CPhill

This is easily solved....just set each factor  =  0   .......so we have

x = 0                                 x + 6  = 0                                x + 9 = 0

                                 subtract 6 from both sides         subtract 9 from both sides

x = 0                                x = - 6                                    x = - 9

And those are the three solutions

Assuming a square pyramid, the base is easy....it's just  5^2   = 25 square units

We need to find the slant height.......this is given by :

√ [5^2 + 2.5^2]  = about  5.6 units

So.....the area of the sides  =

4 * (1/2) * (side length) * (slant height)

4 * (1/2)(5)(5.6)  =  56 square units

So....the total surface area  ≈ [ 25 + 56] square units ≈   81 square units

[43 + 5/12] * 11  =

43* (11)  +  (5/12) * 11   =

473   +  55/12   =

473 +  4 +  7/12   =

477 + 7/12  =

5731 / 12

I think this is supposed to be :

9 * 2x  + 3     ....if so, we have

18x  +  3     [multiplication takes precedence over addition going from right to left ]

In a finite amount of time.......she can never jump one foot forward.......she will always come up just a bit short.....however...if we bend the rules.......the distance she will jump in an infinite amount of time is given by the following infinite sum where a is the distance of the first jump [in feet]  and r is the common ratio beween each jump  = 1/2

distance = a / (1 - r)  =   (1/2) / (1 - 1/2)  =  (1/2) / (1/2)  = 1 ft  exactly !!!!

Very  nice, Landry.......!!!

Just type it in as " 2 + 2/8 "

Remember that the tangent is the relationship y/x......but, at 90°, x =0 and we can't divide by 0

[3 * 6 * 12 - 6] / 4   *  5   =

[18 *  12  - 6] / 4  * 5  =

[216 - 6 ] / 4  * 5 =

[210] / 4 * 5 =

[105/2] * 5  =

[525] / 2 =

262.5

Let us find the angle at Canton - angle NCE - of the triangle formed by  Canton, Elgin and Naples.....using the Law of Sines, we have

sin (angle NCE) / 500  = sin (46)/ 720

sin (angle NCE)  = 500sin(46)/ 720

arcsin (500 sin(46)/720) = angle NCE ≈ 30°

So....the angle at Elgin - angle CEN -   equals   [180 - 46 - 30] °  ≈ 104°

So....the bearing from Elgin to Canton is given by  [ 180 + (104 - 44)]°  ≈ 240°