CPhill

2.  Area of rhombus =

2 * (1/2) * 6^2 * sin (120)   =   36 sqrt (3)/2 = 18 sqrt(3) units^2

3.       (A) A quadrilateral is a parallelogram if it is a rhombus.

The center is (0,0)

The since (8,2) is on the graph....the equation of the asymptote will have a  positive slope...so....

y = (b / a) x        and we know that  (x, y)  = (8, 2)....so....

2 = (b / a) (8)

2/8 = (b / a)

1/4 =  b / a        which implies that

4 /1  =  a / b   =   4

OK, YEEEEEET....I think the radius should be sqrt 225 = 15

Remember that if we draw a perpendicular to the chord from A....this perpendicular will bisect the chord.......call the point where this perpendicular intersects the chord, M

And we will have a right triangle   AMR

MR = 10

AR = 15 = the hypotenuse of AMR

The other leg, AM....is the distance we seek

So

AM =  sqrt [ (15 )^2 - 10^2 ]  =  sqrt [ 225 - 100] = sqrt (125) = 5sqrt(5)

IIf we have non-real roots, the discriminant must be < 0

In the form

ax^2 + bx + c = 0

The discriminant is   b^2 - 4ac

So  b = b     a = 1    and c = 9       ... so.....

b^2 - 4 (1) (9)   < 0

b^2 - 36 < 0

b^2 < 36

The interval that makes this true is

-6 < b < 6

d = kt^2

So

80 = k(4)^2

80 = 16k      divide both sides by 16

5 = k

So....in 8 more seconds....t = 12 and we have

d = 5 (12)^2

d = 5 (144)

d = 720

So....the didtance it falls in the next 8 seconds [ after the first 4] =  720 - 80 =  640 ft

We have this

h = kv^2     so we need to find  "k"

And we know that

8 = k (20)^2

8 = k (400)    divide both sides by 400

8/400 = k  =   1/50

So....when v = 35

h = (1/50) (35)^2   =  24.5 m

Correct!!!!

Correct!!!

Correct....!!!!

Correct, NSS....!!!!

Good job....and thanks, tertre....!!!!