Sorry but grade math is this so I can have a better understanding
Answer:
true
Step-by-step explanation:
Answer:
36.9 m
Step-by-step explanation:
h(t) = -4.9t² + 25t + 6
At t = 3:
h(3) = -4.9(3)² + 25(3) + 6
h(3) = -44.1 + 75 + 6
h(3) = 36.9
After 3 seconds, the ball will have a height of 36.9 meters.
Answer:
16Km due east of school P
Step-by-step explanation:
Given
A school P is 16km due west of a school Q
Thus, we can say that distance PQ = 16 km.
________________________________
now we have to find the bearing of Q from P
As distance is same
distance PQ = distance QP
Thus,
Distance will remain same of 16 km.
For direction,
If Q is west of P, then P will be east of Q
as shown in figure P is west of Q,
now from point P , Q is west P.
Thus,
Bearing of School Q from P is 16Km due east of school P