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1 answer:
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Answer:
y = (5/4)2^x
Step-by-step explanation:
The function value increases by a factor of 40/10 = 4 when x increases by 2. The function can be written as ...
y = (reference value)·(growth factor)^((x -reference)/(change in x for growth factor))
y = 10·4^((x-3)/2) . . . . . . using point (3, 10) as a reference
This can be simplified to ...
y = 10·2^(x -3) = 10/8·2^x
y = (5/4)2^x
Answer:
s = 22.5 m
Step-by-step explanation:
the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.
v (t) = 25 - 5 t
at t = 0 , v = 20 m/s
Let the distance is s.
Let at t = t, the v = 20
So,
20 = 25 - 5 t
t = 1 s
So, s = 25 x 1 - 2.5 x 1 = 22.5 m