The amount of trout y (in tons) caught in a lake from 1995 to 2014 can be modeled by the equation y = -0.08×2 + 1.6x + 10, where x is the n

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The amount of trout y (in tons) caught in a lake from 1995 to 2014 can be modeled by the equation y = -0.08×2 + 1.6x + 10, where x is the number of years since 1995. When were about 15 tons of trout caught in the lake? The year and the year .

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    2021-09-14T17:30:36+00:00

    For this case, the first thing we want to do is substitute the value of y = 15 in the equation.
     We have then:
     15 = -0.08x ^ 2 + 1.6x + 10
     Rewriting we have:
     0 = -0.08x ^ 2 + 1.6x + 10 – 15
     0 = -0.08x ^ 2 + 1.6x – 5
     The solutions are:
     x1 = 3.876275643042054
     x2 = 16.123724356957947
     Nearest whole number:
     x1 = 4
     x2 = 16
     were about 15 tons of trout caught in the lake in the years:
     1995 + 4 = 1999
     1995 + 16 = 2011
     Answer:
     The year 1999 and the year 2011

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