{"id":46812,"date":"2025-07-02T08:38:13","date_gmt":"2025-07-02T08:38:13","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=46812"},"modified":"2025-07-02T08:38:15","modified_gmt":"2025-07-02T08:38:15","slug":"a-can-of-soda-is-placed-inside-a-cooler-2","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-can-of-soda-is-placed-inside-a-cooler-2\/","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 (+) in degrees Celsius is given by the following function, where x is the number of minutes since the can was placed in the cooler: T(x) = -19 + 41e^(0.051x) Find the initial temperature of the soda and its temperature after 18 minutes. Round your answers to the nearest degree as necessary. Initial temperature: Temperature after 18 minutes:<\/p>\n\n\n\n

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

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

Here are the correct answers and the explanation.<\/p>\n\n\n\n

Initial temperature:<\/strong> 22 \u00b0C
Temperature after 18 minutes:<\/strong> -2 \u00b0C<\/p>\n\n\n\n

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

The problem provides a function, T(x) = -19 + 41e^(-0.05x), which describes the temperature of a can of soda in degrees Celsius after a certain number of minutes, x, inside a cooler. We need to find two specific values: the initial temperature and the temperature after 18 minutes.<\/p>\n\n\n\n

1. Finding the Initial Temperature<\/strong><\/p>\n\n\n\n

The “initial temperature” refers to the temperature at the very beginning, when the can is first placed in the cooler. In the context of the function, this corresponds to a time of zero minutes. Therefore, to find the initial temperature, we need to calculate T(0) by substituting x = 0 into the equation.<\/p>\n\n\n\n

The calculation is as follows:
T(0) = -19 + 41e^(-0.05 * 0)
T(0) = -19 + 41e^(0)<\/p>\n\n\n\n

A fundamental rule in mathematics is that any non-zero number raised to the power of 0 is equal to 1. So, e^0 = 1.<\/p>\n\n\n\n

T(0) = -19 + 41(1)
T(0) = -19 + 41
T(0) = 22<\/p>\n\n\n\n

The initial temperature of the soda is 22\u00b0C. Since this is a whole number, no rounding is necessary.<\/p>\n\n\n\n

2. Finding the Temperature after 18 Minutes<\/strong><\/p>\n\n\n\n

To find the temperature after 18 minutes, we substitute x = 18 into the function.<\/p>\n\n\n\n

The calculation is:
T(18) = -19 + 41e^(-0.05 * 18)<\/p>\n\n\n\n

First, we calculate the value of the exponent:
-0.05 * 18 = -0.9<\/p>\n\n\n\n

Now, we substitute this back into the equation:
T(18) = -19 + 41e^(-0.9)<\/p>\n\n\n\n

Using a calculator, we find the value of e^(-0.9):
e^(-0.9) \u2248 0.40657<\/p>\n\n\n\n

Next, we multiply this value by 41:
T(18) \u2248 -19 + 41(0.40657)
T(18) \u2248 -19 + 16.66937<\/p>\n\n\n\n

Finally, we perform the addition:
T(18) \u2248 -2.33063<\/p>\n\n\n\n

The problem asks to round the answer to the nearest degree. The value -2.33063 is closer to -2 than to -3. Therefore, the temperature after 18 minutes, rounded to the nearest degree, is -2\u00b0C.<\/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 (+) in degrees Celsius is given by the following function, where x is the number of minutes since the can was placed in the cooler: T(x) = -19 + 41e^(0.051x) Find the initial temperature of the soda and its temperature […]<\/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-46812","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46812","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=46812"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46812\/revisions"}],"predecessor-version":[{"id":46824,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46812\/revisions\/46824"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=46812"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=46812"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=46812"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}