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\(\pi\)
e
\(\frac{a}{b}\)
!
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→
(
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\(\sqrt{\,}\)
\(a^{b}\)
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\(\div\)
log
ln
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\(\times\)
cos
cos⁻¹
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sin
sin⁻¹
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tan
tan⁻¹
A radioactive substance has an initial mass of 1000 grams. Its half-life is one year, meaning it loses half of its mass every year through nuclear decay. Let \(u_n\) be the mass of the substance after \(n\) years.
Part A: Write the Explicit Formula
The formula for the mass remaining after \(n\) years is:
\(u_n =\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
Part B: Calculate a Future Value
How much of the substance will remain after 5 years? (Round to two decimal places)
\(u_5 = \)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
grams
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