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Simplify as

x^2 + 2x + 27  =  0     complete the square on x

x^2 + 2x  + 1 =   -27 + 1

(x + 1)^2  =  -26       take the negative root  

x + 1   =  -sqrt (-26)

x =  -1 -  i sqrt (26) 

Simplify as

x^2 + 15x  +  k   =  0

The double root occurs when the  discriminant   =   0

So

15^2  - 4 (1)k  = 0

225  - 4k =   0

225  = 4k

k = 225 / 4

(x + 4) / (x + 5)   = ( x - 3) / 2  + (x + 7) / 5

(x + 4) / (x + 5)  =  [ 5(x - 3) + 2(x + 7) ] /  10

(x + 4) / (x + 5)  = [ 5x -15 + 2x + 14] / 10

(x + 4) / (x + 5)  = [ 7x -1 ] / 10

 [ 10 (x + 4)]  - (7x -1) (x+5)] / [10 (x + 5)]     = 0           multiply through by 10(x + 5)

[10x + 40  ] - [ 7x^2 +34x -5 ]   =  0

-7x^2 - 24x + 45  = 0

You can use the quadratic formula to solve this

The roots  of   (x-2)(x + 2)  =     2,-2

The sum  of the roots  of Ax^2  + Bx  + 1 =   -B/A

So  the sum of 2 + -2  = 0

Which means that B  = 0

The product of the roots   of Ax^2 + Bx + 1  = (2(-2))  = - 4  =  C / A  =   1 /A  

So

1/A  = - 4

-1/4  = A

So

A + B =      -1/4 + 0  =  -1/4

5x^2 - 2x + 8 - 2x^2 + 6x + 2 - 2x^2 + 4x - 1     simplify

x^2 + 8x + 9       =    ax^2  + bx + c 

discriminant  =   b^2 - 4ac   =   8^2  - 4(1)(9)  =  64  - 36  =   28

x^2 + 6x + 13  = 2x - 7x + 10          simplify as

x^2 + 11x + 3  =  0        complete the square on x

x^2 + 11x  +  121/4   =  -3 + 121/4

(x + 11/2)^2  =   109/4          take both roots

x + 11/2  =   ±sqrt (109) / 2

x =   -11/2 ± sqrt (109) / 2

2x^2 - 10x + 13  =  -6x^2 - 18x  - 15      simplify as

8x^2 + 8x + 28  =  0

Using the quadratic formula

x  =   [ -8 ±  sqrt  [  8^2  - 4*8*28] ]          -8 ± sqrt [ -832]                 -8 ± 8sqrt (13) i

       ________________________  =   ______________  =          ______________  =

                     2 * 8                                          16                                    16

-1/2 ± sqrt (13) / 2  i

a *  b  =      -sqrt (13) / 4

great job! I was thinking move the -21 to the other side and say b^2-4ac<0

I agree, just simplify and if the denominator is imaginary or complex, say 10i+10, multiply by 10-10i

Solid line  means that we have either   ≤ or ≥  inequality

Two points on the line are  (-3,0) and (0,6)

Slope =  (6 - 0) / ( 0 - -3)  =  6/3 =  2

Using ( 0,6)   the equation of the inequality is either

y ≤  2x + 6    or    y ≥ 2x + 6

(0,0) is in the shaded region  and satisfies the first equation

So

y   ≤  2x + 6

-2x + y  ≤  6

                   B

   45  A        D           C 45

                 12

Angle ABD =   180 - 2(45)  =  90

Then  AB = BC =    AC / sqrt 2  =   12 / sqrt 2 =   6sqrt 2

The area  =  (1/2) (6 sqrt 2)^2  =   18 * 2 =   36

\( a^3 - \dfrac{1}{a^3} if a - \dfrac{1}{a} = 0 \)

a^3  - 1/a^3  =  ( a - 1/a)(a^2 + a^3/a^3 + 1/a^2)

a^3 - 1/a^3  = (0) ( a^2 + 1 + 1/a^2)  =   0

                                        A

                                                   90 X

                                                                5

                                   90 B            15             C

If BXA  = 90  then BXC  = 90

Then triangle BXA is right   and by the Pythagorean Theorem

BC^2  = CX^2  + BX^2

15^2  = 5^2  + BX^2

225  = 25 + BX^2

200 = BX^2

BX = sqrt(200)  =  10sqrt (2)

The line is not solid....so  we will have either   a    <   or >    inequality

Two points crossed by  the "line"  are   (-3,0) and (0, -5)

Slope =  [ -5 -0 ] /[ 0 - -3]  =   -5 / 3

Using  (0, -5)  the equation of the inequality  is either

y  > (-5/3) ( x)  - 5    or   y < (-5/3)x - 5

(0,0) is not in  the shaded area and it satisfies  y > (-5/3)x - 5

So.....the other equation is  correct

y < (-5/3)x -5            (multiply through by 3)

3y  < -5x - 15

5x + 3y <  -15

Here's a similar problem answered by Alan : https://web2.0calc.com/questions/please-help_63707

Here's a similar problem :

https://web2.0calc.com/questions/inequality_131

The shortest altitude will be drawn to the longest side

Using Heron's formula, the semi-perimeter  =  [ 9 + 10 + 15 ] /  2     =  17 

The area =     sqrt [ 17 * (17 - 9) * (17 -10) * (17 -15) ] =  sqrt [ 17 * 8 * 7 * 2 ]  =

sqrt [ 1904]

So  area  =  (1/2) (AB) (altitude drawn to AB)

sqrt [ 1904]  = (1/2)(15) ( altitude drawn to AB)

sqrt [ 1904] / 7.5  = altitude drawn to AB  ≈  5.82 

                            A  60

                           24

              B           D            C

If  BAC  is 60°  and  AD is a bisector of  BAC, then angle  CAD  must = 30°, not 45° 

Total minutes  = Total minutes

4 * 40 =  x *  20

160  = 20x

160 / 20   =  8   min

A =  1  * 4/6   =  2/3

B =  1536

C = (1/8) (1/25)  = 8 / 25

D =  (-4/2) *( - 1*(-8) )  =  -2 * (8)  = -16

E = 3 * sqrt 2 *   4* (1/4) =   3 sqrt (2)  * 1   =  3 sqrt 2

D , C , A, E , B

                 A

         6

      B         8        C  135

Impossible because the greatest angle would be opposite the greatest side

But  the triangle is  obtuse so  ACB  IS  the greatest angle, but it is NOT opposite the greatest side !!

Each member paints   100 / (10 * 10)   =  1 sq ft per minute

So  15 will paint 15 sq ft in a minute

So  15 will take

100 / 15  =   20 / 3   =     6 + 2/3 min  =   6 min 40sec

Each member   must paint    1000 / (10 * 10)  =  1000 / 100  =  10sq ft per minute

So......5 members take

1000 / ( 5 * 10)  =   1000 / 50  =   20 min

Another way to see this is that 5 members must take  twice as long as  10 members !!!!