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1) find the sec θ for and angle whose teminal side contains that poing (-3,4)

 

2) give x-5 is a factor of x^3 + 9x^2 - 37x - 165 what are the other factors

 

3) solve 16^x = (1/4)^(x+4)

 

4) simplify (9x + 27)/x^2 - 6x + 8) ÷ (6x + 18)/(4x^2 - 16)

 

5) simplify (2/x) + 3/(x-1) ÷ 1/(2x-2)

 

6) what is g(f(x)) if f(x) = 2x^2 - x and g(x) = x^(-1/2)

 Jan 31, 2019
 #1
avatar+106539 
0

Gonna eat supper, Blank.....hang on and I'll answer these in just a few minutes...

 

 

cool cool cool

 Jan 31, 2019
 #2
avatar+106539 
+2

1) find the sec θ for and angle whose teminal side contains the point (-3,4)

 

Sec =  r /x 

 

r =  sqrt [ (-3)^2 + 4^2 ]  =  sqrt (25) = 5

 

So....sec  θ  = 5 / - 3

The angle will be in the 2nd quadrant

 

To find this....use the secant inverse....

 

arcsec (5/-3) ≈  126.9° =   θ

 

 

cool cool cool

 Jan 31, 2019
 #3
avatar+106539 
+2

2) give x-5 is a factor of x^3 + 9x^2 - 37x - 165 what are the other factors

 

We can use sythetic division to dwtermine the remaining polynomial

 

 

5  [  1    9     - 37     -  165 ]

             5       70         165

     _____________________

        1   14    33           0

 

The remaining polynomial  is    x^2 + 14 + 33 

 

Factoring this we have that

(x + 11) ( x + 3)  = the remaining factors  !!!

 

cool cool cool

 Jan 31, 2019
 #4
avatar+120 
0

So basically we take the answer of the provided factor and use that as the multiplier in the syntetic divison to get a quadratic function that we use any way to factor down into the final factors? So how would this work if there was a quadratic to the 5th power? The same thing?

BLANK  Jan 31, 2019
 #6
avatar+106539 
0

We would do the same thing.....one thing though....the remaining polynomial may or may not be factorable

 

Fortunately....this one WAS capable of being factored....!!!!

 

[ We might have to use the Rational Zeroes Theorem to find the other possible roots if the remaining polynomial doesn't factor easily ]

 

 

cool  cool cool

CPhill  Jan 31, 2019
 #5
avatar+106539 
+2

3) solve 16^x = (1/4)^(x+4)

 

Note that   16 = 4^2

And 1/4 = 4^(-1)

So  we have

 

(4^2)^x   = (4^(-1) )( x + 4)          using  a law of exponents

 

4^(2x)  = (4)^ (-x - 4)      we have the same bases....so....we can solve for the exponents

 

2x  = -x - 4     add x to both sides

 

3x = - 4     divide both sides by 3

 

x = -4/3

 

 

cool cool cool

 Jan 31, 2019
 #7
avatar+106539 
+2

4) simplify (9x + 27)/x^2 - 6x + 8) ÷ (6x + 18)/(4x^2 - 16)

 

Factor tops/bottoms

 

9 (x + 3)                     6 (x + 3)

__________    ÷   ____________        flip the second fraction and multiply

(x -4)( x - 2)           (x + 4)(x - 4)

 

9(x + 3)                  (x + 4) ( x - 4)

__________    *      _____________

(x - 4) ( x - 2)           6(x + 3)

 

 

9(x + 3) * (x + 4) ( x - 4)

___________________

6  (x + 3) *  (x -4) ( x - 2)

 

 

3 (x + 4)             3x + 4

_______  =       _____ 

2 (x - 2)             2x - 4

 

 

 

cool  cool cool

 Jan 31, 2019
 #8
avatar+106539 
+2

5) simplify (2/x) + 3/(x-1) ÷ 1/(2x-2)

 

Easiest to take this in "pieces"

 

2              3                    2          (x - 1)             2(x - 1)

_   +     ____    =        ____   *   ______  =     _______  

x           (x - 1)               x              3                    3x

 

 

So now...we have

 

2(x - 1)               1                    2(x - 1)          (2x- 2)            2(x - 1) * 2 (x - 1)                4(x - 1)^2

______    +     ______    =   _________  *    ______   =     _________________  =    _________

  3x                  (2x - 2)                3x                   1                         3x                                   3x

 

 

 

cool cool cool

 Jan 31, 2019
 #9
avatar+106539 
+2

6) what is g(f(x)) if f(x) = 2x^2 - x and g(x) = x^(-1/2)

 

g(f(x))   just means that we are putting  "f" into "g"

 

So we have

 

(2x^2 - x)^(-1/2)  =      1 / √[ 2x^2 - x ]

 

And that's it...!!!

 

 

cool cool cool

 Jan 31, 2019

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