Which parabola will have a minimum value vertex? On a coordinate plane, a parabola opens down with x-intercepts at (negative 2.2, 0) and (2.2, 0), and a y-intercept at the vertex (0, 5). On a coordinate plane, a parabola opens down with a vertex at (0, 0). On a coordinate plane, a parabola opens down with x-intercepts at (negative 1, 0) and (2.3, 0), a y-intercept at (0, 2), and has a vertex at (0.75, 2.5). On a coordinate plane, a parabola opens up with x-intercepts at (negative 2, 0) and (2, 0), and a y-intercept at the vertex (0, negative 4).

(D)On a coordinate plane, a parabola opens up with x-intercepts at (negative 2, 0) and (2, 0), and a y-intercept at the vertex (0, negative 4).

Step-by-step explanation:

For a parabola to have a minimum vertex, it must open upward. (a>0).

In the given options, the parabola which opens upward is option D.

On a coordinate plane, a parabola opens up with x-intercepts at (negative 2, 0) and (2, 0), and a y-intercept at the vertex (0, negative 4).

Therefore, it has a minimum vertex with coordinates (0,-4).

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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step-by-step explanation:

what is the figure? attach a picture.