Answer:
v-12
Step-by-step explanation:
1Nonly_n8thanfull
So it seems as if the first problem is asking this:

.
If x=0.15, then 11/15*(0.15) is 0.11 or 11/100.
The second problem is asking:

To solve algebraically, you want to get rid of the denominators under the x's. To do so, you want to multiply by the Least Common Multiple (LCM) of the denominators. For 3 and 4, the LCM is 12. Multiply everything by 12 to get:
3x+6=4x+2
Now, solve for x
Subtract 4x from both sides:
-x+6=2
Subtract 6 from both sides:
-x=-4
Multiply by -1 on both sides:
x=4.
a. The water in the second tank decreases at a faster rate than the water in the first tank. The initial water level in the first tank is greater than the initial water level in the second tank.
Step-by-step explanation:
Step 1:
It is given that the time remaining in first tank is given by the equation y = -10x + 80. We can get the total water in the tank by substituting x = 0 in the equation. The total volume of water in first tank is 80 litres.
Step 2:
The value of y in the equation y = -10x + 80 will be 0 when the tank is fully empty. When y = 0 , 10x = 80, so x = 8. We can conclude that the first tank empties fully in 8 minutes.
In 8 minutes 80 litres of water is emptied from first tank. So the water in the first tank decreases at rate of 80 / 8 = 10 litres per minute
Step 3:
As per the given table for the second tank, 60 litres of water remains when x =0. So the total volume of water in the second tank = 60 litres.
Step 4:
As per the given table for the second tank, the volume becomes 0 in 5 minutes. In 5 minutes 60 litres of water is emptied from second tank. So the water in second tank decreases at rate of 60 / 5 = 12 litres per minute.
Step 5:
The initial volume of water in first tank is higher. The water in second tank decreases at a faster rate than the first tank.
Step 6:
The only correct option is:
a. The water in second tank decreases at a faster rate than the water in the first tank. The initial water level in first tank is greater than the initial water level in the second tank.