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Sagot :
a) 436529 cats b) Approximately 278 years
1) Gathering the data
400,000 cats
Increases yearly by 2.5%
2) Let's write that growth as a function. Note that we must rewrite 2.5% as purely decimal 0.0025. A growth of 2.5 must be written as 1.0025.
Because every time we multiply by 1.0025 we are multiplying the number and 2.5%. Considering that there are currently, in this 1st year 400,000 cats 2036 then this will be 35 years after
[tex]\begin{gathered} y=400000(1.0025)^n \\ y=400000(1.0025)^{35} \\ y=436529.23\text{ }\cong436,529\text{ } \end{gathered}[/tex]So considering we're in the first year, 35 years after in 2036 there'll be 436,529
b) Since n= is the number of years in that function, and y stands for the number of cats.
[tex]\begin{gathered} 800,000=400,000(1.0025)^n \\ \frac{800,000}{400,00}=\frac{400,000}{400,000}(1.0025)^n \\ 2=(1.0025)^n \\ \log 2\text{ =}\log (1.0025)^n \\ 0.3=^{}n1.08\cdot10^{-3} \\ n=\frac{0.3}{1.08\cdot10^{-3}} \\ n=277.8 \\ \end{gathered}[/tex]So, it will take at this rate approximately 278 years for the population of cats doubles.
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