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emmainna [20.7K]
4 years ago
14

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).
Mathematics
1 answer:
mart [117]4 years ago
6 0

Answer:

(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 y=ax^2+bx+c 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).

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Write an equation in the form y = mx + b for the line that has a slope of ¼ and contains the point (12, 3).
Alborosie
\bf \begin{array}{ccccccccc}
&&x_1&&y_1\\
%  (a,b)
&&(~ 12 &,& 3~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies \cfrac{1}{4}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-3=\cfrac{1}{4}(x-12)
\\\\\\
y-3=\cfrac{1}{4}x-3
\implies 
y=\cfrac{1}{4}x
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Step-by-step explanation:

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The width of a rectangle is 5 feet, and the diagonal is 8 feet. Which is the area of the rectangle? (Round to nearest hundredth.
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The \ formula \ for \ the \ area \ of \ a \ rectangle \ is: \\\\A=l\cdot w\\ where \ l \ is \ the \ length, \ w \ is \ the \ width \\\\ w = 5 \feet , \\ diagonal: \ d= 8 \ feet \\ \\ apply \ the \ Pythagorean \ Theorem:

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