Answer:
i. C
ii. C
iii. A
iv. B
Explanation:
Exponential growth / decay can be shown using the formula below;
[tex]y=a(r)^{kt}[/tex]
Where;
[tex]\begin{gathered} y=\text{ final amount} \\ a=\text{ initial amount} \\ r=\text{exponential decay or growth.} \\ t=\text{time} \\ k=\text{constant} \end{gathered}[/tex]
The difference between exponential decay and growth is;
[tex]\begin{gathered} \text{ For exponential growth;} \\ r>0 \\ \text{ For exponential decay;} \\ r<0 \end{gathered}[/tex]
For the given question;
i.
[tex]y=(0.9)^{\frac{t}{2}}[/tex]
the values of r is;
[tex]\begin{gathered} r=0.9 \\ 0.9<1 \\ r<1 \end{gathered}[/tex]
Therefore, the function reveals exponential decay
C.
ii.
[tex]y=(0.81)^{6t}[/tex]
The same rule as in i applies here;
the values of r is;
[tex]\begin{gathered} r=0.81 \\ 0.81<1 \\ r<1 \end{gathered}[/tex]
Therefore, the function reveals exponential decay
C.
iii.
[tex]y=(1.08)^{t+6}[/tex]
the values of r is;
[tex]\begin{gathered} r=1.08 \\ 1.08>1 \\ r>1 \end{gathered}[/tex]
Therefore, the function reveals exponential growth
A.
iv.
[tex]y=(0.85)^t[/tex]
the values of r is;
[tex]\begin{gathered} r=0.85 \\ 0.85<1 \\ r<1 \end{gathered}[/tex]
it is an exponential decay.
The rate of decay is;
[tex]\begin{gathered} (1-r)\times100\text{\%} \\ =(1-0.85)\times100\text{\%} \\ =15\text{\%} \end{gathered}[/tex]
Therefore, the rate of decay of the function is 15%.
B.
v.
[tex]y=(\frac{1}{2})^t[/tex]
The value of r is;
[tex]\begin{gathered} r=\frac{1}{2}=0.5 \\ 0.5<1 \\ r<1 \end{gathered}[/tex]
It is an exponential decay.
The rate of decay is;
[tex]\begin{gathered} (1-r)\times100\text{\%} \\ =(1-0.5)\times100\text{\%} \\ =50\text{\%} \end{gathered}[/tex]
Therefore, the rate of decay of the function is 50%.
D.
