Unit 5
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University of the People *
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MATH 1201
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Mathematics
Date
Jan 9, 2024
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docx
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Uploaded by GrandLapwingMaster302
Calculate the future value of the retirement account with compound interest:
A = P(1 + r/n)^(nt)
A is the future value of the account.
P is the principal (initial deposit) amount.
r is the annual interest rate (as a decimal).
n is the number of times interest is compounded per year.
t is the number of years.
A = 8500(1 + 0.0812/12)^(12*20)
A ≈ $32,947.57
So, the account will be worth approximately $32,947.57 in 20 years with compound interest.
Calculated the deposit using simple interest:
A = P(1 + rt)
A is the future value of the account.
P is the principal (initial deposit) amount.
r is the annual interest rate (as a decimal).
t is the number of years.
A = 8500(1 + 0.0812*20)
A ≈ $17,700.00
So, the account will be worth approximately $17,700.00 in 20 years with simple interest.
To compare the two situations graphically, we can plot the growth of the account over time on a graph. The x-axis represents time (in years), and the y-axis represents the account balance.
For compound interest, the graph will show exponential growth, starting from the initial
deposit of $8,500 and increasing gradually over time until it reaches approximately $32,947.57 after 20 years.
For simple interest, the graph will show linear growth, starting from the initial deposit of
$8,500 and increasing at a constant rate until it reaches approximately $17,700.00 after 20 years.
Graph the function:
prove that a^x = e^(xlna):
y = a^x
Taking the natural logarithm (ln) of both sides:
ln(y) = ln(a^x)
ln(y) = xln(a)
Rewrite y using the definition of the natural exponential function:
y = e^(ln(y))
e^(ln(y)) = e^(xln(a))
e^(ln(y)) is equal to y:
y = e^(xln(a))
Hence, we have proved that a^x = e^(xlna).
determine the time it takes before twenty percent of the 1,000-gram sample of uranium-235 has decayed:
A(t) = A_0 * e^(Kt)
Where:
A(t) is the amount of the radioactive substance remaining at time t.
A_0 is the initial amount of the substance.
K is the decay constant.
t is the time.
200 = 1,000 * e^(Kt)
To find the time t, we need to solve this equation for t. To calculate K, we can use the equation K = ln(0.5) / T, where T is the half-life of the substance.
T = 703,800,000 years
K = ln(0.5) / T
K = ln(0.5) / 703,800,000
K ≈ -9.852e-10
Now, we can substitute K into the equation and solve for t:
200 = 1,000 * e^(-9.852e-10 * t)
0.2 = e^(-9.852e-10 * t)
Taking the natural logarithm (ln) of both sides:
ln(0.2) = -9.852e-10 * t
t = ln(0.2) / (-9.852e-10)
t ≈ 3.1402e+9
So, it will take approximately 3.1402 billion years for twenty percent of the 1,000-gram sample of uranium-235 to decay.
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