The answer is $boxed{bf{option; (c)}}$

c) $-dfrac{128}{2187}$

a) $-5$, 23, $-61$, $-565$, 1703

$$
{text{a) }}{x^5} – 15{x^4} + 90{x^3} – 270{x^2} + 405x – 243
$$

Therefore, the half-life is $4.52$ days, the answer is (a).

a) $10.4%$

[begin{gathered}
{text{We shall calculate how much is the future value}} hfill \
{text{of an $ 800 quarterly payments for 10 years}} hfill \
{text{compounded quarterly at 8% /a}}{text{.}} hfill \
hfill \
R = 800 hfill \
r = 0.08 hfill \
{text{compounded quarterly}} hfill \
i = frac{{0.08}}{4} = 0.02 hfill \
t = {text{ 10 years}} hfill \
m = 10 times 4 = 40 hfill \
hfill \
{text{Using the formula for present value of annnuity}} hfill \
hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} hfill \
FV = 800 cdot frac{{{{left( {1 + 0.02} right)}^{40}} – 1}}{{0.02}} hfill \
FV = 48321.59 hfill \
hfill \
{text{The lump-sum payment is $ 50,000}} hfill \
50,000 – 48321.59 = boxed{{mathbf{$ 1678}}{mathbf{.41}}} hfill \
hfill \
{text{Therefore, Marisa will earn $ 1,678}}{text{.41 more}} hfill \
{text{if she would choose the lump-sum payment}}{text{.}} hfill \
hfill \
{text{The answer is option (b)}} hfill \
hfill \
end{gathered} ]

b) $$1,678.41$

[begin{gathered}
{text{The selling price of the television is $1894}}{text{. The }} hfill \
{text{finance plan comes with downpayment of $150 }} hfill \
{text{and monthly payment of $113 for 1}}{text{.5 years}}{text{.}} hfill \
{text{We shall calculate the annual interest rate}}{text{.}} hfill \
hfill \
{text{The amount that has to be financed is the difference}} hfill \
{text{between the selling price and the downpayment}} hfill \
{text{1894}} – 150 = 1744 hfill \
hfill \
{text{Thus, 1744 is the present value of the annuity with}} hfill \
{text{regular payments of }}R = 113 hfill \
{text{number of years of }}t = 1.5{text{ years}} hfill \
{text{compounded monthly}} hfill \
{text{number of compounding period of }}m = 1.5 times 12 = 1818 hfill \
{text{interest rate per compounding period of }}i = frac{r}{{12}}{text{ }} hfill \
hfill \
{text{Using the formulfor present value of annuity}} hfill \
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
1744 = 113 cdot frac{{1 – {{left( {1 + frac{r}{{12}}} right)}^{ – 18}}}}{{r/12}} hfill \
frac{{1744}}{{113}} = frac{{1 – {{left( {1 + frac{r}{{12}}} right)}^{ – 18}}}}{{r/12}} hfill \
hfill \
end{gathered} ]

Therefore, the annual interest rate is $20.06%$. The answer is option (c)

c) $20.06%$

[begin{gathered}
{text{First, we shall calculate how much is the downpayment}} hfill \
{text{This is equal to the future value of the annuity with}} hfill \
R = $ 300 hfill \
r = 6% /a hfill \
{text{compounded monthly}} hfill \
t = 4{text{ years}} hfill \
i = frac{{0.06}}{{12}} = 0.005 hfill \
m = 4 times 12 = 48 hfill \
hfill \
{text{Using the formula for the future value of annuity}} hfill \
FV = R cdot frac{{{{left( {1 + i} right)}^m} – 1}}{i} hfill \
FV = 300 cdot frac{{{{left( {1 + 0.005} right)}^{48}} – 1}}{{0.005}} hfill \
FV = 16229.35 hfill \
hfill \
{text{Since the selling price is $56,000, the difference between}} hfill \
{text{this and the downpayment is the amount that has to be financed}} hfill \
56,000 – 16229.35 = 39770.65 hfill \
hfill \
{text{We shall determine how long it takes for his monthly payment}} hfill \
{text{to achieve the present value of 39770}}{text{.65}} hfill \
R = 525 hfill \
r = 8% /a hfill \
{text{compounded monthly}} hfill \
i = frac{{0.08}}{{12}} hfill \
m = 12 times t hfill \
hfill \
{text{The formula for the present value of annuity is}} hfill \
hfill \
PV = R cdot frac{{1 – {{left( {1 + i} right)}^{ – m}}}}{i} hfill \
{text{39770}}{text{.65}} = 525 cdot frac{{1 – {{left( {1 + frac{{0.08}}{{12}}} right)}^{ – m}}}}{{0.08/12}} hfill \
hfill \
end{gathered} ]

Notice that if we add interest paid C3 and the principal paid B3$-$C3, the result is B3 which is the regular payment.

Thus, the regular payment must be the made up of the interest and principal paid.

Therefore, the answer is option (b).

Observe that the amount of drug in the body becomes stable at $514,7059$ mg which is at the 15th dose.

a) $t_1=350$, $t_n=350+0.32t_{n-1}$

b) $514.7059$ mg

c) $15th$ dose

[begin{gathered}
{text{Cowan would like to deposit $ 25 monthly}} hfill \
{text{for 18}} – 4 = 14{text{ years at }}r = 6% /a hfill \
{text{compounded monthly}}{text{.}} hfill \
{text{In this case,}} hfill \
R = 25 hfill \
i = frac{{0.06}}{{12}} = 0.005 hfill \
m = 14 times 12 = 168 hfill \
hfill \
{text{a) We need to show that the monthly payments }}R{text{ }} hfill \
{text{form a geometric series}}{text{. We know that geometric}} hfill \
{text{series must have a constant ratio between consecutive}} hfill \
{text{terms}}{text{. Since the monthly payment is constant, it is}} hfill \
{text{a geometric sequence with common ratio of 1}}{text{.}} hfill \
hfill \
{text{The value of the first deposit increases monthly}} hfill \
{text{with common ratio of 1}}{text{.005}}{text{. Therefore, the }} hfill \
{text{general term }}{t_n}{text{ is }} hfill \
hfill \
{t_n} = {t_1} cdot {r^{n – 1}} hfill \
{t_n} = 25{left( {1.005} right)^{n – 1}} hfill \
end{gathered} ]

[begin{gathered}
{text{b)}};{text{When Bart turns 18, 14 years have passed}} hfill \
{text{since Mr}}{text{. Cowan started investing}}{text{. Since he}} hfill \
{text{invests monthly, then he would have}} hfill \
m = 14 times 12 = 168{text{ payments}} hfill \
hfill \
{text{The balance in Bart’s fund after 168 payments is}} hfill \
{text{the sum of this geometric series with}} hfill \
{t_1} = 25{text{ and }}r = 1.005 hfill \
hfill \
{S_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
{S_{168}} = 25 cdot frac{{{{1.005}^{168}} – 1}}{{1.005 – 1}} = $ 6557.62 hfill \
end{gathered} ]

[begin{gathered}
{text{d) If Mr}}{text{. Cowan, deposited $50 instead of $25}} hfill \
{S_n} = {t_1} cdot frac{{{r^n} – 1}}{{r – 1}} hfill \
{S_{168}} = 50 cdot frac{{{{1.005}^{168}} – 1}}{{1.005 – 1}} = $ 13,115.24 hfill \
end{gathered} ]

a) $t_n=25(1.005)^{n-1}$

b) $168$ payments, $S_{168}=$6557.62$

c) see spreadsheet

d) $$13,115.24$