Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

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January 2019, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


16. Savannah just got contact lenses. Her doctor said she can wear them 2 hours the first day, and can then increase the length of time by 30 minutes each day. If this pattern continues, which formula would not be appropriate to determine the length of time, in either minutes or hours, she could wear her contact lenses on the nth day?
(1) a₁ = 120; a_n = a_{n - 1} + 30
(2) a_n = 90 + 30n
(3) a₁ = 120; a_n = a_{n - 1} + 0.5
(4) a_n = 2.5 + 0.5n

Answer: (4) a_n = 2.5 + 0.5n
On the first day, a₁ must be 2 hours or 120 minutes.
In choice (4), 2.5 + 0.5(1) = 3 hours. The initial amount was already too big before anything was added to it.
Choice (1) starts with 120 minutes. Then adds 30 minutes to the previous day's total.
Choice (2) starts with 90 minutes plus an increment of 30 for a total of 120 on the first day
Choice (3) is the same as (1) except converted to hours.

17. If f(x) = a^x, where a > 1, then the inverse of the function is

(1) f^-1(x) = log_x a
(2) f^-1(x) = a log x
(3) f^-1(x) = log_a x
(4) f^-1(x) = x log a

Answer: (3) f^-1(x) = log_a x
In a^x, a is the base. When you take the log, a is the base.
Ex: If a=2, the f(3) = 2³ = 8
f^-1 ( f(3) ) = f^-1(8) = log _{2 8 = 3}

18. Kelly-Ann has $20,000 to invest. She puts half of the money into an account that grows at an annual rate of 0.9% compounded monthly. At the same time, she puts the other half of the money into an account that grows continuously at an annual rate of 0.8%.
Which function represents the value of Kelly-Ann’s investments after t years?

(1) f(t) = 10,000(1.9)^t + 10,000e^0.8t
(2) f(t) = 10,000(1.009)^t + 10,000e^0.008t
(3) f(t) = 10,000(1.075)^12t + 10,000e^0.8t
(4) f(t) = 10,000(1.00075)^12t + 10,000e^0.008t

Answer: (4) f(t) = 10,000(1.00075)^12t + 10,000e^0.008t
This is almost purely a notation problem.
The choices with .9 and .8 should be eliminated immediately.
Similarly, 0.75 is .9 / 12, so that's eliminated.
Because the first account is compounded monthly and there are 12 months in a year, the 0.009 is divided by 12, giving .00075, and the exponent gets multiplied by 12.



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