CPhill

a)  The internal surface  area will just be  the  area  of  the original rectangle  less the area  of  the 4 squares cut away

So we have

(1200)(1000)  -  4 ( 300)^2   =

1200000 - 4 ( 90000)  =

120000 - 360000 =

840,000 mm^2

b )  When the squares are cut out, the height of the box will be  300mm

The width  will  be  (1200 - 2(300) )  =  ( 600)mm

The length will be  (1000 - 2(300))  -  (400)mm

So  the  volume will be

(300 * 600 * 400)  =  72,000,000  mm^3

f(8)  means that we just sub 8 in for x  and evaluate

3* sqrt  [ 2(8)  - 7 ]   - 8   =

3* sqrt [ 16 - 7 ]   -  8

3* sqrt [ 9 ]  -  8

3 *( 3)  -  8  =

9  -  8   =

1

So f(8)   =  1

Not too tricky,huh  ????

The time it takes to reach  the crossbar  is given by the horizontal  component of motion

42 yds =  126 ft

126  = 70 cos (50°) * t

126 / [ 70 cos (50°) ]  =  t =   2.8  sec

The height at any time  is

h  = -16t^2   + 70sin (50°)t   + initial height

At t = 2.8  we have 

h = -16(2.8)^2  + 70 sin (50°)(2.8)  +  0 

h = 24.7 ft

The ball clears the crossbar by about  24.7 - 10  =  14.7 ft  =   15 ft

1 mm  =  .001m

The  volume  of water  =

3.6m * 11.2m * 3(.001)m   =   

[ 3.6  * 11.2  * .003] m^3  =  .12096 m^3

(.12096 ) (1000L)   =   120.96 L

Here you  are

The resultant  vector  = A + B  = (-2, - 1)

THX, hectictar  and geno.....I learned something new today    !!!!!

Very nice, X^2    !!!!!!

sin ( arcsin (1/2)  + arrcos(2/3) )

arcsin (1/2)  = A

So

sin A  = 1/2      and cos A  =  sqrt ( 2^2 - 1^2) / 2 =  sqrt (3) / 2

arccos (2/3)  = B

So

cos B  = (2/3)    and sin B =   sqrt  ( 3^2 - 2^2) / 3  =  sqrt (5) /3

sin (A + B)   =   

sin Acos B  +  sin B cos A   =

(1/2)(2/3)  + (sqrt (5) / 3) (sqrt (3) / 2)   =

[ 2 +  sqrt (15)  ]  /  6

x = 4 - t   ⇒   t  =  4 - x        (1)

y = (t - 1)^2       (2)

Put  (1)  into (2)

y  = ( 4 - x - 1)^2

y = ( 3 - x)^2

y = ( x - 3)^2

y = a ( x -3)^2  +  0

In the  form   y  = a(x - h)^2  + k       

a = 1     and the  vertex  is  (h,k)  = (3,0)

The initial point is  when t = -2

x  = (4 - -2)   = 6         y =  (-2 - 1)^2  = (-3)^2    = 9  ⇒   (6,9)

When t  = 2

x = (4 - 2) =  2          y  =  (2 - 1)^2   = 1^2   =  1    ⇒  (2, 1)

Since  (6.9)  is to  the right of (2,1)   the motion is  right to left

[x + y]  / 2   = √2 * √(xy)        square both sides

[ x^2 + 2xy + y^2 ] / 4  = 2 (xy)

x^2  + 2xy  + y^2  =  8 (xy)

x^2  + y^2  =  6(xy)      divide through by   x^2

1  + (y^2) / (x^2)  =  6(y/x)       rearrange  as

(y^2/x^2)  - 6(y/x)  +  1  =   0

(y/x)^2  - 6(y/x) + 1  =  0

Let   (y/x)   =  a

a^2   - 6a  + 1    =  0

a^2  - 6a    =   -1       complete the square on a

a^2 -6a + 9  = -1 + 9

(a - 3)^2   = 8       take both roots

a  -3   =  ±√ 8

a - 3  =  ±2√2

a =  3 ± 2√2

(y/x)  =    3 ± 2√2