Wednesday, April 18, 2018

Algebra 2 Problems of the Day

Continuing with daily Algebra 2 questions and answers.

More Algebra 2 problems.

January 2018
17. The function below models the average price of gas in a small town computations. since January 1st.
G(t) = -0.0049t4 + 0.0923t3 - 0.56t2 + 1.166t + 3.23, where 0 ≤ t ≤ 10.

If G(t) is the average price of gas in dollars and t represents the number of months since January 1st, the absolute maximum G(t) reaches over the given domain is about

(1) $1.60
(2) $3.92
(3) $4.01
(4) $7.73

Answer: (3) $4.01
Graph the function and use "maximum" to find the highest value, which you should see is just above $4.00.
See the graph below:

At approximately t = 1.6, G(t) = 4.01, approximately.



18. Written in simplest form, (c2 - d2) / (d2 + cd - 2c2), where c =/= d, is equivalent to
(1) (c + d) / (d + 2c)
(2) (c - d) / (d + 2c)
(3) (-c - d) / (d + 2c)
(4) (-c + d) / (d + 2c)

Answer: (3) (-c - d) / (d + 2c)
The numerator, (c2 - d2), is the difference of two perfect squares, and factors into the conjugates, (c + d)(c - d).
Note that all four choices have (d + 2c) as the denominator, which makes factoring (d2 + cd - 2c2) that much easier into (d + 2c)(d - c).
(c - d) / (d - c) = -1, which reduces the fraction to (-1)(c + d) / (d + 2c).
Distribute the -1, and you get choice (3).



Comments and questions welcome.

More Algebra 2 problems.

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