Solve the problem.A lake is stocked with 670 fish of a new variety. The size of the lake, the availability of food, and the number of other fish restrict growth in the lake to a limiting value of 4187. The population of fish in the lake after time t, in months, is given by the function, . Find the population after 10 month(s).
Category: Uncategorized
Convert to an exponential equation.log464 = t
Convert to an exponential equation.log464 = t
Solve the logarithmic equation. log (x + 10) – log (x + 4) =…
Solve the logarithmic equation. log (x + 10) – log (x + 4) = log x
Solve the equation for x by first rewriting both sides as po…
Solve the equation for x by first rewriting both sides as powers of the same base. 8x-1 = 642x
Write the expression in expanded form.
Write the expression in expanded form.
Solve the problem.How long will it take for the population o…
Solve the problem.How long will it take for the population of a certain country to double if its annual growth rate is 4.9%? Round to the nearest year.
Solve the logarithmic equation. log ( 4+ x) – log (x – 3) =…
Solve the logarithmic equation. log ( 4+ x) – log (x – 3) = log 2
Solve the problem.The number of bacteria growing in an incub…
Solve the problem.The number of bacteria growing in an incubation culture increases with time according to n(t) = 9000(5)t where t is time in days. Find the number of bacteria when t = 0 and t = 4.
Determine the graph of the function.f(x) = 2.548x
Determine the graph of the function.f(x) = 2.548x
Solve the problem. A box contains a radioactive substance. …
Solve the problem. A box contains a radioactive substance. The number of kilograms, r(t), at time t years is given by r(t) = 2-0.002588t. How long will it take until only one-half kilogram of the radioactive substance is left in the box?