CPhill

AC is the diagonal  = 2

x is the side of the square =  diagonal / sqrt 2 =   2 / sqrt 2 =  sqrt 2

This is the graph of the sec  whose period is  twice as long with an amplitude of 3  shifted up one unit

See here : (the sec is in red....our graph is in blue )

Here :  https://web2.0calc.com/questions/help-plz_85848

x(x + 6) > 16 - 7x + 20

x^2 + 6x + 7x - 36 > 0

x^2 + 13x - 36 > 0

x^2 + 13x  > 36     complete the square on  x

x^2 + 13x + 169/4 > 36 + 169/4

( x + 13/2)^2  > 313/4

x + 13/2 >   sqrt (313)  / 2               x + 13/2 < -sqrt(313)/2

x >  [ sqrt (313) - 13 ]  / 2                x <  [ -sqrt (313) - 13 ] / 2

Solution

(-inf , [ -sqrt (313)  -13 ] / 2 )   U  ( [sqrt (313 - 13 ] / 2 ,   inf )

Simplify as

3a + 4b = 7

3a + 4b  =  k

Identical equations produce infinite  solutions, so  k   = 7

3 / (4 + 2 + 3 + 7)  * 240  =

(3 / 16)  * 240  =

3 * (240 / 16)   =

3 * 15  = 

45 blue

The area of parallelogram  MBND  = (1/2)AB * BC = (1/2)(1)(1)  = 1/2

The shaded area is (1/2) of this =  (1/2)(1/2)  = 1/4

Here : https://web2.0calc.com/questions/help-with-counting_71

This will have a domain of  all real numbers if  the discriminant of the quadratic in the  denominator is < 0

So

b^2  - 4*18 < 0

b^2  - 72 < 0

b^2  < 72

b =  8

No. of possible identifiable sequences  =   (No. of rolls)!  / ( multiple occurences of any number(s)) ! =

  7!  / 2!  =  2520