{"id":46825,"date":"2025-07-02T08:41:33","date_gmt":"2025-07-02T08:41:33","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=46825"},"modified":"2025-07-02T08:41:34","modified_gmt":"2025-07-02T08:41:34","slug":"a-can-of-soda-is-placed-inside-a-cooler-3","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-can-of-soda-is-placed-inside-a-cooler-3\/","title":{"rendered":"A can of soda is placed inside a cooler."},"content":{"rendered":"\n

A can of soda is placed inside a cooler. As the soda cools, its temperature T (x) in degrees Celsius is given by the following function, where x is the number of minutes since the can was placed in the cooler.
Find the temperature of the soda after 12 minutes and after 18 minutes. Round your answers to the nearest degree as necessary. Temperature after 12 minutes: \u00b0C Temperature after 18 minutes: \u00b0C<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Correct Answer:<\/strong><\/p>\n\n\n\n

Temperature after 12 minutes: 11<\/strong> \u00b0C
Temperature after 18 minutes: 8<\/strong> \u00b0C<\/p>\n\n\n\n

Explanation<\/h3>\n\n\n\n

The problem provides a function, T(x) = -4 + 26e^(-0.045x), which models the temperature of a can of soda in degrees Celsius after a certain amount of time. In this function, T(x) represents the temperature, and x represents the time in minutes that the soda has been in the cooler. To find the temperature at specific times, we need to substitute the given number of minutes for x in the equation and solve for T(x).<\/p>\n\n\n\n

Temperature after 12 minutes:<\/strong><\/p>\n\n\n\n

First, we calculate the temperature after 12 minutes by setting x = 12.<\/p>\n\n\n\n

    \n
  1. Substitute\u00a0x = 12\u00a0into the function:
    T(12) = -4 + 26e^(-0.045 * 12)<\/li>\n\n\n\n
  2. Calculate the value of the exponent:
    -0.045 * 12 = -0.54<\/li>\n\n\n\n
  3. The equation becomes:
    T(12) = -4 + 26e^(-0.54)<\/li>\n\n\n\n
  4. Using a calculator, find the value of\u00a0e^(-0.54):
    e^(-0.54) \u2248 0.5827<\/li>\n\n\n\n
  5. Now, substitute this value back into the equation:
    T(12) \u2248 -4 + 26 * 0.5827
    T(12) \u2248 -4 + 15.1502
    T(12) \u2248 11.1502<\/li>\n\n\n\n
  6. The problem asks to round to the nearest degree. Since the decimal is .15, which is less than .5, we round down to\u00a011 \u00b0C<\/strong>.<\/li>\n<\/ol>\n\n\n\n

    Temperature after 18 minutes:<\/strong><\/p>\n\n\n\n

    Next, we follow the same process for 18 minutes by setting x = 18.<\/p>\n\n\n\n

      \n
    1. Substitute\u00a0x = 18\u00a0into the function:
      T(18) = -4 + 26e^(-0.045 * 18)<\/li>\n\n\n\n
    2. Calculate the value of the exponent:
      -0.045 * 18 = -0.81<\/li>\n\n\n\n
    3. The equation becomes:
      T(18) = -4 + 26e^(-0.81)<\/li>\n\n\n\n
    4. Using a calculator, find the value of\u00a0e^(-0.81):
      e^(-0.81) \u2248 0.4449<\/li>\n\n\n\n
    5. Substitute this value back into the equation:
      T(18) \u2248 -4 + 26 * 0.4449
      T(18) \u2248 -4 + 11.5674
      T(18) \u2248 7.5674<\/li>\n\n\n\n
    6. Rounding to the nearest degree, the decimal is .56, which is greater than or equal to .5, so we round up to\u00a08 \u00b0C<\/strong>.<\/li>\n<\/ol>\n\n\n\n

      <\/p>\n\n\n\n

      \\<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      A can of soda is placed inside a cooler. As the soda cools, its temperature T (x) in degrees Celsius is given by the following function, where x is the number of minutes since the can was placed in the cooler.Find the temperature of the soda after 12 minutes and after 18 minutes. Round your […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-46825","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46825","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=46825"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46825\/revisions"}],"predecessor-version":[{"id":46838,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46825\/revisions\/46838"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=46825"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=46825"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=46825"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}