2(m - 5) + 6PEMDASu multiply 2 into (m-5)2*m and 2*-5that would be 2m-10 + 6now subtract common variables-10 + 6 = -42m - 4
The maximum value can be determined by taking the derivative of the function.
(dh/dt) [h(t)] = h'(x) = -9.8t + 6
Set h'(x) = 0 to find the critical point
-9.8t + 6 = 0
-9.8t = -6
t = 6/9.8
Plug the time back into the function to find the height.
h(6/9.8) = -4.9(6/9.8)^2 + 6(6/9.8) + .6
= 2.4
And I don't understand your second question.
Is this a riddle?? Or are you really doing this :-(
Answer:
The total distance traveled by the particle is S = 30.
Step-by-step explanation:
Given that velocity,
v(t) = 2t + 1
To find the total distance travel, we integrate the velocity function, v(t), to obtain the distance function s(t), and evaluate the resulting distance at the interval given. That is at t = 0 to t = 5.
Integrating v(t) with respect to t, we have
s(t) = t² + t + C.
At t = 5
s(5) = 5² + 5 + C
= 25 + 5 + C
= 30 + C
At t = 0
S(0) = 0 + 0 + C
= C
The required distance is now
S(5) - S(0)
= 30 + C - C
= 30.
