Find the average distance each data value in the set is from the mean. Round
1 answer:
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Answer:
(a) at t=2, the particle is moving toward the origin
(b) a(t) = 70(8 -9t^2)/(3t^2 +8)^3; a(2) = -0.245
(c) the particle approaches x = 1/3 as t gets large
Step-by-step explanation:
(a) The function x(t) is negative for -3 < t < 3, so at t = 2, the particle is to the left of the origin.
The velocity of the particle is given by the derivative of the position function:
Then, at t=2, the expression is positive, indicating the particle is moving toward the origin.
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(b) The acceleration is the derivative of the velocity, so is ...
Then at t=2, the acceleration is ...
a(2) = 70(8 -9·4)/(3·4+8)^3 = 70(-28)/8000
a(2) = -0.245
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(c) As t gets large, the value of x(t) approaches t^2/(3t^2) = 1/3.
x = original price, or 100%
if we take off 25% from 100% what's left is 75%, that's the discounted price.
let's then take 10% of that.