CPhill

Draw  a  perpendicular line  from the top left vertex of the parallelogram to the base.

So...we have a right triangle  with a leg  of 5.6 in and a hypotenuse of 10 in

The sine of the angle at the  bottom left  vertex  =   5.6 / 10

And since we have a parallelogram, the angle at the top right vertex  is the same, so its sine is the same

So...using the Law  of Sines we have

9.8/ sin (90)  =    h / (5.6 / 10 )

9.8  =  10h / 5.6   rearrange as

9.8 * 5.6  / 10   =   h   =  5.488 in

The center is ( -2, -1) and the radius is 7

So....using   (x - h)^2  + ( y - k)^2  = r^2  where   (h , k) is the center and r is  the radius, we have

( x - -2)^2  + ( y - - 1)^2  = 7^2

(x + 2) ^2  + (y + 1)^2  = 49         expand   and simplify

x^2  + 4x + 4  + y^2   + 2y  + 1  = 49

x^2 + y^2 +  4x   + 2y  +  5 = 49       subtract 49 from  both sides

x^2 + y^2  + 4x  + 2y  - 44  =  0      ⇒   Answer  "B"

The directional cosine  is

-3  / sqrt  [ 7^2  + 4^2 + (-3)^2  ] =

-3 / sqrt  [ 49 + 16 + 9  ] =

-3  / sqrt [ 74 ]

Julius...I believe the answer  is  "C"....a force  of 23 N  in the SouthEast direction will bring the system  into equilibrium

We can solve this :

20 cos (theta)  = 12

cos (theta)  = 12/20

cos (theta)  =  3/5

arccos (3/5)  =  53.13°  = theta

Rate  of change  =

[ Profit in year  4  - Profit  in year 2  ]  /   [ 4 - 2 ]  =

[217.2  - 238.4  ] /  2  =   -  10.6

Profits decreased by an average of $10,600  from year 2  to year 4

We have

cos (theta)  =  dot  product  /  [ product  of the vector  magnitudes ]

cos (theta)   = 8  /  [ 4 * 5  ]

cos (theta)  = 8 / 20  =   2/5

To find theta

arccos ( 2/5)  = theta  =   66.42°

The scale factor  from the smaller figure  to the larger  is  12/ 9 =  4/3

So...the area of the larger figure  is  = 

Area  of smaller figure *  (scale factor)^2  =

27 *  ( 4/3)^2  =

27 / 9  *  16  =

3 * 16  = 

48 cm^2

Let the first vector lie along the  x axis

Its components are :

5cos 0°   =  5

5 sin 0°   =  0

For the second vector we have

So..its components are

7 cos 50°     and

7 sin 50°

So....the sum of the components  is

5 + 7 cos 50°  =  5 + 7 cos 50°    and

0 + 7 sin 50°   =   7sin 50°

So...the magnitude of the resultant  vector  is  given by     sqrt  [ (  5 + 7 cos 50°)^2  +  (  7sin 50°)^2 ]   ≈  10.91 N

This is a hexagon with an apothem  of   5sqrt(3)  cm

The area is given by  :  6 * (1/2) (side length)^2  *  sqrt (3)  / 2

We can find the side length  as   2 *  5*sqrt(3)  / sqrt(3)  = 2 * 5  = 10 cm

So....the area is

6* (1/2)  10^2 * sqrt(3)  / 2  =

3 * 100 * sqrt(3)  / 2  =

150sqrt(3)  cm^2  ≈   259.8 cm^2