If sin(t) = 3/5 and t is in the 2nd quadrant, find cos(t).
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A linear equation can be written as:
y = a*x + b
where a is the slope and b is the y-intercept.
If we know that the line passes through the points (x₁, y₁) and (x₂, y₂) the slope is:
The answers are:
a) y = (-12 L/min)*t + 9,000 L
b) 6,840 L
a) Here we know that we can model this situation with a linear equation, we know that the leak starts at 6:00 am, then we define t = 0 as 6:00 am
We know that at 6:42 am (at t = 42 mins) the tank has 8,496 liters.
And at 7:15 am (at t = 75 mins) the tank has 8,100 liters.
Then we have two points of the line:
(42 mins, 8,496 liters)
(75 mins, 8,100 liters)
Then the slope is:
So the line will be something like:
y = (-12 L/min)*t + b
To find the value of b, which is the initial volume of water in the tank, we use the point (75 mins, 8,100 liters)
This means that if t = 75 mins, then we have y = 8,100 L
So we can replace these two on the above equation:
8,100 L = (-12 L/min)*75 mins + b
8,100L = -900L + b
8,100L + 900L = b = 9,000 L
Then the linear equation is:
y = (-12 L/min)*t + 9,000 L
b) Now we want to know how much water is in the tank at 9:00 am.
Notice that 9:00 am is 3 hours after 6:00 am, our t = 0.
Then 9:00am is equivalent to t = 180 mins.
So to know the volume of water that there is in the tank at 9:00 am, we just need to evaluate the linear function at t = 180 mins.
y = (-12 L/min)*180 mins + 9,000 L = 6,840 L
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