CPhill

Slight mistake  in (3)

It should be

2x - y - 4 ≥  0 

Sorry, MS!!!

Ok, man!!!

OK.....I'll see what I can do.....

3) We have the points (2,0)  and (3,2)...and a sloid line

This is easier

slope  [  2-0] / [ 3 -2] =  2/1  = 2

The equation is 

 y = 2 ( x -2)

y = 2x - 4

Rearrange  as

2x - y - 4   = 0

2) We have two points on this   inequality  (-1,0)  and  (0,1)

Slope  [ 1 -0 ] / [ 0 - -1] =   1/1  = 1

If we had a solid line the equation would be

y= x + 1

But we have a dotted line

So

The equation is either

 y < x + 1                     or  y  >  x + 1

To see which one is correct, pick a point in the shaded area  and  test it  in  the equations  ..I'll pick (0,2)

2 < 0 + 1

Not true

So the equation is

y > x + 1

Rearrange  as

x - y + 1 <  0

See the graph here : https://www.desmos.com/calculator/arefq6sleo

Don't know, MS.....look back at your earier  question .....I'll answer what I can

1)  The terms are already in decreasing order

We know the opposite side and  hypotenuse....we  can use the arcsin to find x

arcsin (6/11)  ≈  33°

x - 2y  =7  →    2y  = x - 7    →  y = (1/2)x - 7/2   the slope of  t  =  (1/2)

The slope of a line perpendicuar to this   = -2

And it contains the point ( 4, -4)

Equation of p

y = -2(x - 4)  -4

y = -2x + 8 - 4

y = -2x + 4

x intersectionof these lines

(1/2) x  - 7/2 =  -2x + 4

2x + (1/2) x  = 4 + 7/2

(4/2)x + ( 1/2) x  = 8/2 + 7/2

(5/2) x  = 15/2        

x  = (15/2) ( 2/5)   =  30/10   = 3

See the graph, here :  https://www.desmos.com/calculator/s0denwkyj8

(x , y)   reflected in  the  x  axis  →  (x, -y)

(x, -y)   rotated 90°  counter-clockwise  →  (y, x)

(y , x)   reflected in the line y = x  →  (x , y) = image point

The constant term is

C(5,2)  (  10x^3)^2  * [ -1/ (2x^2) ]^3   = [ 10 * 100x^6 ] *[ -1/8 ] [ 1/x^6 ]   = -(1000/8)  = -125