CPhill

Agreed........I'll try to answer your second problem with extensive explanations

9 / (1 + cos x  )    + 9 / ( 1 - cos x)

We can do this

9 ( 1 - cos x)                            9 ( 1 + cos x)

_________________     +    ___________________

(1 + cos x) (1 - cos x)             (1 - cos x) ( 1 + cos x)

9 - 9 cos x  +  9  +  9 cos x 

______________________

   1  - cos^2 x

   18

_______

 sin^2 x

  18   *            1

               _______   

               (sin x)^2

18  *    [  1  / sin x ] ^2  

18  *  [ csc x ] ^2

18 csc^2 x

I DID look at the graph to make sure that I gave you the correct answers......but...it is also helpful to think about WHY  the graph looks as it does.....

Yeah I know.....we are using a lot of   exponential rules here

Here are a few that I employed

^m √ [ a^n ]    converts  to exponential form   (a)^(n/m)

And

a^(m)  *  a^(n)  =   a^(m + n)

And

Remember.... we can take  something out of a radical  when  we have  (a)^(n/m)  and n  ≥ m

For instance...let's suppose that we have     ³ √ [ 4^5]

In  exponential form....we have  (4)^(5/3)

To  determine how many powers of 4 come out of the radical and how many stay in.....write the exponent as a mixed number

4^(1 +2/3)

The whole number tells us that 1 power of 4  comes out of the radical  and  the numerator of the fraction tells us that 2 powers of 4 stay inside the radical...so we have

4 ³√ [4^2 ]    =    4 ³√ [ 16 ]

But note that  16  =  2^4....so...we can apply this rule once more

We  have    4 ³√ [2^4 ]    =   4 (2)^(4/3)

To determine how many powers of 2 that we can take out of the radical and how many we can leave in...write this as

4 (2)^(1 + 1/3)....so....we can take 1 power of 2 out of the radical and 1 power of 2 stays in

So..putting this all together, we have

(4 * 2) ³√2     =   8 ³√2

Hang in there, RP......you will begin to understand this

y  =    1  / [ (x + 1) (x - 1) ]  =   1 / ( x^2  - 1 )

The  vertical asymptotes  are the x values that make the denominator  = 0

These are the lines x  = -1   and x  = 1

Since we have a lower degree polynomial over a higher degree polynomial...we will have a horizontal asymptote of 0   [ this will always be true in a  lower / higher   situation ]

The range is a little difficult

Note that when x is 0, the function will reach its maximum negative value  of -1

When  x  is either a large negative  or positive value   1 / [ x^2  - 1 ]  will be almost  0

As x  moves  closer  to the left   vertical  asymptote  x  = -1  from the left side  or  closer to the right vertical asymptote x = 1 from the right side...the function aprroaches infinity

For example.....let x  =   -1.0001   or   1.0001  and we get the large positive value  of ≈ 5000

And this will increase as we get closer to each of these asymptotes in this manner

As the function moves closer to the left asymptote of x  = -1 form the right or closer to  the right vertical asymptote of x = 1 from the left, the function approaches negative infinity

For example when  x  =  -.9999  or   x  =.9999 the function  value   ≈ -5000

And this will decrease as we get closer to these aymptotes in this manner

So....the range is  ( -infinity , 1 ] U  (0 , infinity )

First one.....convert to exponential form

 5 ( 4n)^(1/4)

__________        

2  (27n)^(1/4)

We can "cancel" the n's and get

5 (4)^(1/4)

_________         multiply top/bottom by   (27)^(3/4)

2 (27)^(1/4)

5 ( 4 )^(1/4) (27)^(3/4)

__________________

2 (27)^(1/4) 27^(3/4)

5 [ 4 (27)^3 ] ^(1/4)

_________________

  2 * 27

5 [ 4 (3^3)^3 ] ^(1/4)

________________

   54

5 [ 4 * 3^9 ]^(1/4)

______________        

    54

Note :  (3^9)^(1/4)  =  3^(9/4)  =  3^(2 + 1/4)...we can take 3^2  out of the radical  and leave one power of 3 inside

5 * 3^2   [ 4 * 3 ]^(1/4)

_________________

    54

5 * 9 [ 12]^(1/4)

_____________

54

5 [12]^(1/4)

_________

     6

5 ⁴√12

______

    6

Let  pure water  = 0%  acid

Let the amount of water to be added  = W....so we have

(90 cc) * (5% acid  solution)  +  W *( 0%  acid)  = ( Final amt) * (3% acid)

(90) (.05)  +  W (0) =  ( 90 + W) (.03)      simplify

4.5   =   2.7 +  .03W      subtract 2.7 from both sides

1.8  = .03W    divide both sides by .03

1.8 / .03  =  W   =  60 cc   of water 

We can make this a little more simple than by testing all possible roots right off

One possible rational root is 1

If we can add the coefficients and  constant of the polynomial and get 0, then 1 is a root....and this is true

So  ( x - 1)  is a factor

We can find the remaining polynomial with some division

             x^2  + 8x + 15

x - 1  [  x^3 + 7x^2  + 7x  - 15  ]

            x^3  - 1x^2

           ___________________

                      8x^2  + 7x - 15

                      8x^2  - 8x

                     _____________

                               15x  - 15

                               15x -  15

                               ________

So the remaining polynomial is

x^2 + 8x  + 15

The possible rational roots are   ±1  ±3  ±5  ±15

However...instead of testing these to see which produce a 0...note that the polynomial can be factored as

(x + 3) ( x + 5)

So....the factors are   ( x - 1) ( x + 3) ( x + 5)

x^3 + 3x^2 - 25x - 75

Factor by grouping

x^2 ( x +3) - 25 (x + 3)

(x^2 - 25) ( x + 3)

(x + 5) (x - 5) ( x + 3)    are the binomial factors

For the second one we have

x^4 + 5x^3 -x^2  - 5x      factor by grouping

x^3 ( x + 5) - x (x + 5)

(x^3 - x) ( x + 5)

x ( x^2  - 1) (x + 5)

(x - 0 ) ( x + 1) (x - 1) (x + 5)

These are true

f(x)  =    when  x  = -5

f(x)  ivided by ( x + 5)  as a remainder of 0      [ because ( x + 5) is a factor ]

cos  [ 2 arcsin ( x) ]

Let  arcsin (x)  be θ

So...using a trig identity....  we have

cos ( 2θ)  =    1  - 2sin^2(θ)  =  1 - 2 [ sin (θ) ]^2

The sin of θ  =  x / 1  =  x

So...

cos (2θ)  =  1 - 2sin^2 ( θ)  =  1 -  2 [ sin (θ)]^2  =   1  - 2 [x]^2  =  1  - 2x^2