Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Using the inverse function, it is found that the equation is:
[tex]P(t) = \frac{e^{0.1t}}{50} + 20[/tex]
It's initial value is 20.
---------------------------
We want to find the inverse of:
[tex]y = P^{-1}(t) = 10\ln{50t - 1000}[/tex]
To do this, we exchange y and t, and isolate y.
Then:
[tex]10\ln{50y - 1000} = t[/tex]
[tex]\ln{50y - 1000} = \frac{t}{10}[/tex]
[tex]e^{\ln{50y - 1000}} = e^{0.1t}[/tex]
[tex]50y - 1000 = e^{0.1t}[/tex]
[tex]50y = e^{0.1t} + 1000[/tex]
[tex]y = \frac{e^{0.1t} + 1000}{50}[/tex]
[tex]y = \frac{e^{0.1t}}{50} + \frac{1000}{50}[/tex]
[tex]P(t) = \frac{e^{0.1t}}{50} + 20[/tex]
It's initial value is:
[tex]P(0) = \frac{e^{0.1(0)}}{50} + 20 = 0.02 + 20 = 20.02[/tex]
Rounding, 20.
A similar problem is given at https://brainly.com/question/23950969
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.