If a cup is a regular sized cup that is 2 deciliters, and 1 liter equals 10 deciliters, that means that 1 liter = 5 cups, which means that 3 liters = 15 cups. If it is a cup that has 0.33 dl, then the correct number would be 9 cups.
LCM = 1260
The lowest common factor is 1260
Hope this helps! :)
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Answer:
a) E = 6500 -50d
b) 5000 kWh
c) the excess will last only 130 days, not enough for 5 months
Step-by-step explanation:
<u>Given</u>:
starting excess (E): 6500 kWh
usage: 50 kWh/day (d)
<u>Find</u>:
a) E(d)
b) E(30)
c) E(150)
<u>Solution</u>:
a) The exces is linearly decreasing with the number of days, so we have ...
E(d) = 6500 -50d
__
b) After 30 days, the excess remaining is ...
E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days
__
c) After 150 days, the excess remaining would be ...
E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system
The supply is not enough to last for 5 months.
Answer:
V'(t) = 
If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.
Step-by-step explanation:
Given:
V = 
, where 0≤t≤40.
Here we have to find the derivative with respect to "t"
We have to use the chain rule to find the derivative.
V'(t) = 
V'(t) = 
When we simplify the above, we get
V'(t) =
If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.
