A model airplane is shot up from a platform 1 foot above the ground with an initial upward velocity of 56 feet per second. The h
eight of the airplane above ground after t seconds is given by the equation , where h is the height of the airplane in feet and t is the time in seconds after it is launched. Approximately how long does it take the airplane to reach its maximum height?
2 answers:
Answer:
Option B. 1.8 seconds
Step-by-step explanation:
h=-16t^2+56t+1
This is a quadratic equation, and its graph is a parabola
h=at^2+bt+c; a=-16, b=56, c=1
Like a=-16<0 the parabola opens downward, and it has a maximum value (height) at the vertex, at the abscissa:
Replacing the known values:
Approximately 1.8 seconds.
Answer: It takes approximately 1.8 seconds the airplane to reach its maximum height.
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Answer:
Inequalities are,
y ≥ 4x + 2
y ≥ 2
Step-by-step explanation:
Solid yellow line of the graph attached passes through two points (0, -2) and (1, 2).
Let the equation of this line is,
y = mx + b
Slope of the line =
m =
m = 4
Y-intercept 'b' = -2
Equation of the line will be,
y = 4x - 2
Since shaded area is on the left side of this solid line so the inequality representing this region will be,
y ≥ 4x - 2
Another line is a solid blue line parallel to the x-axis.
Shaded region (blue) above the line will be represented by,
y ≥ 2
Therefore, the common shaded area of these inequalities will be the solution of the given inequalities.