Register
The temperature of a pot of water was 180.3°F and it cooled at a rate of -2.5°F for a minute after 20 minutes what was the temperature
62.0k views
3 votes
The temperature of a pot of water was 180.3°F and it cooled at a rate of -2.5°F for a minute after 20 minutes what was the temperature

User Kiran K Telukunta
by
8.0k points

1 Answer

5 votes
Temperature after 20 minutes is found by multiplying-2.5 by 20 since it goes down by 2.5 degrees for one minute. -2.5 * 20 = -50. Adding this to 180.3 will give the temperature 130.3 degrees.
User Lars Graubner
by
7.7k points
← Prev Question Next Question →

Related questions

Ask a Question
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.5m questions

12.2m answers

Other Questions

How do you can you solve this problem 37 + y = 87; y =
What is .725 as a fraction
A bathtub is being filled with water. After 3 minutes 4/5 of the tub is full. Assuming the rate is constant, how much longer will it take to fill the tub?
Write words to match the expression. 24- ( 6+3)
A dealer sells a certain type of chair and a table for $40. He also sells the same sort of table and a desk for $83 or a chair and a desk for $77. Find the price of a chair, table, and of a desk.
Link Copied!