Pre-Calc Homework Solutions 15

Pre-Calc Homework Solutions 15 - Section 1.3 22 15 x 1 2 3...

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22. 23. Let t be the number of years. Solving 500,000(1.0375) t 5 1,000,000 graphically, we find that t < 18.828. The population will reach 1 million in about 19 years. 24. (a) The population is given by P ( t ) 5 6250(1.0275) t , where t is the number of years after 1890. Population in 1915: P (25) < 12,315 Population in 1940: P (50) < 24,265 (b) Solving P ( t ) 5 50,000 graphically, we find that t < 76.651. The population reached 50,000 about 77 years after 1890, in 1967. 25. (a) A ( t ) 5 6.6 1 } 1 2 } 2 t /14 (b) Solving A ( t ) 5 1 graphically, we find that t < 38.1145. There will be 1 gram remaining after about 38.1145 days. 26. Let t be the number of years. Solving 2300(1.06) t 5 4150 graphically, we find that t < 10.129. It will take about 10.129 years. (If the interest is not credited to the account until the end of each year, it will take 11 years.) 27. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve A (1.0625) t 5 2 A , which is equivalent to 1.0625 t 5 2. Solving graphically, we find that
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