Algebra Unit 5 WA
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University of the People *
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1101
Subject
Mathematics
Date
Jan 9, 2024
Type
docx
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Uploaded by DeaconFieldMagpie39
1.
A retirement account is opened with an initial deposit of $8,500 and earns
8.12% interest compounded monthly. What will the account be worth in
20 years? What if the deposit was calculated using simple interest? Could
you see the situation in a graph? From what point one is better than the
other?
A = P (1+ r/n)^nt
A = 8500 (1+ 0.0812/12)^12*20
A = 42,888.18
After 20 years the account is worth 42,888.18 dollars
Calculated using simple interest:
A = P (1+ rt)
A = 8500 (1+ 0.0812*20)
A = 22304
After 20 years, if the deposit was calculated using simple interest, it would
only be 22304 dollars.
The compounded interest is clearly better than simple interest.
2.
Graph the function
and its reflection about the line
y=x on the same axis, and give the x-intercept of the reflection. Prove that
. [Suggestion: type
{- 7 < x < 2}
{0 < y <
7} in desmos, and then type its inverse function.]
The x-intercept is (5, 0).
e^xlna = e^lna^x. It can also be written as e^xlna = e^lna^x.
3.
How long will it take before twenty percent of our 1,000-gram sample of
uranium-235 has decayed? [See Section 6.6 Example 13]
The decay equation is
, where t is the time for the decay,
and
K
is the characteristic of the material. Suppose
T
is the time it takes for
half of the unstable material in a sample of a radioactive substance to decay,
called its half-life. Prove that
. What is
K
for the uranium-235?
Show the steps of your reasoning.
A(t) = 800, because its decayed 20 percent
0.8 = e^Kt
T = 703,800,000
K = ln(0.5)/T
K = -9.8486385416 * 10^-10
t = ln(0.8)/K
t = 226572993.182 Million years
A(t) = A0e^Kt
A(t) = Ao/2
Ao/2 = Aoe^Kt
1/2 = e^Kt
K = ln(1/2)/T
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