Please help and Imma be postin these all night I literally suck at algebra two so just check my profile if your good at math :)

The range of the equation is y ≥ 9 ⇒ C

Step-by-step explanation:

The quadratic equation y = ax² + bx + c, represents by a parabola of vertex (h, k), where h = and k is the value of y at x = h

∵ The equation is y = x² + 9

→ Compare it with the form above to find a and b

∴ a = 1 and b = 0

∵ a is a positive number

∴ The parabola has a minimum vertex

→ By using the 1st rule above

∴ The range is y ≥ k

→ Use the rule of h above to find it

∵ h = =

∴ h = 0

→ To find k substitute x by the value of h and y by k

∵ k = (0)² + 9

∴ k = 0 + 9

∴ k = 9

∴ The range of the equation is y ≥ 9

the answer is 2/54

step-by-step explanation:

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.