<p>^ can I get something clarified? The CB official practice test is actually… easier than the actual thing? O_O</p>
<p>i have sum qs:
- the number of roots a polynomial has can be determined by the highest degree?
i.e. x^3+2x-1 has a total of 3 roots (imaginary and/or real)?
in other words: for an nth degree polynomial, are there “n” roots? - have there been questions abt the rational root test?</p>
<p>thanks</p>
<p>not even reviewed in my princeton book so i doubt</p>
<p>Anyone?
40. In an arithmetic sequence, a5 = a10 – 3 and a3 = –2. Between which two consecutive terms does 0 lie?</p>
<pre><code>(A) a4 and a5
(B) a5 and a6
(C) a6 and a7
(D) a7 and a8
(E) a9 and a10
</code></pre>
<p>Was in Sparknotes but their explanation is pretty confusing.</p>
<p>ok so each “jump” to the next term is 0.6 (3/5 <– 5 terms to the 10th term from the 5th)
a3= -2
a4 = -1.4
a5 = -.8
a6= -.2
a7= .4</p>
<p>answer is C (a6-a7)
right?</p>
<p>Yeah, thanks. Couldn’t figure out how the explanation got 0.6 for the difference</p>
<p>I have trouble understanding math questions online… :S</p>
<p>Anyone know which math practice exams are the most reflective of the actual exam?</p>
<p>Pretty much any question on this test can be done on a TI84.</p>
<br>
<br>
<p>This is a even function
(1,2)(4,7)(-1,2)(0,4)(-4,7)</p>
<p>okay if it was</p>
<p>(1,2)(4,7)(-1,2)(-4,7)</p>
<p>then i’d say its even</p>
<p>where the heck does the (0,4) come in?</p>
<p>ohh because if its on the yaxis it doesnt matter
like x^2 (it has point (0,0))
it is is still an even function if it has apoint on the yaxis</p>
<p>i’ve been doing the PR and Barron’s practice tests without graphing calculator
and it seems to be ok…
i’ve been getting 800 in PR and like 700 in Barrons lol.
but my real question is. DO WE REALLY NEED GRAPHING CALCULATORS like really really need them?</p>
<p>okay so this is even
(1,2)(4,7)(-1,2)(0,4)(-4,7)</p>
<p>would this still be even if i did this?
(1,2)(4,7)(-1,2)(1,4)(-4,7)</p>
<p>^ I don’t think so; even functions are symmetrical about the y-axis. That is, the graph on the left side of the y-axis is the mirror image of the one on the right. You’ll notice that (1,2) is the mirror image of (-1,2); (4,7) corresponds to (-4,7). (0,4) doesn’t need a partner because it’s on the y-axis. Therefore, by changing that to (1,4), you move it to the right of the y-axis. Now that it does not have a “partner” on the left-hand side of the y-axis, the function is not perfectly symmetrical about the y-axis, and is thus not even. Make sense? :)</p>
<p>Side note: anyone know where I can find a realistic online Math II practice test? I haven’t studied for this other than looking over Sparknotes’ online stuff [which looks suspiciously easy] and want to know if I should do more. I’ve been an A-student in Pre-Cal, Trig, Geometry and most of it makes sense to me :/</p>
<p>Is this a typo in the Sparknotes practice test?</p>
<p>[URL=<a href=“http://img4.imageshack.us/my.php?image=73650343.jpg][IMG]http://img4.imageshack.us/img4/1941/73650343.jpg[/IMG][/URL”>ImageShack - Best place for all of your image hosting and image sharing needs]
http://img4.imageshack.us/img4/1941/73650343.jpg
[/URL</a>]</p>
<p>The answer is A, but I don’t see why it’s not B. Sparknotes explanation for this problem states that both graphs have a range of (-5,5). Someone clarify? </p>
<p>Thanks.</p>
<p>p.s. Yes, I was desperate enough to printscreen the test question</p>
<p>How is princeton review score range compared to what people get on the actual test?</p>
<p>and also, someone please solve this: </p>
<p>"If n is an integer, what is the remainder when </p>
<p>3x (2[1]) - 4x (2[2]) + 5x(2[3]) - 8 </p>
<p>is divided by x+1"?</p>
<p>The answer key says -20</p>
<p>but I have no idea why! If you plug in x = -1, then… aren’t you just ignoring the (2[4]) and everything?</p>
<p>Thanks</p>
<p>For the range question; I think the answer has to do with the fact that B) has horizonal asymptotes [it seems] at y = ±5 but A) only has it at +5?</p>
<p>Forgotten all of polynomial division, I should review that.</p>
<p>Whenever you are dividing a polynomial you can just plug it in.</p>
<p>If you’re dividing a polynomial by x - a then x - a is a factor if and only if P(a) = 0
You can also just plug it in to find the remainder - if you were dividing by x - 3 just evaluate P(3).</p>