Algebra help again -_-?

Question

1. use the quadratic formula to find a solution for -2x^2 - 3x + 9 = 0

2. what is the distance between (7, -3) (-1, 4)

3. what is the standard form of an equation that passes through the points (14, 8) (3, 11)

4. What is the solution of the inequality 8 <or= -3x + 1 <or= 11

5. simplify the expression (5x^2 y^-3/8x^3y) (4x^-2y^3/10xy^-4)

6. what is the vertex of the graph of the equation y = x^2 + 4x - 12 =

7. find the solutions for x in the equation x^2 - 8x -24 = 20

8. find the solutions for x in the equation x^2 - 2/3x + 1/9 = 0

Answer 1

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Answer 2

-2x^2 - 3x + 9 = 0

x=3/2

x= -3

(7, -3) (-1, 4)

the distance is:

d=√(7+1)^2+(-3-4)^2=√(64+49)=√113= 10.63

the points(14, 8) (3, 11)

the equation of the line passing through these two points is:

(y-8)/(11-8)=(x-14)/(3-14)

(y-8)/3=(x-14)/(-11)

y-8=(3x-42)/(-11)

-11y+88=3x-42

3x+11y-130=0

8 ≤-3x + 1 ≤ 11

8-1≤ -3x ≤ 11-1

-7/3 ≥ x ≥ -10/3

(5x^2 y^-3/8x^3y) (4x^-2y^3/10xy^-4)=

(5/8)x^(2-3)y^(-3-1) * (4/10)x^(-2-1)y^(3+4)=

(5/8)x^(-1)y^(-4) * (2/5)x^(-3)y^(7)=

(1/4)x^(-1-3)y^(-4+7)=

(1/4)x^(-4)y^(3)=

y^3/4x^4

y = x^2 + 4x - 12

V(-b/2a ; -Δ/4a)

-b/2a= -4/2= -2

-Δ/4a= -(b^2-4ac)/4a= -64/4= -16

V( -2; -16)

the graph of the equation is the parable.

x^2 - 8x -24 = 20

x^2-8x-44=0

x = 4 - 2√15

x = 4+ 2√15

x^2 - 2/3x + 1/9 = 0

x=1/3