Answer:
Step-by-step explanation:
When you have something like that, just think that the first number goes on top and the second number goes on bottom.
The number of gallons of water in a large tank at time (t) minutes is given by the function W(t)=160,000-8,000t+t². Find the ave
rage rate of change of the number of gallons of water in the tank from t=10 to t=10.5 minutes. Interpret the sign of your answer in the context of this problem.
1 answer:
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.
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Hey there :)
- tan²x + sec²x = 1 or 1 + tan²x = sec²x
sin²x + cos²x = 1
Divide the whole by cos²x
so 
and
so 
Therefore,
tan²x + 1 = sec²x
Take tan²x to the other side {You will have the same answer}
1 = - tan²x = sec²x or sec²x - tanx = 1
- tan²x + sec²x = 1 or 1 + tan²x = sec²x
sin²x + cos²x = 1
Divide the whole by cos²x
and
Therefore,
tan²x + 1 = sec²x
Take tan²x to the other side {You will have the same answer}
1 = - tan²x = sec²x or sec²x - tanx = 1