The number of gallons of water in the tank at t=10 is
... W(10) = 160,000 -10(8000 -10) = 80100
The number of gallons of water in the tank at t=10.5 is
... W(10.5) = 160,000 -10.5(8000 -10.5) = 76110.25
The rate of change over the interval is
... (W(10.5) - W(10))/(10.5 - 10) = (76110.25 - 80100)/(0.5) = -7979.5
The average rate of change in the number of gallons of water in the tank over the interval is -7979.5 gal/min.
The sign is negative, so the amount of water is decreasing.
Answer:
a₆ = 12
Step-by-step explanation:
The nth term of an arithmetic sequence is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Given a₂ = 5 and a₁₀ = 19 , then
a₁ + d = 5 → (1)
a₁ + 9d = 19 → (2)
Subtract (1) from (2) term by term to eliminate a₁
8d = 14 ( divide both sides by 8 )
d = 1.75
Substitute d = 1.75 into (1)
a₁ + 1.75 = 5 ( subtract 1.75 from both sides )
a₁ = 3.25
Then
a₆ = 3.25 + 5(1.75) = 3.25 + 8.75 = 12
Hello,
Please, see the attaced file.
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