Answer:Depending on how high u do it either by 0,25,50 or 0,50,100
Step-by-step explanation:
(3c)3 beacause c(3)is multiplies by an numerical number
3/5a = 1/4....multiply both sides by 5/3, cancelling out the 3/5 on the left
a = 1/4 * 5/3
a = 5/12
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
![\frac{L}{min}](https://tex.z-dn.net/?f=%5Cfrac%7BL%7D%7Bmin%7D)
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:
Step-by-step explanation:
((14×3) + 1 )/3 = (43/3)