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Klio2033 [76]
3 years ago
5

The rabbit population in an isolated forest rises and falls depending on the population of predators. Within the year 2014201420

14, the population, ppp, in thousands of rabbits, mmm months after January 111, 201420142014 is: \qquad p = 0.05(m-1.5)(m-8.5)+10p=0.05(m−1.5)(m−8.5)+10p, equals, 0, point, 05, (, m, minus, 1, point, 5, ), (, m, minus, 8, point, 5, ), plus, 10 The population reached ten thousand some time in February. At what other time, given as months after January 111, 201420142014, did the rabbit population reach ten thousand?
Mathematics
1 answer:
WITCHER [35]3 years ago
5 0

Answer:

Step-by-step explanation:

If within the year 2014, the population p of the rabbits m months after January 2014 is modeled by the equation p = 0.05(m-1.5)(m-8.5) and the population reach 10,000 some time in February, to determine the time, given as months after January 1 2014, that the rabbit population reach ten thousand, we will substitute p = 10 into the modeled equation and get the value of m as shown;

p = 0.05(m-1.5)(m-8.5)+10 \\substitute \ p = 10\\\\ 10 = 0.05(m-1.5)(m-8.5)+10\\\\10-10 =  0.05(m-1.5)(m-8.5)\\\\0 =  0.05(m-1.5)(m-8.5)\\\\0/0.05 =  (m-1.5)(m-8.5)\\\\0 =  (m-1.5)(m-8.5)\\ (m-1.5)(m-8.5) = 0\\ m-1.5 = 0 \ and \ m-8.5 = 0\\m = 1.5 \ and \ 8.5\\

Hence the population of the rabbit reach 10,000 after 1.5 months and 8.5 months

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<u />

<u>Step 2: Solve for </u><em><u>r</u></em>

  1. Substitute:                    77 = \frac{4}{3} \pi r^3
  2. Isolate <em>r</em> term:               \frac{77}{\frac{4}{3} \pi} = r^3
  3. Isolate <em>r</em>:                        \sqrt[3]{\frac{77}{\frac{4}{3} \pi}}  = r
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  5. Rewrite:                         r = 2.63917 \ cm

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<em>Differentiate the Volume Formula with respect to time t.</em>

  1. Define:                                                                                                            V = \frac{4}{3} \pi r^3
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