CPhill

3Ax - 24 = 5x - 9 + x        simplify the right side

3Ax - 24  =  6x - 9

This will have no solution when    3A  = 6  ⇒   A   = 2

To see why this is so....

3(2)x   - 24   =   6x - 24    and this can never equal  6x - 9

y = 2x^2 - 6x + 3

y = x - 2     

Set the y's equal  and solve for  x

2x^2 - 6x + 3  =  x - 2         subtract x , add 2    from/to each side

2x^2 - 7x + 5  = 0          factor, if possible

(2x -5) ( x - 1)  = 0

Set each factor  to 0   and solve  for  x

2x  - 5  = 0                   x  - 1  =0

2x  = 5                         x = 1

x = 5/2 = 2.5                x  = 1

So    y  = x - 2 = 2.5 - 2  = 0.5        or   y = x - 2 =   1 - 2  = -1

So  {x,y}   =  { 2.5, 0.5}     or  { 1, -1 }         

( ( 5p+1)-2p*4) ( 3 ) + ( 4-1/3) ( 6p-9) 

[ (5p + 1) - 8p ] (3) +  ( 12/3 - 1/3) (6p-9)  =

[ -3p + 1 ] (3)  +   (11/3) [ 6p - 9)  =

[-9p + 3 ]  +  (66/3)p - 99/3  =

-9p + 3  +  22p   - 33  =

13p -  30

Using the Law of Sines....we have that

(1/2) side length of  the square  / sin (67.5°)   = 2 / sin (90°)          .....multiply both sides by 2.....

side length of the square =  4 sin (67.5°)  =  4 √ [  (1  - cos (135°)  ) / 2 ] =  4 √  [ (1 + √2 /2) / 2 ]  =  4 √  [ (2 + √2)/4 ] 

So....the area of the square is the square of this  =   16 (2 + √2)/4  =  4 (2 + √2) units^2

Note that the max value for the sine  = 1  and the minimum value is -1.....therefore....

The high point on the graph  is  12 (1)  + 6  = 18

And the low point is 12(-1) + 6  =   -6

To find the midline.....add these values and divide by  2  =   [ 18 + -6 ] / 2 =  12 /  2  =  6

So.....the midline is y = f(x)  = 6

Nicely done, heureka!!!!!!

Note that  

Any x in  (-inf, 1)  will produce  negative / negative  = positive

Any x in (1, 2)  will produce negative / positive  = negative

Any x in (2, 3)  will  produce  pos / pos  = pos  = true

Any x in (3, 4)  will produce neg/ pos  = neg  = false

Any x in (4, 5)  will produce pos/ pos  = pos  = true

Any x in (5, 6) will produce pos/ neg  = neg  = false

Any x in (6, inf) will produce pos/ pos  = pos  = true

So.....the intervals that make this true  are  (-inf, 1) U (2,3) U (4,5) U (6, inf)

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

Well......if it makes you feel any better.....I have to review some of them, too.....particularly the hyperbola  !!!

I think you just forgot to include the equilateral triangles of side length 1

50  of side length 1

16 + 16 =  32 of side length  2

9 + 9  =  18 of side length 3

4 + 4  =  8 of side length 4

1 + 1  = 2  of side length 5

So

50 + 32 + 18 + 8 + 2  =   110

Since the max height occurs at t = 2

Just stick 2 into the function for t   and evaluate

So....you are evaluating

-16(2)^2  + 64(2)  + 31      =   max height

Can you take it from here, Logic ????