Answer:
see attached
Step-by-step explanation:
The equation is in the form ...
4p(y -k) = (x -h)^2 . . . . . (h, k) is the vertex; p is the focus-vertex distance
Comparing this to your equation, we see ...
p = 4, (h, k) = (3, 4)
p > 0, so the parabola opens upward. The vertex is on the axis of symmetry. That axis has the equation x=x-coordinate of vertex. This tells you ...
vertex: (3, 4)
axis of symmetry: x = 3
focus: (3, 8) . . . . . 4 units up from vertex
directrix: y = 0 . . . horizontal line 4 units down from vertex
So i'm assuming your eqtn is h = -16t^2 + 27t + 10
and we're looking for t = ? when h = 0
=> 16t^2 - 27t - 10 = 0
(16t + 5)(t - 2) = 0
t = 2 and t = -5/16
answer 2 seconds
Answers:
Row 1: No, No, No
Row 2: Yes, Yes, Yes
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Explanation:
Use the vertical line test. If it is at all possible to draw a single vertical line through more than one point on the relation curve, then the relation is not a function. This is because a function is only possible when any x input leads to exactly one y output.
So that's why graphs 1,2,3 are not functions, while graphs 4,5,6 are functions.
Answer:
the answer is C
Step-by-step explanation:
X=2.6
step by step explanation: