tap a was turned on to fill a rectangular tank of 50cm by 40cm by 28cm with water at a rate of 6 liters per minute. after 2 minutes, tap b was turned on to drain water from the tank at a rate of 2 liters per minutes. 6 minutes after tap b was turned on, both taps were tuned off. find the depth of water left in the tank

Tap a filled for 8 mins=8x6=48 lit.

Tap b drained for 6 mins=6x2=12 lit.
Water remaining=48-12=36 lit=36000ml
Ht of water in tank=36000/(50x40)=36/2=18 cm

A full tank will hold 56 L

Using the calculations of anonymous of 36 L of water after both taps are turned of,

h/28 = 36/56
h = 28(36/56) = 18

the water is 18 cm deep

Trig aplication meseramants width of lake

Haw can solev this qen trig application measruments width of lake?

To find the depth of water left in the tank, we need to consider the flow rates and the time for each tap.

First, let's calculate the initial volume of water in the tank:
Volume = Length x Width x Height
Volume = 50 cm x 40 cm x 28 cm
Volume = 56,000 cm³

Now, let's calculate the amount of water flowing in and out for each tap:

Tap A: Flow rate = 6 liters per minute
After 2 minutes, the amount of water filled by tap A = Flow rate x Time = 6 liters/min x 2 min = 12 liters

Tap B: Flow rate = 2 liters per minute
After 6 minutes, the amount of water drained by tap B = Flow rate x Time = 2 liters/min x 6 min = 12 liters

Since the amount of water filled by tap A and drained by tap B is the same, the net change in volume is zero.

Final volume = Initial volume + amount of water filled - amount of water drained
Final volume = 56,000 cm³ + 12 liters - 12 liters = 56,000 cm³

Now, let's calculate the depth of water left in the tank:
Depth = Final volume / (Length x Width)
Depth = 56,000 cm³ / (50 cm x 40 cm)
Depth = 56,000 cm³ / 2,000 cm²
Depth = 28 cm

Therefore, the depth of water left in the tank is 28 cm.