Its not a full question.... send the full question
Answer:
The liters that the tank will contain at 5:11 PM that day are:
Step-by-step explanation:
Firstly, you must identify the outlet flow of the water pumped from the tank, for this, you must subtract the last volume given from the first volume:
- 19,140 L - 8,097 L = 11,043 L
And the minutes that passed from the first volume until the last volume given (18 minutes from 4:47 PM to 5:05 PM), so, you must divide that two values to obtain the outlet flow:
- Outlet flow =
- Outlet flow =
- Outlet flow = 613.5
Now, you must see the next hour given (5:11 PM), if you see, from 5:05 PM to 5:11 PM has passed 6 minutes, taking into account this, you replace the equation of outlet flow to clear the volume:
- Outlet flow =
- Volume = Outlet flow * time
And replace the values to obtain the new volume pumped:
- Volume = 613.5 * 6 min
- Volume = 3681 L.
At last, you must subtract these liters from the last volume identified in the tank:
- New Volume in the tank = 8097 L - 3681 L
- New Volume in the tank = 4416 L
The volume in the tank at 5:11 PM is <u>4416 Liters</u>.
Answer:
A
Step-by-step explanation:
=2x+7-x^2+2
=-x^2+2x+9
Answer:
-8/45
Step-by-step explanation:
we can solve a/b by plugging in the given values and we have the new expression:
4/5 / -9/2
to solve, we can use multiply the two fractions together using KCF which stands for: Keep Change Flip. we keep the first fraction as is, change the division sign into a multiplication, and flip the 2nd fraction to its reciprocal (numerator becomes denominator and denominator becomes numerator)
4/5 / -9/2 becomes:
4/5 × 2/-9 < simply multiply the fraction
4 × 2 = 8
5 × -9 = -45
8/-45 is our new fraction. we cannot simplify this any further so this is our answer which can also be written as -8/45