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What is (f−g)(x)?

Rate  of change  =

f(9)  - f(5)               14  -  (-2)              16

_______      =      _________   =    _______     =   4

  9   -  5                  9  -     5                  4

(f  / g ) (x)  =

f(x)  / g(x)  =

[ 4x^2 + 6x ]  /  [ 2x^2 + 13x + 15 ]

First one.......this function is symmetric with the origin....so....odd

Second one  .... this function is not symmetric to the origin or the y axis....so neither odd nor even

Third one....this function is symmetric to the y axis.....so....even

They want you  to  find the x values that make

x^2 - 6x  + 8  =    x - 2

Pick  a few values for x

                   f(x)                                g(x)

x = 0       0^2 - 6(0) + 8  = 8          0  - 2    =  -2       not true

x  =1       1^2  - 6(1) + 8  = 3         1 - 2  =  - 1         not true

x = 2        2^2 - 6(2) + 8 = 0           2 - 2  =  0         true  

x = 3       3^2  - 6(3) + 8 = 8           3 - 2  = 1          not true

x  = 4      4^2  - 6(4) + 8 = 0           4 - 2  =  2         not true

x = 5       5^2  - 6(5) + 8 = 3           5 - 2  = 3          true

So  x  = 2   and x = 5   are the solutions

help please I don't understand. 

(f * g) (x)  =

f(x) * g(x)  =

[ x^3  - 4x  + 2  ] * [ x^2 + 2 ] =

x^2 [ x^3 - 4x + 2]  +  2 [ x^3 - 4x + 2  ]  =

x^5 - 4x^3 + 2x^2  + 2x^3 - 8x + 4       combine like terms

x^5 - 2x^3 + 2x^2 - 8x + 4

Graph the functions on the same coordinate plane.

(f + g) (x)  =

f(x)  + g(x)  =

[8x^2 + 6x + 6]   + [ x^3 - 3x^2 - 9 ]       combine like terms

x^3 + 5x^2  + 6x - 3

If even

f(x)  = f(-x)

f(-x)  =  -2(-x)^3 + 3(-x)^2   =   2x^3 + 3x^2   ....so....not even

If odd

f(-x)  =  - f(x)

-f(x)  =  2x^3 - 3x^2  .......so ....not odd

Neither even or odd 

And if you want to find AC   ...

cos BDC  =  12/20  = 3/5

cos BDA  = 5/13

sin BDC = 4/5

sin BDA = 12/13

So  cos  (BDC +  BDA)  =  cosBDCcosBDA - sinBDC sinBDA =

                                            (3/5)(5/13) - (4/5)(12/13)  =

                                             15/65  -  48/65 =

                                              -33/65

So....using the  Law of Cosines

AC^2  = AD^2 + DC^2  - 2(AD * DC) cos ( BDC + BDA)

AC^2  = 5^2   +  20^2  - 2(5 * 20)(-33/65)

AC^2  =  425   + 200(33/65)

AC^2  = 6845 / 13

AC  ≈ 22.946

1 - kx  =  - 3y         sub (4, -3) into the equation

1 - k(4) = - 3(-3)     simplify

1 - 4k  =  9       subtract 1 from both sides

-4k  =  8         divide both sides by - 4

k  =  -2

10xy + 14x + 15y  = 166     simplify as 

5y ( 2x + 3)  =  166 - 14x

Since the left side is divisible by 5, then so is the right side

When  x  =  4

5y (2*4 + 3)  =  166 - 14 * 4

5y (11)  =  166 - 56

55y  = 110

y  =  2

So  x  + y  =   4 + 2  =  6

Midpoint  of  (2, 5) , (4,9)  =

[  (2 + 4) / 2, (5 + 9) / 2 ]  =   ( 3, 7)

So......  x  = 3, y  = 7

And  x +  y  =   10   =    b

Independent

P(A l B)  =  P(A)

\((5+3i)(a+bi)=32-8i\\ 5a+5bi+3ai-3b=32-8i\\ 5a-3b+(5b+3a)i=32-8i\\ \)

\(so\\ 5a-3b=32 \qquad and \qquad 3a+5b=-8 \\ 25a-15b=160 \quad and \qquad 9a+15b=-24 \\ add\\ 34a=136\\ 17a=68\\ a=\frac{68}{17}\\ a=4 \)

\(3a+5b=-8\\ 3*4+5b=-8\\ 5b=-20\\ b=-4\\ \)

Ramanujan pick          \(4-4i\)

The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12, then what are the x-intercepts of the graph of f?

\(f(x)=ax^3\;\;so\;\; d=0\\ f(x)=ax^3+bx^2+cx\\~\\ f(-1)= -a+b-c=15\\ f((1)=a+b+c=-5\\ f(2)=8a+4b+2c=12 \)

\(\begin{bmatrix} 1 && 1 && 1&&|&&-5\\ -1 && 1 && -1&&|&&15\\ 8 && 4 && 2 && | && 12 \end{bmatrix}\\~\\ *\text{row 2 replaced with row 2 + row 1}\\~\\ \begin{bmatrix} 1 && 1 && 1&&|&&-5\\ 0 && 2 && 0&&|&&10\\ 8 && 4 && 2 && | && 12\\ \end{bmatrix}\\~\\ *\text{row 3 replaced with row Row 3 - 8* row 1 }\\~\\ \begin{bmatrix} 1 && 1 && 1&&|&&-5\\ 0 && 2 && 0&&|&&10\\ 0 && -4 && -6 && | && 52\\ \end{bmatrix}\\~\\ \)

2b=10

b=5

-4b-6c=52 becomes

 -20-6c=52

-6c=72

c=-12

a+b+c=-5   becomes

a+5-12=-5

a=2

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

roots are  x=0 and 

\(x = {-5 \pm \sqrt{25+96} \over 4}\\ x = {-5 \pm 11 \over 4}\\ x=0\qquad x=1.5\qquad x=-4\)

They are the x intercepts.

Thank you so much!

I'm done for today, I will continue on the rest of your problem tommorow. 

I know how to do them but I'm just too tired. 

We first need to simplify both inequalities. 

4x + 3 < 25

Subtracting 3 from both sides:

4x < 22

Dividing both sides by 4:

x < 11/2

-7n + 5 < 24

Adding both sides by 7n and subtracting both sides by 24 (to avoid dividing by a negative number):

7n > -19

Dividing both sides by 7:

n > -19/7

n must be greater than -2 and -5/7 but also smaller than 5 and 1/2

The integers in between are: 

-2, -1, 0, 1, 2, 3, 4, and 5

There are 8 integers 

Hi gueesstt, to make this a square of a bunomial, we must take the square of one half of the coefficient of the term where the degree of x is 1. 

So the coefficient of the term where the degree of x is 1, is 5/3.

We must take the square of one half of that.

First, one half of 5/3 is 5/6.

Then we take the square of that, which is 25/36. 

That is the number that makes the quadratic a square of a binomial. 

Hey there,

Try substituting 24n+51 into m for 2m

Then, use the Euclidean Algorithm and repeatedly subtract 3n from 2(24n+51)

Or check the NT message board. I had the same problem :)

Since all four girls are next to each other, we can treat them like one person. 

So it is basically a six person permutation.

\(6!=720\)

There are 720 ways to arrange the boys and girls, but we can also change the order the 4 girls sit in. 

\(4!=24\)

In total, there are \(24\cdot720=17280.\) ways to arrange everyone in a row. 

I hope this helped,

Gavin.

I am still struggling to find the values...

You didn’t specify what we need to find, but assuming you want BC, here is my solution:

Using the Pythagorean Thereom, BD^2 = AB^2 - AD^2 = 13^2 - 5^2 = 12^2 

BD = 12

Using the same method, BC^2 = DC^2 - BD^2 = 20^2 - 12^2

You take the square root of that and you have your answer.