Uche pumps gasoline at a rate of 18\,\dfrac{\text{L}}{\text{min}}

Question

What is Uche's pumping rate in \dfrac{\text{mL}}{\text{s}}

Answer 1

First, we need to convert the rate from liters per minute to milliliters per second.

To do this, we start by converting from liters to milliliters:
1 liter is equal to 1000 milliliters.

So, 18 liters per minute is equal to:
18 liters * 1000 milliliters = 18000 milliliters per minute

Next, we need to convert from minutes to seconds:
1 minute is equal to 60 seconds.

So, 18000 milliliters per minute is equal to:
18000 milliliters / 60 seconds = 300 milliliters per second

Therefore, Uche's pumping rate is 300 \, \dfrac{\text{mL}}{\text{s}}.

Answer 2

-\dfrac{2}3p+\dfrac{1}5-1+\dfrac{5}6p

Answer 3

To simplify the expression, we first combine like terms:

We have:
-\dfrac{2}{3}p + \dfrac{1}{5} - 1 + \dfrac{5}{6}p

Combine the p terms:
-\dfrac{2}{3}p + \dfrac{5}{6}p

To combine these terms, we first find a common denominator for the coefficients of p, which is 6 (the least common multiple of 3 and 6):
-\dfrac{4}{6}p + \dfrac{5}{6}p

Now, combine the p terms:
-\dfrac{4}{6}p + \dfrac{5}{6}p = \dfrac{1}{6}p

Now, bring down the remaining constants:
\dfrac{1}{5} - 1

Combine the constants:
\dfrac{1}{5} - \dfrac{5}{5} = \dfrac{1}{5} - \dfrac{5}{5} = \dfrac{-4}{5}

Putting it all together, we have:
-\dfrac{2}{3}p + \dfrac{1}{5} - 1 + \dfrac{5}{6}p = \dfrac{1}{6}p + \dfrac{-4}{5}