Answer:
10
Step-by-step explanation:
Ground level is where h = 0, so solve the equation ...
h(x) = 0
-5(x -4)^2 +180 = 0 . . . . substitute for h(x)
(x -4)^2 = 36 . . . . . . . . . . divide by -5, add 36
x -4 = 6 . . . . . . . . . . . . . . positive square root*
x = 10 . . . . . . add 4
The object will hit the ground 10 seconds after launch.
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* The negative square root also gives an answer that satisfies the equation, but is not in the practical domain. That answer would be x = -2. The equation is only useful for time at and after the launch time: x ≥ 0.
The value of when
comes to be
.
Given that trigonometric ratio:
The tangent of an angle is the ratio of the opposite side(to that angle) to the adjacent side(to that angle).
So, for the given problem
Opposite side to = 11
Adjacent side to = 60
So,
Therefore, the value of when
comes to be
.
To get more about trigonometric ratios visit:
Answer:
» Collect like terms, r terms on the left hand side by subtracting r from both sides and adding st to both sides
» On the left hand side, factorise out r