CPhill

1). The sum of the angle measures of a triangle  = 180°

This implies that :

x + 2x  + 5x   = 180

8x  =  180       divide both sides by 8

x  = 180/8  =  22.5” 

2)   The angle marked  x°  is supplemental to the 75°  angle....this is because a transversal that cuts parallel  lines makes corresponding angles equal

Therefore....

x  +  75  =  180          subtract 75  from both sides

x  = 105°

Set the functions equal

(1/2)x^2 + x + 6   = x^2 + 4x + 6  rearrange as

(1/2)x^2  + 3x   =  0

Factor

x [ (1/2)x + 3 ]  = 0 

Setting both factors equal to 0  and solving for x, we have that

x   =  0        and   x  =  -6

c  = 0     and a  = - 6

So   c  - a   =  0 - -6   =  6

Adam, let me know if you don't know how to graph these....I'm using  pi/6  as θ

a)  -θ  =  -pi/6

b)  pi + θ   =  pi  + pi /6  =   6pi / 6  + pi/6   =  7pi/6

c) pi - θ   =  pi - pi/6  =  6pi/6  - pi/6  = 5pi/6

a)  csc θ > 0   and  cos θ < 0

csc is related to the sine.....so  the sine θ > 0   in Q1, Q2

And the cos θ < 0   in  Q2  and Q3

So.....the intersection of these is   Q2  ⇒  ( pi/2, pi)

b )   tan θ < 0   and  secθ > 0

tan θ < 0   in Q2 , Q4

An sec is related  to the cos.....so  cos θ > 0  in Q1 , Q4

So.....the intersection of these is  Q4 ⇒  (3pi/2, 2pi )

724 = 15(a^5/2) + 12    subtract  12 from both sides

712  = 15a^(5/2)   divide both sides by 15

712 / 15  = a^(5/2)       take each side to the 2/5 power

[ 712/15 ] ^(2/5)  =  a  ≈  4.6833

\( $x = \frac{a \pm b\sqrt{c}}{d}$\)

x + y   = 4       (1)

3xy  = 4  ⇒   y  = 4/ [ 3x]     (2)

Sub (2)  into (1)

x + 4 / [3x]  =   4            multiply through by  3x

3x^2  + 4  =  12x        rearrange as

3x^2  - 12x  + 4  =  0

x =  [ 12 ±√ [ 144 - 48] ] / 6

x  =   [ 12 ±√ 96 ] / 6 

x = [  12 ± 4√6 ] / 6      =   [ 6 ± 2√6 ] / 3

So  

A =  6     B  = 2   C   = 6   D   =  3

And their sum is 

17 

-1 / x   =  -7

Note  that we can write this as

-1 / x   =  - 7 / 1          flip the fractions over

x / -1   =   1 / -7           multiply both sides by -1

x / 1  =  1 / 7

x   =  1 / 7

I assume we want the positive square root...so we can  write

√ 121y²           √ 121        √ y²                         11 y 

 _____  =      _____    *   ____       =            ______

√ 64x⁶              √ 64          √ x⁶                        8 x³

The only thing that might give you trouble is this :

  √ x⁶    =   ( x⁶ )^1/2    =  x³

Did you understand, Manuel   ????

2x^3+13x^2+23x+12  =  0

Let's see if we can find an alternative way to write this.

A little trial and error produces  a way to break up 13x^2   as  2x^2 + 11x^2

So we have

2x^3  + 2x^2  + 11x^2 + 23x + 12      factor by grouping

2x^2 ( x + 1)  +  (11x + 12) ( x + 1)        the common factor is  x + 1

So we have

(x + 1)  [  2x^2 + 11x + 12 ]  =  0        factor the second polynomial

( x + 1)  ( 2x  + 3) ( x + 4)  =  0

Setting each factor to o and solving for x  produces

x + 1  = 0             2x + 3  = 0           x  + 4  = 0

x = -1                    2x  =  -3              x  = -4

                               x  = -3/2

The  zeroes are in red

BTW  - the first factoring trick won't always work...but....it is something to try

125x^3  - 8

This is the difference of cubes

The  factorization  of  a^3  - b^3  =  ( a - b) (a^2  + ab  + b^2)

"a"  is   the cube root of the first term  = 5x

"b" is the cube root of the second term  =  2

So we have

( 5x  - 2)   ( [ 5x ]^2  +  [ 5x * 2 ] + 2^2 ]  =

(5x - 2) (25x^2  + 10x  + 4 )