CPhill

f (∛( -8) ) +  f (- pi) + f (√50 )  + f(9/2)

Edit

Sorry... I didn't see the absolute bars  in the first function....so ....the answer is

 l-2 l  +  (-3)^2   +  (8)^2  +  l 4  l  =

2  + 9   +  64   +  4   =

79

a₃₁  =   104    ......represents  the number of boys in Town A  who play soccer 

Girls  enrolled in basketball in Town B  =  b₁₂  =  94

B   -   A    =

6       14

8       -7

6        2

c₂₂  means that there are 7 fewer girls enrolled in lacrosse in Town B than in Town A

1.125 A  =

108        90

 72         99

117       135

4x - y + z = -5    ⇒   8x - 2y + 2z  =  -10    (1)

2x + 2y + 3z =10       (2)

5x - 2y  + 6z = 1        (3)

Add (1)  and (2)

10x + 5z  = 0        ⇒    2x  + z  =  0   ⇒   z  =  - 2x     (4)

Add (2)  and (3)

7x + 9z  = 11    (5)

Sub (4)  into (5)

7x + 9 (-2x)  =  11

7x - 18x = 11

-11x  =  11

  x  =  -1

So.......z   =  -2(-1)  =  2

And using  4x - y + z = -5   to find y, we have

4(-1) - y +  2  =  -5

-4   -  y   +  2   = - 5

-2 - y   =  - 5

-y  =  - 3

y  =  3

So    {x, y, z }  =   ( -1, 3,  2 }

√ [ 5x + 4 ] - √x  =  4          rearrange as

√[5x + 4 ]  =  4 + √x          square both sides

5x + 4  =  16  + 8√x  +  x

4x - 12  =  8√x           divide both sides by 4

x -  3  =  2√x             square both sides again

x^2  - 6x  +  9   =   4x

x^2  -  10x  +  9   =   0           factor

(x - 9) ( x - 1)   = 0

Setting both factors to 0  and solving for x leads to two possible solutions :

x  = 1     and  x   = 9

Checking these in the original equation.....only x  = 9   will be a true solution

Second one

a. (7,6)  and (1,12)  best model the data

b.   Slope  =  [ 12 - 6 ]  / [ 1 - 7 ]  =  6 / -6  =  -1

c.  Point-slope form     y - 6  =  -1  (x - 7 )

d. Slope-intercept form      y  =  -1x  +  13

e.  Least squares  equation  :  

y = -1.0357142857143x + 13.142857142857

Notice that this is very close to the slope-intercept form !!!! 

First one

a.

Order the data from low to high

61, 71, 73, 76, 78, 85

Minimum  =  61

Median  = Q2 =  [73 + 76 ] / 2  =  149/2  =  74.5

Q1   = [61 + 71 ] / 2  =  132/2  =  66

Q3 = [ 78 + 85 ] / 2  =  76.5

Max  =  85

b.   Mean   =  [ 61+ 71+ 73+ 76 + 78+ 85] / 6   =  444/6  =  74

c.      61  -  74          - 13                169

         71 -  74           - 3                      9

         73 -  74           - 1                      1

         76 -  74             2                       4

         78  - 74             4                     16

         85  - 74            11                   121      

                                           Sum      320

d. Standard  deviation  =   √ [320 / 5 ]  =  √64   =  8

Thanks, Julius....!!!

Since we have a "<"  sign, the graph will have a dashed line....thus....only the top graphs are possible

Pick   (0,0)  and put it into the inequality....if it makes the inequality true, then the solution area will contain this point

18(0)  - 3(0)  <  -21

0  <  - 21

This is not true, so (0,0) isn't in the solution area....so......the top left graph is correct

This figure is an irregular pentagon.......the sum of the interior angles  =  540“

So....we have that

100  +  102  + 108 + 121  +  x  =  540

431  +  x    =  540          subtract   431 from both sides

x  = 109°

Here's one  method :

All the graphs of the lines are the same.....just the solution areas are different......so....let's just pick a point  and see if it solves the original system

(0,0)  is one I  like to try, first......so we have

(0)  + (3/2)(0)   >  -  3   ⇒    0  >  -3     

0  -  5   ≤ - 6(0)    ⇒    -5  ≤  0

Note that both of these are true.....so  ...(0,0)   is in the solution region.....the only graph that has this point in the solution area is  the lower right graph