<p>would 3 or 4 wrong still be an 800 , also what is the max possible wrong (approx) with no omits you can have for an 800
thanks</p>
<p>no it should be 800</p>
<p>also does anyone remember that problem with like the workers, and how we had to find an expression for the increase in number of years? wat did u put?</p>
<p>I think the maximum is ~4 wrong (or 5 omits, no wrong)</p>
<p>A(1.05^t)
its an exponential decay/growth graph.
you use those type of stuff in calc alot</p>
<p>1.20^t (A)…something like that</p>
<p>3 wrong, 2 omits…there goes my 800…</p>
<p>The symmetry one was the -tan x one. Everything else was symmetric to either the x or y axis.</p>
<p>The factory workers one was A(1.05)^t = P I believe.</p>
<p>The odd number product was 1/4 (1/2 to get an odd m, 1/2 to get an odd p, the only way to get an odd number from multiplication is odd x odd)</p>
<p>No, 3 wrong and 2 omit is a definite 800!</p>
<p>I think i can confirm the symmetry one. I mentally flipped all the others and knew that they were not the same as the original version. I physically flipped the tanx-like graph and that seems right. I saw the proctor looking at me…</p>
<p>ok good i got the same for the exponential growth one. well i have three wrong right now(number 49, ax2+bx+c, and the range one) i hope i dont get any more wrong, i really want that 800!!</p>
<p>3 wrong + 2 omit would be the same as 6 omits, putting you at risk for a 790.</p>
<p>isn’t the raw score 43 = 800?</p>
<p>Also remember that a lot of junior kids will take this test and probably 2 more like history and physics or a language just because the school recommends it and they just had a year of class on it. They mostly likely won’t prepare outside of their class material (it’s near the finals anyways) so the average score should be less than that of may or march, maybe?</p>
<p>lol ^ I’m pretty sure that one was the tanx-like graph too.</p>
<p>I have a question about the rectangle. I think they gave you a rectangle with and coordinates (6,0) and (0,4) on the rectangle. Then they asked you to find c of y = mx + c (so basically the y-int) of the line that would divide the rectangle into two congruent parts. Unless I’m talking about the wrong question, everyone else has said -1.</p>
<p>Wouldn’t it be 0? if the rectangle lies on the x and y axes … wouldn’t c just cross the rectangle at (0,0) and (6,4) and make two triangles?</p>
<p>Also, I forget which # this was, but it had something like
I. (sinx)^2 + (cosx)^2 = 1
II. 2(sinx)^2 + (cosx)^2 = 1
III. (sinx)^2 + 2(cosx)^2 = 2</p>
<p>Was the answer E (II & III)? (I’m not really sure what the first conditional was, exactly … :P)</p>
<p>yes i put 2 and 3</p>
<p>got that too ^
no to 0, because it’d have to cross at 6,6 not 6,4</p>
<p>I don’t know about that…
There were lots of sophomores in my math prep class that were getting like 46’s as their raw score haha.
I think the book we used prepared us pretty well, too, because I didn’t find today’s test too hard</p>
<p>No as it’s slope 1.</p>
<p>First conditional was (sinx)^2 + (cosx)^2 = 2.</p>
<p>0 is incorrect, but II and III is correct.</p>
<p>Astropuff, what was the question asking?
Which ones are wrong, or which ones are right?</p>
<p>I think only roman numberal I is correct…</p>
<p>m^91 + 2m^90
m^91=(m^90)(m^1) right; now you cam factor</p>
<p>(m^90)(m^1) + 2m^90
=(m^90)(2+m)</p>
<p>ok so both brackets need to be a square
m^90=(m^45)^2 so thats good</p>
<p>to make 2+m a square, the only answer choice was 34 as 2+34=6^2</p>