Solve the problem.
-A company models its yearly expenses in millions of dollars using the equation $f ( t ) = 0.05 t ^ { 3 } - 0.6 t ^ { 2 } + 1.25 t + 2.5$ where $t = 0$ represents 1986 . The company's account manager decides to adjust the model so that $t = 0$ corresponds to 1991 rather than 1986 . To do this, she obtains $g ( t ) = f ( t + 5 )$. Use the Binomial Theorem to express $g ( t )$ in descending powers of $t$.
A) $0.05 \mathrm { t } ^ { 3 } + 0.15 \mathrm { t } ^ { 2 } - 1 \mathrm { t } + 0$
B) $0.05 \mathrm { t } ^ { 3 } - 1.35 \mathrm { t } ^ { 2 } + 11 \mathrm { t } + 2$
C) $0.05 t ^ { 3 } + 0.15 t ^ { 2 } + 11 t + 6$
D) $0.05 t ^ { 3 } - 1.35 t ^ { 2 } - 2.25 t + 12$ 3 Find a Particular Term in a Binomial Expansion
Write the first three terms in the binomial expansion, expressing the result in simplified form.

Question

Quizplus · Accepted Answer

Using the Binomial Theorem, we have:$$g(t) = f(t+5) = 0.05(t+5)^3 - 0.6(t+5)^2 + 1.25(t+5) + 2.5$$$$= 0.05(t^3 + 15t^2 + 75t + 125) - 0.6(t^2 + 10t + 25) + 1.25(t+5) + 2.5$$$$= 0.05t^3 + 0.75t^2 + 3.75t + 6.25 - 0.6t^2 - 6t - 15 + 1.25t + 6.25 + 2.5$$$$= 0.05t^3 + 0.15t^2 - 1t + 0$$

Solve the Problem f(t)=0.05t3−0.6t2+1.25t+2.5f ( t ) = 0.05 t ^ { 3 } - 0.6 t ^ { 2 } + 1.25 t + 2.5f(t)=0.05t3−0.6t2+1.25t+2.5

Solve the Problem $f ( t ) = 0.05 t ^ { 3 } - 0.6 t ^ { 2 } + 1.25 t + 2.5$