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kirill [66]
3 years ago
8

A 100-gallon fish tank fills at a rate of x gallons per minute. The tank has already been filling for 5 minutes. The

Mathematics
1 answer:
vagabundo [1.1K]3 years ago
6 0

Answer:

was shifted 5 unit right side then stretched vertically by the factor 100, Shifted 5 unit right side then stretched vertically by the factor 100

Step-by-step explanation:

Here, parent function,

Transformed function,

Since, if a function f(x) gives cf(x), where c is any number,

Then the function was vertically stretched by factor c,

While, the function gives f(x-c),

Then the function was shifted c unit right.

Hence, the graph of  was shifted 5 unit right side then stretched vertically by the factor 100 to create the graph of

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Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and t
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Answer:

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.  

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Step-by-step explanation:

Use the initial condition when dp/dt = 1, p = 10 to get k;

\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}

Seperate the differential equation and solve for the constant C.

\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2}  }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}

You have 100 rodents when:

100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months

You have 1000 rodents when:

1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months

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3 years ago
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