2(i).
(3y-20)+(4x-5)+(x+y-10)+(2x+5)=360 (Angles sum of it trapezium)
3y-20+4x-5+x+y-10+2x+5=360
7x+4y=390
(Shown)
2(ii).
(3y-20)+(2x+5)=180 (Interior Angles, Parallel Lines)
2x+3y=195
(Shown)
Hope it helps : )
The answer is 76
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Answer:
At the time of launch height of the object was 60 meters.
Step-by-step explanation:
An object was launched from a platform and its height was modeled by the function,
h(x) = -5x² + 20x + 60
Where x = time or duration after the launch
At the time of launch, x = 0
So, by putting x = 0 in this equation,
h(0) = -5×(0) + 20×(0) + 60
h(0) = 60
Therefore, at the time of launch height of the object was 60 meters.
0.19992 because u need to times the digits
