120k views
4 votes
Clearlake holds 1.3x10 to the 15th power gallons of water. If woodland removed water from the lake and it never rained again. How many days would the water last? Woodland uses 1.2 x 10 to the 9th power of water per day.

2 Answers

4 votes

Final answer:

The water in Clearlake would last approximately 1.3 million days if Woodland continued to use water at a rate of 1.2 x 10^9 gallons per day and it never rained again.

Step-by-step explanation:

To calculate how many days the water in Clearlake would last if Woodland removed water from the lake and it never rained again, we need to divide the total volume of water in the lake by the amount of water used per day by Woodland.

The total volume of water in Clearlake is 1.3 x 1015 gallons. Woodland uses 1.2 x 109 gallons of water per day. To find the number of days the water would last, we divide the total volume by the daily usage: Number of days = Total volume of water / Daily usage

Plugging in the values, we get: Number of days = 1.3 x 1015 gallons / 1.2 x 109 gallons per day

Using scientific notation, this calculation simplifies to: Number of days = 1.3 x 106 days

Therefore, the water in Clearlake would last approximately 1.3 million days if Woodland continued to use water at a rate of 1.2 x 109 gallons per day and it never rained again.

User Rewbs
by
7.9k points
3 votes
The water would last a very little amount of water
User Vivek Tyagi
by
8.7k points