题目
MAT136H5 S 2025 - All Sections 2.8 Preparation Check
多重下拉选择题
Read Example 2.46 in the textbook Links to an external site. . Suppose a patient is given 300 mg of an anti-inflammatory medicine. The amount of medicine in the patients' body decreases exponentially, and it has a half-life in the body of 42 hours. a) Use the half-life formula to find the decay constant k. [ Select ] k = ln(2) / (42 * 300) k = 42 * ln(2) k = e^42 k = 42 * 300* ln(2) k = ln(2) / 42 k = (300* ln(2)) / 42 b) With this value for k, what is the correct model for this situation? [ Select ] y = 42 e^(-300 * ln(2) * t) y = 300 e^(-42/(ln(2)) * t) y = 300 e^(-(ln(2)/42) * t) y = 42 e^(-300 *42 *ln(2) * t) c) How much of the medicine is left after 6 hours? (Round to the nearest mg.) [ Select ] 245 mg 267 mg 272 mg 211 mg 198 mg d) After how many hours is there only 10% of the medicine remaining? (Round to the nearest hour.) [ Select ] 56 170 122 136 140
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思路分析
We will examine each part and each option carefully, explaining the reasoning for why an option is reasonable or not, and tying it to the relevant formulas.
For part a, the question asks to use the half-life formula to find the decay constant k.
Option a option: k = ln(2) / (42 * 300)
This expression multiplies 42 by 300 in the denominator, which does not align with the standard half-life decay constant formula. The half-life t1/2 relates to k via t1/2 = ln(2)/k, so solving for k gives k = ln(2)/t1/2, not involving the initial amount (300) at all. Thus this option incorrectly incorporates the amount and the product 42*300, making it dimensionally inconsistent with the usual derivation.
Option b option: k = 42 * ln(2)
Here, k is formed by multiplying the half-life by ln(2). This would yield a value with units inverse of time squared or something inconsistent with the standard k dimension (per time). The correct relationship is k = ......Login to view full explanation登录即可查看完整答案
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