CPhill

-4.9t^2  + 14t - .04  =  4

-4.9t^2  + 14t  -.04 - 4  = 0

-4.9t^2 + 14t - 4.4  = 0           multiply through by -10

49t^2  - 140t + 44  = 0

Solving this with the quadratic formula  we get

t ≈ .36 sec

and

t ≈ 2.5 sec

It is above 4 m  for  about      2.5 - .36  =  2.14 seconds =  214/100 =   107/50 sec

Vertex form

y = a ( x - h)^2  + k        where ( h, k)  = the vertex   and (x,y)  is another point on the graph

7 = a ( 5 - 4)^2  + 0

7  = a(1)^2

7  = a

Slope   =  ( -12 - 7) /( 3 - - 1)  = -19/ 4

So  using the slope and (3, -12)

y = (-19/4) ( -6 - 3) - 12

y = ( 171) / 4  - 12

y = (171)/4  - 48/4

y = 123/4

23 * 158 =

3634 =

16  =

7

80mi / 65mi/hr =   16/13 hrs

(16 /13) * 60 min  ≈  73.85 min ≈  74 min =   1hr 14 min after 10:32 AM  =

11:46 AM

F     8     D

      15

A             I

Min possible distance =  sqrt [ 15^2 - 8^2 ]  = sqrt[ 225 - 64 ] = sqrt [ 161] ≈ 12.69 m

2x + 5     must   = 1    .....so

2x = 1 - 5

2x = -4

x = - 2  .... so

3(-2) + 9 = 

3

Let the length = x  and the width x - 60

perimeter =  2x + 2 (x - 60)  = 500

2x + 2x - 120  = 500

4x  = 500 + 120

4x   = 620

x =  620 / 4  =   155

x - 60  = 155 - 60  =   95

Area = 155 * 95  =  14725 ft^2

345,600  =

3456 x 100  =

576 x 6 x 100 =

24^2 x 2 x 3 x 10^2 =

6^2 x 4^2 x 2 x 3 x 2^2 x 5^2  =

2^2 x 3^2 x 2^4  x 2 x 3 x 2^2 x 5^2=

2^9 * 3^3 * 5^2

abc  = 9 * 3 * 2  =   54

arctan x + arctan y =   pi / 4

This will be true  when either

x  = 0   and y = 1

or

x = 1  and  y = 0

So

xy + x + y =   1