A company deposits $5,000 into an account that earns interest. The rate at which the value changes is given by dA/dt=0.0225A, where A(t) is the account value, in dollars, after t years. What will be the value of the account after 10 years if no additional deposits or withdrawals are made?

  $6261.61

Step-by-step explanation:

The solution to the differential equation is the exponential function ...

  A(t) = 5000e^(0.0225t)

We want the account value after 10 years:

  A(10) = 5000e^(0.225) = 6261.61

The value of the account after 10 years will be $6,261.61.

The rate of change equation basically tells you that interest is compounded continuously. After working interest problems for a while you know the formula for that is the exponential formula A = A0·e^(rt).

Or, you can solve the differential equation using separation of variables:

  dA/A = 0.0225dt

  ln(A) = 0.0225t +C . . . . integrate

  A(t) = A0·e^(0.0225t) = 5000·e^(0.0225t) . . . . solution for A(0) = 5000

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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