Friday, April 14, 2017

Algebra 2 Problems of the Day

The following problems were taken from the ALGEBRA II (Common Core) Regents Exam given on Friday, January 27, 2017.
Previous problems can be found here

Part 1

13. The price of a postage stamp in the years since the end of World War I is shown in the scatterplot below.

The equation that best models the price, in cents, of a postage stamp based on these data is

(3) y = 1.43(1.04)x

Looking at the graph, you can see that it has an exponential shape, starting somewhat flat and then increasing more quickly. This eliminates choices (1) and (4). (And, really, a sine graph would be a silly choice here.) If you put the other two into your graphing calculator and expand the size of the window, you will see that choice (2) increases much more quickly than the graph in the question. Choice (3) is a better fit.
If you have a problem expanding the window size, you can also check the Tables of Values to see that (2) shoots up to quickly.

14. The eighth and tenth terms of a sequence are 64 and 100. If the computations. sequence is either arithmetic or geometric, the ninth term can not be

(1) -82

In an arithmetic sequence, the common difference can be found be subtracting the 8th term from the 10th term and dividing by 2 (which is 10 - 8). If we do that, we get (100-64)/2, which is 18.
64 + 18 = 82, which eliminates choice (4). (At this point, you might suspect that -82 is likely the answer.)
In a geometric sequence, the common ratio can be found by dividing the 10th term by the 8th term and then taking the square root (because the ratio is applied twice to get from the 8th term to the 10th term).
So (100/64)^.5 = +1.25, which are the common ratios. It is important to remember that the square root may positive or negative.
Then 64 * 1.25 = 80, and 80 * 1.25 = 100
And 64 * (-1.25) = -80 and -80 * 1.25 = 100.
This eliminates choices (2) and (3).

Continue to the next problems

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