Wednesday, May 22, 2019

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.

More Algebra 2 problems.

January 2019, Part I

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


19. Which graph represents a polynomial function that contains x2 + 2x + 1 as a factor?


Answer: (1) see graph
Factor x2 + 2x + 1 into (x + 1)2.
This means that -1 is a both a zero on the graph and a turning point, which is shown in Choice (1).





20. Sodium iodide-131, used to treat certain medical conditions, has a half-life of 1.8 hours. The data table below shows the amount of sodium iodide-131, rounded to the nearest thousandth, as the dose fades over time.

Number of Half Lives12345
Amount of Sodium Iodide-131139.0069.50034.75017.3758.688

What approximate amount of sodium iodide-131 will remain in the body after 18 hours?

(1) 0.001
(2) 0.136
(3) 0.271
(4) 0.543

Answer: (3) 0.271
Note that the problem says that a half-life is 1.8 hours and that the top row of the table is the number of half-lives, not the number of hours.
There are 10 half-lives in 18 hours, because 18 / 1.8 = 10.
If you continue the table by entering 8.688 into your calculator and dividing by 2 five more times, you will get 4.344, 2.172, 1.086, 0.543, 0.271.

Also, you could put y = 139(1/2)x into your graphing calculator and check the Table of Values. Note: if you do this, you want to look at x = 9. Otherwise, you have to play with your original equation -- using 139 * 3 = 278 as the initial amount, or using (x - 1) as the exponent, with parentheses.





21. Which expression(s) are equivalent to (x2 - 4x) / 2x, where x =/= 0?

I. x / 2 - 2
II. (x - 4) / 2
III. (x - 1) / 2 - 3 / 2


(1) II, only
(2) I and II
(3) II and III
(4) I, II, and III

Answer: (4) I, II, and III
Test-taking tip: All four choices include choice II, so there is no need to check it. It's correct.
That being said, you can check to see if I or III are equivalent to II without using the original expression.

If you split the fraction (x - 4) / 2 = x / 2 - 4 / 2 = x / 2 - 2, which is I, so we can eliminate choices (1) and (3).
If you look at choice III, you can combine the fractions because they have a common denominator of 2:
(x - 1) / 2 - 3 / 2 = (x - 1 - 3) / 2 = (x - 4) / 2, which is II. II and III are the same.
So the answer is choice (4).



Comments and questions welcome.

More Algebra 2 problems.

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