CPhill

-x^2  + 5x  =  3       rearrange as

x^2  - 5x  +  3   =  0

a  = 1    b  = -5    c  = 3

So....the solutions are

-(-5) ± √  [  (-5)^2  -  4(1)(3)  ]

_______________________

           2(1)

5  ± √[ 25  -  12  ]

______________

           2

5/2   ± √13/2

(-5i)(-3i)  =

15i^2  =

15(-1)  =

-15

17

x  +  y   =  4  ⇒   y  =  4 - x     (1)

x^2  + y^2  =  16       (2)

Sub   (1)  into (2)  and we have that

x^2  + (4 - x)^2   =  16

x^2  +  (x - 4)^2  = 16

x^2  + x^2  - 8x  +  16  =  16      subtract 16 from both sides

2x^2  - 8x  = 0     factor

2x ( x   - 4)  = 0

Setting each factor to 0  we have that x  = 0   and x  = 4

And when x  = 0   , y  = 4 - 0  = 4

And when x = 4, y  = 4 - 4  = 0

So.....the solutions are    (0,4)  and (4,0)

18.....easy,  NSS

Just change the signs  on the constants in the parentheses and we get

3, -2 , 2

If   5 + √3   is a root then so is  5 - √3

There is no x^2 term......and the sum of the roots  = 0/a  =  0

So

 (5 + √3)  + (5 - √3)  + R  = 0   where R is the remaining root

10  + R  =  0    ⇒  R   = -10

And  ....let  R1, R2  and R3   represent the roots...and we have that

R1*R2   +  R1*R3  + R2*R3   =   b  / a   =  b   .....so.....

-10(5 + √3) + -10(  5 - √3)  +   (5 + √3) ( 5 - √3)  =  b

-50 + -50  + 22   =  b   ⇒   -78

And the product of the roots  =  -c   ....so....

-10(22)  =  -c

220  =  c

So   ...b +  c  =  -78   +  220   =  142

y = (x - 2)^2  + 3

If this is rotated 180° about its vertex, the vertex does not change..........the new function is

y  =  - (x- 2)^2  + 3

Shifting this to the left 3 units left and 2 units down  results in the function

y  =  - (x + 1)^2   +  1

To find the zeroes, we have

-(x + 1)^2  +  1  =  0

-(x +1)^2  =  -1

(x + 1)^2  =  1

x + 1  =  ±1

So

x +  1  =  1 ⇒  x  = 0

x + 1  =  - 1  ⇒  x  = -2

So    a  +  b  =    0   +   -2    =    - 2

Here's the graph :  https://www.desmos.com/calculator/zbkwtqadyr

A)

C  =  .65p

For 2 lbs., we have

C  =  .65 (2)  =  $1.30

B)    C  = .60p

If we add  the coefficients  and the constant term and get 0,  then 1 is a root

So....this is true

Using synthetic division, we can reduce the polynomial......

1   [   1      3       3       - 3      -  4  ]

                 1       4        7          4

      ________________________

        1       4      7         4         0

So......the remaining polynomial is

x^3  +  4x^2  +  7x  +  4

Note  that  if we sub  - 1  into  this,  we get  0

So......the rational  roots are    1  and -1

We have this :

H  = kv^2         and we need to find the proportionality constant, k.....so we have

30  =  k (30)^2

30  =  k (900)     divide both sides by 900

(1/30)  =  k

So....at 20m/s, H  =

(1/30)*(20)^2  = 400/30  = 40/3  ≈  13.33 m

When  H  = 15m,  v  =

15  =  (1/30) v^2      multiply both sides by 30

450  =  v^2         take the positive root of both sides

sqrt (450)  =  v  =  3sqrt(50)  ≈  21.2 m/s

21.   Rearrange as

x^4 + 27x^2 - 324  =  0       factor as

(x^2  -  9)  ( x^2  + 36)  = 0

(x + 3) ( x - 3)  ( x + 6i) (x - 6i)  = 0

Set each factor to 0 and solve for x and we get the solutions  x = -3, 3, - 6i, 6i

22.  ( s +  3v)^5   =

s^5   +  5*s^4*(3v)  + 10*s^3*(3v)^2  +  10*s^2*(3v)^3  + 5*s*(3v)^4  + (3v)^5  =

s^5  + 15s^4v + 90s^3v^2  + 270s^2v^3 + 405sv^4  + 243v^5

Thanks, alan  !!!!

The first is still the best.....but not very good...!!!