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And the question is ...

idk how you got that but thanks

Two points on the line are  (2, 0) and (0, -3)   --->   m  =  ( - 3 - 0 ) / (0 - 2 )  =  3/2

The y-intercept is -3   --->  the equation is  y  =  (3/2)x - 3

Since the line is not shaded but the graph is shaded below that line:  y  <  (3/2)x - 3

--->   2y  <  3x - 6    --->     -3x + 2y  <  -6     --->     3x - 2y  >  6

E(3) is 5. 

um there is no graph...

Second question:

f(x) = 1.8 in three places:

1)  -1 <= x <= 1   --->   y = 2x    --->   1.8  =  2x   --->   x  =  0.9

2)  1 <= x <=2   --->   y = -x + 3   --->   1.8  =  -x + 3   --->   -1.2  =  -x   --->   x  =  1.2

3)  2 <= x <= 4   --->   y  =  2x - 3   --->   1.8  =  2x - 3   --->   4.8  =  2x   --->   x  =  2.4

First question:         f(x)  =  ( 3x + 2 ) / 5     --->     y  =  ( 3x + 2 ) / 5   

To find the inverse function:

Step 1:  interchange  y  and  x:          x  =  ( 3y + 2 ) / 5

Step 2:  solve for y:                          5x  =  3y + 2

                                                   3y + 2  =  5x

                                                         3y  =  5x - 2

                                                           y  =  ( 5x -2 )  / 3

Step 3:  replace  y  with  f^-1(x):     f^-1(x)  =  ( 5x - 2 ) / 3

Find  f^-1(4)  --->   f^-1​(4)  =  ( 5·4 - 2 ) / 3   =   18 / 3  =  6

Find  [ f^-1(4) ]^-1  =  [ 6 ]^-1  =  1/6

Area  =  [ (3x + 6)(2x + 4) ] - [  (x - 3)(x - 1) ]

         =  [ 6x² + 12x + 12x + 24 ] - [ x² - x - 3x + 3 ]

         =  [ 6x² + 24x + 24 ] - [ x² - 4x + 3 ]

         =  6x² + 24x + 24 - x² + 4x - 3

         =  5x² + 28x + 21

The gcd is 2.

Who says I don't try and work on these problems?

PR = 5*sqrt(3).

Erm I am good

(e_1,e_2) = (4,6).

The solution is (x,y,z) = (4/5,-4,-4/3).

maybe im dumb or smth but how do i find those 3 values? I don't know how to graph them

Which of the following is the graph of the quadratic function? y=x^2+4-12   

You mistyped the question.  It should be 4x not just 4. 

x² + 4x – 12  factors to (x + 6)(x – 2)  

Therefore y equals 0 when x equals –6 and when x equals +2 

Find which curve crosses the x-axis (i.e., y=0) at –6 and +2   

The print is too small for me to see well, but I think it's showing that on Graph A. 

_.

The equation of the parabola is y = 1/3*x^2 - x + 4/3, so abc = (1/3)(-1)(4/3) = -4/9.

maybe im dumb or smth but how do i find those 3 values? I don't know how to graph them

You should check out the book "How to Win Friends and Influence People".  No, really, you should.

(a) we can first note that 147 is divisible by 49 (7*7*3) and a^4 can come out a^2 so out of the square root comes 7a^2. Your answer is \(\boxed{7a^2\sqrt3}\)

(b) 80 is divisible by 16 (4*4*5) so the answer is \(\boxed{4\sqrt{5n}}\)

(c) 75 is divisible by 25 (5*5*3) and x^4 can come out x^2 so your answer is (don't forget the -7) \(\boxed{-35x^2\sqrt{3}}\)

There are only two pairs (b,c): (4,14) and (46,8)

that's a wrong answer

maybe post your solution next time