A large pipe can fill a tank in 6 hours less than it takes the small pipe. Working together, they can fill it in #4# hours. How long would it take the small pipe to fill the tank if it was working alone?
Solve this equation.
We can now eliminate the denominators.
Hopefully this helps!
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Let ( x ) represent the number of hours it takes for the small pipe to fill the tank alone.
The large pipe takes ( x - 6 ) hours to fill the tank alone.
When they work together, their combined rate is the sum of their individual rates. Therefore, the combined rate is ( \frac{1}{x} + \frac{1}{x-6} ).
Given that they can fill the tank together in 4 hours, their combined rate is ( \frac{1}{4} ).
Setting up the equation ( \frac{1}{x} + \frac{1}{x-6} = \frac{1}{4} ), solve for ( x ) to find the time it takes for the small pipe to fill the tank alone.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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