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.

*19. Which graph represents a polynomial function that contains
x*^{2} + 2x + 1 as a factor?

*
*

**Answer: (1) see graph **

Factor x^{2} + 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 Lives | 1 | 2 | 3 | 4 | 5 |

Amount of Sodium Iodide-131 | 139.00 | 69.500 | 34.750 | 17.375 | 8.688 |

*
*

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 (x*^{2} - 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.

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