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ty to the guy who answered!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Guest May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
edited by Guest  May 2, 2018
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8+0 Answers

 #1
avatar+86643 
+1

Here's a few

 

Task 2

 

y  = x^2       (quadratic)

y = x            (linear)

 

Sub  the   second equation into the first for y

 

x  = x^2       subtract x from  both sides

 

0  = x^2  - x      

x^2  - x   =  0   factor

x ( x - 1)  =0

Set each factor to 0  and solve for  x

 

x  = 0        x  - 1  =  0   

                add 1 to  both sides

 

x = 0          x  = 1

 

So....when x  = 0, y  = 0

And when x  = 1, y  = 1

So....the solutions are  (0,0)  and (1,1)

 

Here's a graph ;

 

https://www.desmos.com/calculator/ldpimov7kd

 

cool cool cool

CPhill  May 2, 2018
 #2
avatar
0

ty very much

Guest May 2, 2018
 #3
avatar+86643 
0

Task 3

 

y  =   (x + 3)

        ______

         (x - 2) 

 

The horizontal asymptote  occurs at  y  = 1

The vertcal asymptote occurs at  x  = 2

The x intercept is  at  x  = -3

The y intercept is  at   (-3/2)

 

Here's a  graph : 

 

https://www.desmos.com/calculator/4l3egciy58

 

cool cool cool

CPhill  May 2, 2018
 #4
avatar
+1

thanks again

Guest May 2, 2018
 #5
avatar+86643 
+1

Task 5

 

Let  the polynomial  be    x^2  - 4x  + 4

Let the  linear binomial be    ( x - 2)

 

 

Part 1

 

               x  -  2

x - 2   [   x^2  -  4x  + 4  ]

              x^2   -2x

             _____________

                       -2x   + 4

                       -2x   + 4

                       _______

                                 0

 

Part 2

 

f(a)   = f(2)  =   2^2  - 4(2)  + 4   =    4 - 8  + 4   =  0

 

Part 3 

 

Since  we obtained a "0"  with polynomial long division......then x  = 2   is  a  root. The Remainder Theorem says that if f(a)  = 0, then  "a" is a root.  This is borne out by the fact that f(2) = 0  where  "a"   = 2.

 

 

 

cool cool cool

CPhill  May 2, 2018
 #6
avatar+86643 
+1

Task 6

 

Let  N  = 37

We can write this as  (x + y)  =  (49-  12)

 

So....note that

 

(49 -  12) ^2  =    49^2  - 2(49*12) + 12^2  =   2401  - 1176  + 144    = 1369  =  372 

 

 

cool cool cool

CPhill  May 2, 2018
 #7
avatar
+1

helpful, thank you, you helped me learn a bit too :D

Guest May 2, 2018
 #8
avatar+86643 
+1

Task 4

 

x + a      =    b

____           ___

  ax               x

 

We can "engineer"  an extraneous solution, as follows :

 

Cross-multiply ;

 

x (x + a)  =  bax    simplify

 

x^2  + ax  = bax

 

x^2  + ax - bax  = 0              let  a  =  1, let  b  = 5....so we have

 

x^2  + 1x  - (5)(1)(x)  = 0

 

x^2 + 1x  - 5x  = 0

 

x^2  - 4x  = 0      factor

 

x(x  - 4)  = 0

 

Set both factors to 0   and solve for  x

 

x  = 0           x  - 4  =0

                      x  = 4

 

So.....note that the original problem becomes ;

 

x + 1            5

____   =      __

   1x              x

 

Note that if  x  = 4

 

4 + 1             5

_____   =    ___       which is true

   1*4             4     

 

 

 

However...if x  =  0   ....both denominators are undefined because we cannot divide  by 0....thus....x  = 0  is an extraneous solution....!!!!!

 

 

 

cool cool cool

CPhill  May 2, 2018

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