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x           x^2 - 6x + 8          x - 2

0                 8                      -2

1                 3                      -1

2                 0                       0 

So      f(x) = g(x)    at    (2 , 0)

Thank you very much! This helped me understand it easier than how my professor explained it. 

Yes! thank you very much

Let's look at (d) for another example

(f ° g ) (3)

We put 3 into g  and get g(3) = 2

Then...we put this result into f and get f(2)   = 16

The notation is probably what is giving you trouble

( f ° g) (0)    means this   ⇒   f ( g(0) )

We first evaluate  g(0)   = 4

So...now we have

f ( g(0) ) =   f(4) = -6

Another way to see this is to work from right to left

So  put 0 into g  and we get  g(0) =  4

Then...put this into f  and we have f(4)   = -6

Does that make sense ???

\((f\circ g)(x) = f(g(x))\\ -\\ \begin{matrix} x &g(x) &f(g(x)) \\ 0 &4 &-6\\ 1 &3 &13 \\ \vdots \end{matrix}\\ \text{see how it works?}\)

Never heard of this but....the law says that : In the short run, the unemployment rate decreases about 1 percentage point for every 3 percent increase in real GDP

So....it seems that if the unemployment rate decreases by 4 %.....then the GDP should rise by 4(3%) = 12%

So....I suppose  that "C"  comes closest

3.2 / 74  ≈ .0432  ≈   4.32% unemployment

2. 5 / 74 ≈  .0337 ≈   3.37% unemployment

So

4.32% - 3.37% ≈  0.95%   =  1%  (rounded)

So.... it decreased about 1%  ⇒   "B"

82 + x

_____    ≥     90             multiply both sides by 2

    2

82 + x  ≥  180            subtract 82 from both sides

x ≥ 98

She must make at least a "98"

There are two scores to average to 90 %

(82 + x )/2 = 90

82+x = 180

x = 98             She has to make at least 98 %  on the second test.

(red  edited)

We have the form

ax^3 + bx^2 + cx + d

If f(0) = 0.....then d = 0       and we can solve this system

a(-1)^3 + b(-1)^2 + c(-1)  =  15

a(1) ^3 + b(1)^2 + c(1) = - 5

a(2)^3 + b(2)^2 + c(2) = 12             simplify

-a + b - c   =     15            (1)

a + b + c  =     -5             (2)

8a + 4b + 2c = 12       ⇒ 4a + 2b + c = 6      (3)

Add (1) and (2)

2b = 10

b = 5

Add (1) and (3)

3a + 3b = 21

a + b = 7

 a + 5 = 7

a = 2

And using (1) to find c  we have

-2 + 5 - c = 15

3 - c = 15

3 - 15 = c

-12 = c

So...the polynomial is

2x^3 + 5x^2 - 12x 

To find the x intercepts (roots) we have

2x^3 + 5x^2 - 12x = 0         factor

x (2x^2 + 5x^2 - 12) = 0

x ( 2x  - 3) (x + 4) =  0

Setting each factor to 0  and solving for x produces the x intercepts of

x = 0       x = 3/2       and      x =  -4

\(f(x) = a x^3 + b x^2 + c x + d\\ f(0)= 0 \Rightarrow d = 0\\ \text{using the rest of the values we obtain 3 equations}\\ a + b + c =-5\\ -a + b -c =15\\ 8a+4b + 2c = 12\)

\(\text{adding the first two}\\ 2b = 10,~b = 5\)

\(a+c=-10\\ 8a+2c = -8\\ \text{subtract twice eq 1 from eq 2}\\ 6a = 12,~a=2\\ c= -12\)

\(f(x) = 2x^3 + 5x^2 -12x\)

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\(\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)} = \\ \dfrac{4x^2+6x}{2x^2+13x+15}\)

b )    2x^4 - 50  =  0        divide through by 2

x^4 - 25 =  0           we can factor this as

(x^2 - 5) (x^2 + 5) = 0

This implies that

x^2 - 5 = 0

x^2 = 5      take both roots

x = ±√5

Also

x^2 + 5 =  0

x^2 = -5      take both roots

x = ±√-5

x = ± i√ 5

And those are the four solutions

a)  x^4 - x^2 - 20 =  0

We can actually factor this one in a straightforward manner thusly

(x^2 - 5) ( x^2 + 4) = 0

This means that

x^2 - 5 = 0     add 5 to both sides             

x^2 = 5   take both roots

x = ±√5

Also

x^2 + 4 = 0

x^2 = -4        take both roots

x = ±√[ -4] 

x = ± 2i

And those are the four solutions.....

Okay Then....

x^3 - 4x^2 + 2x - 5 and x^2 + tx - 7

So  we have

(x^3 - 4x^2 + 2x - 5)  (x^2 + tx - 7 )   =

    x^5 - 4x^4 + 2x^3 - 5x^2

          + tx^4 - 4tx^3 +2tx^2  - 5tx

                    -7x^3  + 28x^2  - 14x  + 35

If we have no x^2 term   this means  that

( - 5 + 2t + 28)x^2   =   0x^2

Which implies that

-5 + 2t + 28 = 0

23 + 2t = 0

2t = -23        divide both sides by  2

t = -23/2

Remember that the sum of the probabilities = 1

So we have

n + 1/3 + 1/8 + 1/6 =  1

n + 8/24 + 3/24 + 4/24 =  1

n +  [ 8 +  3 + 4 ] / 24 = 1

n + [ 15/24] =  1

 n +   5/8  =   8/8        subtract   5/8 from both sides

n =  3/8

Not too bad , GM

1)     ( x - 3) ( x -  -5)  =

       (x - 3) ( x + 5) =

        x^2 -3x + 5x - 15 =

       x^2 + 2x  - 15

2 )   ( x - 0)  ( x - 1) ( x - 8)  =

          x ( x - 1) (x - 8) =

         x [ x^2 - 1x - 8x + 8 ]  =

         x [ x^2 - 9x + 8 ] =

          x^3 - 9x^2 + 8x

The answer is 1,369.86. I did this by caculator.

To expand on EP's answer

If   a + bi    is a root of a polynomial......then so is  a - bi

These are known as "conjugate pairs"  

Thus....if we have complex roots of a polynomial, they will occur in pairs

So   the conjugate of  0 + 2i    is    0 - 2i

Taking EP's answer from here

9x-22    +  34  +  5x-7   +  61 = 360    simplify

14x + 66 = 360          subtract 66 from both sides

14x = 294

x = 21 

So

GK = 9(21) - 22  =  167°

HJ = 5(21) - 7 =  98°

HGJ = 51 + 167 + 34 = 252°

GKJ = 167 + 34 = 201°

To have real coeeficients there has to be a    -  i    root also

Quadratic formula:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)      

     shows that to have an i root the discriminant has to be NEGATIVE...and the disriminant is a + or - term....so there will be ANOTHER -2i root to go with the +2 i root

That's what I did but my answer didn't seem right