CPhill

2. If you throw a ball at 30 degrees into the air with a velocity of 18 m/s. How long is it in the air? 

We are using :

t  = 2 v₀ sin θ / g 

t =  2* 18m/s * sin (30)  / [9/8m /s^2]  =

36 * (1/2)  / 9.8    ≈  1.837 s

Here's a graph : https://www.desmos.com/calculator/2i436tqxnr

- [(+11.1)2 + (-6.1)2 + (-5.25) 2] ½  =

- [ 11.1 - 6.1 - 5.25 ]   =

- 11.1 + 6.1 + 5.25  =  .25   =  1/4

Unless this is supposed to be ;

- [ (11.1)^2 + (-6.1)^2 + (-5.25)^2] ^(1/2)  ≈ -13.7107

a + b  = 110

a - b  =    16       add the equations

2a  = 126      divide by 2

a  =  63    and

a - b  = 16

63 - b  = 16

63 - 16  = b   =   47

If the slope is  3.1,  this equals  3.1 / 1

So......everytime x changes by 1, y changes by 3.1

And x changes by 5  from  f(5) to f(10).....so....y changes by  5(3.1)  = 15.5

And adding that to its original value of 2 ....[ remember....(5.2)  is the original point ] we have

15.5 + 2  = 17.5

slope = change in y / change in x  =   7.8 / 2.4   =  3.25  =  3 + 1/4  = 13/4

25 *  $10.10  =

25 * $10   = $250    +    25 * .10  =  $2.50      =    $252.50

Sin^-1  and  Arcsin are the same thing.........the inverse sine

We have the point (5,2) on the graph

If the slope is  3.1,  f(10)  =  (10, 17.5)  

If the slope is meant to be 3:1, then f(10)  =  (10, 17)

A   =    A *  1

B                B

Just finished "Anne of Green Gables" not too long ago, hectictar.....very good!!!

"The Count of Monte Cristo" is also a fav of mine, TheMathGuy.......