9514 1404 393
Answer:
a) about -1.03 cubic feet per hour
b) R'(t) = -6/5t·cos(t^2/25); the rate of flow into the reservoir
c) 0.859 and 5.972 hours
d) 9.838 hours
Step-by-step explanation:
The rest of the problem statement is in the first attachment. See the second attachment for the graphing calculator input/output.
a) The average rate of change on the interval [0, 8] is ...
(R(8) -R(0))/(8 -0) ≈ -1.03004 . . . cubic feet per hour
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b) The derivative can be found using the chain rule.
R'(t) = -15(2t/25)cos(t^2/25)
R'(t) = -6/5t·cos(t^2/25)
The derivative of the volume is the rate of change of volume, that is, the flow rate into the reservoir.
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c) To find the time values when R' = 'average rate of change', we defined the constant a1 to be that average rate of change. The graphing calculator shows the curve R'(t)-a1, which has zeros at the times of interest. The times for which the instantaneous rate of change equals the average rate of change are t = 0.859 and t = 5.972 hours.
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d) The time of interest is beyond the domain of the function. We cannot use the given function to determine the time of interest.
However, if we extend the domain to include the time of interest, we find it is t = 9.838 hours.
Answer: There is 40% of team's win that Ryan score a goal.
Step-by-step explanation:
Since we have given that
Team Won Team lost
Ryan scored 6 4
Did not score 9 11
(by Ryan)
-------------------------------------------------------------------------
Total 15 15
Percentage of team's win that Ryan score a goal is given by
Hence, there is 40% of team's win that Ryan score a goal.
Answer:
x=3 MT=36 MH=18
Step-by-step explanation:
The two triangles are the same, so 7x+8=10x-1
3x=9
x=3
MT=12x3=36
MH=36/2=18
9514 1404 393
Answer:
a) average rate = (total distance)/(total time)
b) Rave = 2·R1·R2/(R1 +R2)
c) cheetah's average rate ≈ 50.91 mph
Step-by-step explanation:
a) Let AB represent the distance from A to B. Let t1 and t2 represent the travel times (in hours) on leg1 and leg2 of the trip, respectively. Then the distances traveled are...
First leg distance: AB = 70·t1 ⇒ t1 = AB/70
Second leg distance: AB = 40·t2 ⇒ t2 = AB/40
The average rate is the ratio of total distance to total time:
average rate = (AB +AB)/(t1 +t2)
average rate = 2AB/(AB/70 +AB/40) = 2/(1/70 +1/40) = 2(40)(70)/(70+40)
average rate = 560/11 = 50 10/11 . . . mph
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No equations are given, so we cannot compare what we wrote with the given equations. In each step of the solution, we have used the rules of algebra and equality.
b) For two rates over the same distance (as above), the average is their harmonic mean:
average rate = 2r1·r2/(r1+r2)
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c) The cheetah's average rate was 50 10/11 mph ≈ 50.91 mph.
