More Algebra 2 problems.
June 2017, Part I
All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.
16. For x ≠ 0, which expressions are equivalent to one divided by the sixth
root of x?
Which explanation is appropriate for Miles and his dad to make?
1) I and II, only
2) I and III, only>
3) II and III, only
4) I, II, and III
Answer: 4) I, II, and III
"One divided by the sixth root of x" would be the same as "the negative sixth root of x".
The sixth root can be expressed with the fractional exponent 1/6.
Therefore, III is correct.
Again, keeping in mind that the sixth root is exponent 1/6, and third root is exponent 1/3, you can see that choices I and II are both the same. So either they are both correct, or they are both incorrect. This means that choice 4 is the answer.
However, let's show our work:
Because we are dividing the same base, we can subtract the exponents (1/6) - (1/3) = (-1/6). This means that choices I and II are equivalent to III, which is correct, so all three are correct.
17. A parabola has its focus at (1,2) and its directrix is y = -2. The
equation of this parabola could be
1) y = 8(x + 1)2
2) y = 1/8(x + 1)2
3) y = 8(x - 1)2
4) y = 1/8(x - 1)2
Answer: 4) y = 1/8(x - 1)2
The standard form of a parabola is y = (1/(4p))(x - h)2 + k, where the focus is (h, k + p) and the directrix is y = k - p. The vertex is (h, k), directly in the middle, and that must be (1, 0), which makes h = 1, k = 0 and p = 2. That makes 1/(4p) = 1/(4*2) = 1/8. Choice 4.
18. The function p(t) = 110e0.03922t models the population of a city, in
millions, t years after 2010. As of today, consider the following two
statements:
II. The population increases continuously by approximately 3.9% per year.
1) I, only
2) II only
3) both I and II
4) neither I nor II
Answer: 2) II only
The rate is given as 0.03922, which is approximately 3.9%.
However, 110 million was the population in 2010, not the "current" population -- which is odd phrasing since no specific time was given. Were students to assume 2017, and seven years after 2010? Thankfully, the Regents do not (usually?) use imprecise questions like this when the information matters!
Comments and questions welcome.
More Algebra 2 problems.
I get thrown off by the wording in question 18: "The population increases continuously by approximately 3.9% per year". I understand that population grows continuously, and that Pe^kt is the model for continuous compounding. What hangs me up is that after a year, the population has grown not by 3.922% but by e^0.03922 - 1 = 4.00%, which I believe would be called an "effective" annual interest rate if we were dealing with a bank account.
ReplyDeleteIs there a way to more clearly distinguish between a continuously compounded growth or decay rate and the effective rate?