<p>While preparing for my subject tests, I only used SparkNotes (as test prep). I could not find information comparing their practice tests to the real thing, so here is what I think. (I got 2x 800).</p>
<p>The physics practice tests are relatively hard (more EM than the actual test), but are less 'wordy'. My scores were 720-740-750, and I reviewed by skimming their book. So it's a good preparation. Not all of their answers are correct though.</p>
<p>The real math test was slightly harder than the practice tests. Though that might vary from test to test. I hope someone else can help comparing, because the subject is not very hard for me (I'm a math major). I had 800s on the practice tests, but the real test seemed to have more trick questions. A graphing calculator is <em>very</em> nice to have.</p>
<p>If you have any specific questions, please ask. I know I wanted to do so.</p>
<p>I want to do the Physics practice tests from Sparknotes so badly but it isn’t letting me register. It says my cookies aren’t turned on, even when they are. What should I do?</p>
<p>I think the sparknotes math 2 tests were good preparation for the real thing. My scores were 700,760,780, and the computer screwed up the last two so I never got a score. The sparknotes ones were a little harder though. And I used the two college board tests and got a 760,800. I got an 800 on the real thing. So to answer your question, yes you can, as I believed I was well prepared.</p>
<p>Sparknotes is harder than the real thing, in my experience. Got 800 in the real one, but I never completed a full Sparknotes practice test. I used Princeton Review instead, which I thought was fantastic.</p>
<p>I was doing the second sparknotes practice test, but I didn’t understand four questions. Can somebody explain the answers to me please? :)</p>
<ol>
<li><p>Suppose the graph of y=f(x) has a vertical asymptote at x=4 and a horizontal asymptote at y=-1. If g(x)= 1/2f(x), which lines are the asymptotes of the graph of g(x)?
The answer is x=4, y=-1/2.</p></li>
<li><p>f(x)=3x^3+4x^2=x. What are all the values of x such that f(x) is greater than or equal to 0.
The answer is x is greater than or equal to -1 and less than or equal to -1/3 or x is greater than or equal to 0.
(I did this on my TI-84 and got an answer of only x is greater than or equal to 0…)</p></li>
<li><p>A two-sided coin is flipped four times. Given that the coin landed heads up more than twice, what is the probability that it landed heads up all four times?
The answer is 1/5.</p></li>
<li><p>In an arithmetic sequence, a (subscript 5)=a(subscript 10) -3 and a (subscript 3)=-2. Between which two consecutive terms does 0 lie?
The answer is a (sub 6) and a (sub 7). </p></li>
</ol>
<p>If you can do only a few of the four problems, that’s still ok. Thanks so much in advance!!</p>
<ol>
<li><p>Graph shrinks in the y-direction (If something happens when y = f(x) = -1, then it will happen when y = g(x) = 1/2 f(x) = -1/2) and does not change in the x-direction.</p></li>
<li><p>f(x) = 3x^3 + 4x^2 + x = x (x + 1)(3x + 1) so roots, or x-intercepts, are located at -1, -1/3, and 0. Because a function on the xy-plane can only change signs when it crosses the x-axis, the function can only change signs at these three roots, and thus we have four intervals to consider for different signs: (-infty,-1), (-1, -1/3), (-1/3, 0), and (0, infty). Our function has no double root; thus, at each x-intercept the graph will change sign, and each contiguous interval will have different signs. At (0, infty), f(x) > 0; it then follows that f(x) < 0 at (-1/3, 0) and f(x) > 0 again at (-1, -1/3) and so on. With equality condition we add the endpoints to our intervals where f(x) > 0 to get [-1, -1/3]∪[0, infty).</p></li>
<li><p>There are only four flips, so it is easier just to write out the possibilities with more than two heads (H for heads, T for tails). There are only 5: HHHT, HHTH, HTHH, THHH, and HHHH. We are looking for probability of HHHH, which is one out of five.</p></li>
<li><p>Let common difference d. Then we have
a<em>5 = a</em>10 - 3
=> (a<em>0 + 5d) = (a</em>0 + 10d) - 3
=> 5d = 3, or d = 3/5.
we know a<em>3 = -2 and thus a</em>n = -2 + 3/5 * (n-3).
and if we let a<em>n = 0, we can solve for 0 = -2 + 3/5 * (n-3)
(we can do this since arithmetic sequence is a linear function and thus always increasing or decreasing)
=> n = 6 1/3.
Thus it lies between a</em>6 and a_7.</p></li>
</ol>
<p>Sparknotes for Math II is definitely worth it. saves some cash too. Also, chemistry is good too if you are planning on taking it. I got like 690 on the practice test for Math II and ended up with 740 on the real thing. Chemistry i got like 590 but ended up with 690. perhaps the subject tests this year were easier or something but a 100 difference is awesome especially if you’re already a hardcore chem student which i never was.</p>