Algebra 2 Problems of the Day

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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.


4. When a ball bounces, the heights of consecutive bounces form a geometric sequence. The height of the first bounce is 121 centimeters and the height of the third bounce is 64 centimeters. To the nearest centimeter, what is the height of the fifth bounce?
(1) 25
(2) 34
(3) 36
(4) 42

Answer: (2) 34
We know a₁ = 121 and a₃ = 64.
The common ratio, r = 64 / a₂ or a₂ / 121
Then r² = (64 / a₂) (a₂ / 121) = 64 / 121
And r = SQRT(64/121) = 8/11

To get from a₃ to a₅, you need to multiply by the common ratio two more times (or multiply by r²).
64 * (8/11)*(8/11) = 33.851... = 34

5. The solutions to the equation 5x² - 2x + 13 = 9 are
(1) 1/5 + SQRT(21)/5
(2) 1/5 + SQRT(19)/5 i
(3) 1/5 + SQRT(66)/5 i
(4) 1/5 + SQRT(66)/5

Answer: (2) 1/5 + SQRT(19)/5 i
If 5x² - 2x + 13 = 9
then 5x² - 2x + 4 = 0
If you graph this, you will see that there are no real roots, and you can eliminate (1) and (4).
Calculate the discriminate, b² - 4ac = (-2)² - 4(5)(4) = 4 - 80 = -76
SQRT(-76) = SQRT(-1 * 4 * 19) = 2i * SQRT(19), which elminates choice (3).

x = (-b + SQRT (b² - 4ac) ) / (2a)
x = ( -(-2) + SQRT (-76) ) / (2*5)
x = ( 2 + 2i SQRT (19)) / (10)
Split the fraction
x = 2/10 + 2i SQRT (19) / 10
x = 1/5 + SQRT(19)/5 i

6. Julia deposits $2000 into a savings account that earns 4% interest per year. The exponential function that models this savings account is y = 2000(1.04)^t, where t is the time in years. Which equation correctly represents the amount of money in her savings account in terms of the monthly growth rate?
(1) y = 166.67(1.04)^0.12t
(2) y = 2000(1.01)^t
(3) y = 2000(1.0032737)^12t
(4) y = 166.67(1.0032737)^t

Answer: (3) y = 2000((1.0032737)^12t
If you take the 12th root (1/12 power) of 1.04, you get 1.00327373978...
Conversely, if you raise 1.0032737 to the 12th power, you will get 1.039999... or 1.04.



Comments and questions welcome.

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