A drone is monitoring the atmospheric conditions above a farm field. The drone hovers 5 meters above the crop line. Suddenly, it rises to approximately 5.9 meters (which takes 1.9 seconds) to avoid colliding with the sprinkler system. Based on this information, which equations could model the height, y, of the drone as a function of time, x?

The correct options are;

g(x) = -0.27·x² + x  + 5

h(x) = 2·㏒(x + 1) + 5

Step-by-step explanation:

To answer the question, we substitute x = 1.9 seconds into the given options as follows;

1) For f(x) = √(1.6·x) + 5

When x = 1.9 seconds, we have;

y = f(1.9) = √(1.6×1.9) + 5 = 6.74 which is not equal to the given height of 5.9 meters

Therefore, f(x) = √(1.6·x) + 5 does not model the height of the drone y as a function of time, x

2) For g(x) = -0.27·x² + x  + 5

When x = 1.9 seconds, we have;

y = g(1.9) = -0.27×1.9^2 + 1.9  + 5 = 5.93 meters, which is approximately 5.9 meters to one place of decimal

Therefore, the function, g(x) = -0.27·x² + x  + 5,  approximately models the height of the drone y as a function of time, x

3) For h(x) = 2·㏒(x + 1) + 5

When x = 1.9 seconds, we have;

y = h(1.9) = 2·log(1.9 + 1) + 5 = 5.92 meters,

The function, h(x) = 2·㏒(x + 1) + 5,  approximately models the height of the drone y as a function of time, x

4) For j(x) = -∛(-1.4·x - 1) + 5

When x = 1.9 seconds, we have;

y = j(1.9) = -∛(-1.4×1.9 - 1) + 5 = 6.54 meters

The function j(x) = -∛(-1.4·x - 1) + 5 does not model the height of the drone y as a function of time, x

5) For k(x) = -1.2·x³ + 2.6·x² - 0.5·x + 5

When x = 1.9 seconds, we have;

y = k(1.9) = -1.2×1.9^3 + 2.6×1.9^2 - 0.5×1.9 + 5 = 5.21 meters

Therefore, the function, k(x) = -1.2·x³ + 2.6·x² - 0.5·x + 5, does not model the height of the drone y as a function of time, x

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step-by-step explanation:

351°f

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