Let the angle of elevation is x and the height of the rocket from the ground is y
tanx = y/15
by differentiating both sides with respect to T
sec²x·dx/dt = (dy/dt)/15
at y = 30 , the hypotenuse of the triangle = 15√5
sec²x=(15√5/15)²=5
5 dx/dt = 11/15
dx/dt = 11/75 rad/sec
Answer:
26-33-40-47-54-61-68-75-82-89-96
I answer is 54a hope this help
She doodle on 5 of the paper strips
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7
7/2 = 3 1/2
Combine like terms.
2x + 6 + 6x - 1
8x + 5
