

Equating real and complex parts

2a  = 3 - b     ⇒    2a + b  = 3    ⇒  4a + 2b  = 6    (1)

(2b + 1)  =  a ⇒     a - 2b =  1   ⇒    a  - 2b  =  1      (2)

Add (1)  and (2)  and we get that

5a = 7

a = 7/5

And

2(7/5) + b  = 3

14/5 + b = 15/5

b = 15/5 - 14/5

b = 1/5

So

z = a + bi

z  = (7/5)  + (1/5) i

The original time  is   D / R  = 200/25  = 8 hrs

So.....he Total Distance is given by 200 + 100T     

Where T is the additional time driving at 100 km / hr

So    Total Distance / Total Time = 40

[200 + 100T ]

___________  =   40      cross-multiply

   8  +  T

200 + 100T  = 40 [ 8 + T]    simplify

200 + 100T  = 320 + 40T

60T  =  120      divide both sides by  60

T  =120 /60   =  2 additional hrs of driving at 100 km.hr  = (2) 100  = 200 additional km 

A. Determine the position and velocity functions for the book.

B. Determine the average velocity of the book on the interval [0, 1].

C. Find the instantaneous velocities when t = 0 and t = 1.

D. At what time is the instantaneous velocity of the coin equal to the average velocity of the book found in part B?

E. What is the name of the theorem that says there must be at least one solution to part D?

F. Find the velocity of the book just before it hits the ground.

A.  

The position function is   :  s(t)  =     -16t^2 -10t + 378 

The velocity function isthe derivative of this = s'(t)  =  -32t - 10

B.

The  average velocity on [ 0, 1]  =

[ (-32(1) - 10 )  - (-32(0) - 10 ]            -32

_______________________ =      ______  =    -32 ft/s

           1  - 0                                         1

C. When t  = 0,  the instantaneous velocity  is  -32(0) - 10  =  -10ft/s    [which we would expect ]

     When t = 1, the instantaneous velocity is  -32(1) - 10  =  - 42ft/s

D. I am taking "coin" to mean "book"

We want to know when 

-32  =  - 32t - 10      add 10 to both sides

-22 = -32 t      divide both sides by -32

-22/-32  = t  =  11/16 sec = .6875  sec

E.  Average Value Theorem  [ or, Mean Value Theorem ]

F.  We need to solve this to find the  time it takes to hit the ground

-16t^2 - 10t + 378  = 0

16t^2 + 10t - 378  =  0        divide through by 2

8t^2 + 5t - 189  = 0

Using the quad formula

t =   -5 ±√ [ 5^2 - 4*8*189 ]            -5 ±√ [ 25 + 6048]

      ___________________  =   _________________  ≈  4.558 sec  or  - 5.183 sec

              2 * 8                                         16

Take the positive time  and the instantaneous velocity  when the book hits the ground is :

-32(4.558) - 10   ≈  -155.86 ft/s

Let x +2  = a

Let y + 2  = b

So   we have that

1/a  +  1/b  = 1/3

(a + b) / ab  = 1/3

3(a + b)  =  ab

3a + 3b  = ab

ab - 3b  =  3a

b (a - 3)  = 3a

b  =      3a

          ______

           (a - 3)

a       b

4      12

6       6

12     4

These are the only integer solutions for  a, b

When 

a = 4, x = 2    and  when b = 12, y = 10    so    x + 2y  = 22

a = 6, x = 4    and  when b = 6, y = 4    so  x + 2y =  12

a = 12, x =10  and when b = 4, y = 2   so  x + 2y = 18

So....the minimum is when  x = y = 4  and the minimum of x + 2y  is  12   [ as the guest found !!! ]

We have that  the sides are

2 , 2r , 2r^2          so.....by the Pythagorean Theorem

2^2  + (2r)^2  = (2r^2)^2

4  + 4r^2  = 4r^4         divide through by 4

1  + r^2  = r^4        rearrange

r^4 - r^2  - 1  =   0             let  r^2   = m    and we have

m^2 - m - 1  = 0            complete the square on m

m^2 - m + 1/4  = 1 + 1/4

(m - 1/2)^2 = 5/4           take the positive root

m - 1/2  = √5 / 2

m = [ 1 + √5 ] / 2 =  r^2      ⇒  this is  known as  the irrational number  "Phi"

So  

m = r^2 = Phi

√m  = r  =  √Phi

So.....the side lengths are

2 ,  2 √Phi , 2Phi

So....the hypotenuse is    2Phi  units in length

Proof :

2^2  + (2√Phi)^2 = (2Phi)^2

4 + 4Phi  =  4Phi^2        divide through by 4

1 + Phi   = Phi^2     which is an identity

We have this

In the first mixture, we have  (1/3) litre of apple juice  and  (2/3) litre of water

In the second mixture, we have (1/4) litre of apple juice and  ( 3/4) litre of water

When we combine these  we get

(1/3 + 1/4)  =  7/12 litres  apple  juice     and

(2/3 + 3/4)  = 17/12  litres of water

So....the ratio is

(7/12) /(17/12) =   7 / 17   =   7 :  17     ratio of apple juice to water

y  =  variable cost  +  fixed cost

The variable cost is the price for  x sweatshirts =  7.50x

The fixed cost is the shipping cost  = 4.99

So

y =  7.50x + 4.99

Maybe some other way to do this.....but....see  the following image :

We have a circle with a radius of 13   [ any radius would actually be Ok  ]

Let  a  have the endpoints  (13,0)   and  (13√3/2, 13/2)

So the length of  a  is √ [ ( 13√3/2 - 13)^2 + (13/2)^2 [   =  13/2 ( √6 - √2)

Let b have the endoints ( 13√3/2, 13/2)  and ( -13/2, 13√3/2)

So  the length of b is √ [  (-13/2 - 13√3/2)^2 + (13/2 - 13√3/2)^2 ]   = 13√ 2

So

a  +  b  =  (13/2) [ √2 + √6  ]  =  c =  6.5 [ √2 + √6 ]

So.......using the Law of  Cosines

( 6.5 [ √2 +√6] )^2  =  13^2 + 13^2 - 2 (13)(13) cos (degree of arc subtended by c )

Call the degree of arc subtended by  c  = angle A

So

( 6.5 [ √2 +√6])^2   =  338 - 338 cos (A)

[ ( 6.5 [ √2 +√6])^2 - 338 ]/ -338  = cos A

Using the arccos to find  A we have

arccos ( [ ( 6.5 [ √2 +√6])^2 - 338 ]/ -338 )  =  A  =  150°

So..... chord c subtends an arc of 150°