Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Hard math question. Please help!?

At the fair, Lisa stumbles upon a game called Cool Down. In this game, there is a container with 8 cups of hot water (at a temperature of 50 degrees Celcius). The object is to determine how many cups of cold water (at temperature of 20 degrees Celcius) to add to the container to reduce the temperature.

a. Find an equation expressing the temperature (degrees Celcius) of the water in the container as a function of the volume (cups) of cold water added.

PLEASE HELP IT WOUKD BE GREATLY APPRECIATED!

2 Answers

Relevance
  • 8 years ago

    Let x = volume of water at 20C

    8*50 + X*20 = (8 + X)T

    Where T is the new and lower Temperature

    T = (400 - 20X)/(8 + X)

    T = 20(20 - X)/(8 + X)

  • Anonymous
    5 years ago

    I have been surfing the internet more than 2 hours today seeking the answers to the same question, and I haven't found any interesting discussion like this. It is pretty worth enough for me.

Still have questions? Get your answers by asking now.