CPhill

x^2 + 2x + 5  = 0

Make sure we are using the correct values  (and formula)

a = 1   b  = 2   and c  = 5

So we have

[ - b ±√ ( b^2 - 4ac) ]

________________      =

        2a 

[ -2 ±√ (2^2 - 4*1 * 5) ]

_________________        =

            2*1

 - 2 ±√  [4 - 20 ] 

_____________       =

          2

-2 ±√ (-16)

_________       =

       2

-2 ± 4i

______      divide everything by 2

    2

-1 ± 2i

Let's  check this in the quadratic  ...one root is 2i - 1

(2i - 1)^2  + 2 (2i - 1) + 5  =

(2i - 1) (2i - 1) + 2 (2i - 1) + 5

4i^2 - 4i + 1 + 4i - 2 + 5

- 4 + 1 - 2 + 5 =

-3 + 3  =   

0

We don't need to check the other root....if one complex root makes a quadratic = 0, then its conjugate will, too !!!

The distance traveled by the plane forms the hypotenuse of a right triangle

This equals 300 * 10  = 3000ft

The altitude , A,  can be found as

sin (5°)  = A / 3000    multiply both sides by 3000

3000* sin(5°)  = A ≈  261.47 ft

And call the distance [ horizontally ] from the takeoff point , D

So  we have

cos ( 5°) = D / 3000

3000cos(5°) = D ≈  2988. ft

         A

         6

B  8   K     6       C

Since AK is an altitude....we can use the Pythagorean Theorem here

AB  = √ [ BK^2 + AK^2 ]  =  √ [ 8^2 + 6^2]  = √100 =  10

And  AC  = √ AK^2 + KC^2 ]   =  √ [ 6^2 + 6^2] = √ [ 2 * 6^2 ]  = 6√2

So....the perimeter is

AB + BK + KC + AC   = 

10 + 8 + 6 + 6√2  =

24 + 6√2   units  ≈  32.49 units

The triangle on the lft is a 30-60-90 right triangle

The altitude   = 8 *  √3/2  =  4√3  = √48

Since the triangle on the right is a 45 - 45 - 90 right triangle

The altitude  = the base  = √48 =  length of the legs

So..by the Pythagorean Theorem

x  = 

√ [ ( √48 )^2 + ( √48)^2 ]  = 

√96  =

√16 * √6  =

4√ 6

LOL!!....when you think about it....most of us are all just "Partial Mathematicians"  ...!!!!

f(x)   =  ax + b    for f(x), write y

y = ax + b      subtract b rom both sides

[y - b]  =  ax   divide both sides by a

[ y - b] / a  =  x      "swap' x and y

[x - b] / a  = y  

(1/a)x - b/a  = y  =   f^-1(x)

So g(x)  = f^-1(x) - 3  =   (1/a)x - b/a - 3  =  5x - 4

This implies that

(1/a) = 5       and    -b/a - 3  = 4

So

a = (1/5)      and 

-b / (1/5) - 3  =  -4     add 3 to both sides

-b / (1/5)  =  -1

-b =  -1 (1/5)

-b = -1/5

b = 1/5

So

5a + 5b  =

5(1/5) + 5(1/5)  =    1 + 1   =   2

What scatterplot did you generate???

We have a horizontal line.....therefore....the slope   = 0

This can be confirmed by the "0"  in the linear function

y = 0r + 40

If r = 0.....there is no correlation

A

10             M                                      cos ABM

B        24             C

Since BM is an angle bisector of  right angle ABC....then angle ABM  = (1/2)(90°) = 45°

So....the cos ABM  = cos (45°)   =  1 / √2