<p>Wasn’t the center of the circle (-2, 1)?</p>
<p>Because the equation was: (x+2)^2 + (x-1)^2 = #</p>
<p>^ Yes I meant that lol, its still me shahdin, dont ask y i have 2 accounts one of them got well banned for a few days since I got an infraction for using offensive words in a post lol.</p>
<p>And I remember it to be choice B. Do you remember it also being choice B?</p>
<p>Also you agree with my other answers also right?</p>
<p>I’m so happy, I’m in algebra II and I took the maths II subjects just for fun (crazy) and I missed some questions but all the others I answered I got almost everything right…
For the answer it 2450$ I used the calculatore with a linear something like that. There was something about that in the Barron’s book at the end.
For the other question something like 1.10+pi
And the others also were right</p>
<p>Does anyone remember the question that was like what does the value of x as it approaches 1 in the equation (x^3-1)/(x-1) or something like that? was it 1 or was it like “no values are approached?”</p>
<p>wait…
“for the ax^2 + Bx + c one, doesn’t B have to be non zero? I think it said two distinct real roots, which means b^2-4ac must be greater than zero. or am I forgetting the exact question?”</p>
<p>I got that a had to be nonzero…because b^2 - 4(ac)>0 because then it would have to be a line…or am i remembering the one that was ax^2+bx+c=d+e?</p>
<p>satmath, u are correct. if A is zero, then you get a linear equation, leading to only 1 solution. A must be nonzero to be a quadratic equation. determining whether or not the equation has two distinct zeros comes after establishing the fact that the equation is quadratic</p>
<p>@satmath09: The question was (x^2-1)(x-1)/(x-1) and the answer for that question was that the limit was 2.</p>
<p>to satmath: In order to have two distinct zeros you really have to have a quadratic function first. Therefore, a has to be nonzero. Even if b^2-4ac is not greater than 0 due to b, there was another factor of x in the other equation. Since you have to get it to =, the true equation was really:</p>
<p>a^2 + bx - dx +c -e = 0
(not sure on d and e, but I do remember there were factors of x and a constant on the other side).</p>
<p>Therefore, the “bx” is actually (b-d)x and u cannot determine it since u get to pick only 1 variable.</p>
<p>Therefore, u have to pick a to be nonzero to ensure that it is quadratic.</p>
<p>Also, the answer is 2 for the limit question. I dont remember the precise question but the answer was 2, and it was C or D i think?</p>
<p>hmm…does anyone recall all the answer choices for the limit question? I think it was something like</p>
<p>b) 1
and e) it does not approach a single function</p>
<p>How many C’s were there for porblems 42-50? 
If you guys remember :D</p>
<p>"Wasn’t the center of the circle (-2, 1)?</p>
<p>Because the equation was: (x+2)^2 + (x-1)^2 = # "</p>
<p>i thought it was (x-2) + (y+1)…i remember i had to change the Y sign i think, not sure. anyone else?</p>
<p>^ we prolly got it right, we all understood the concept. all i remember is picking b for that question</p>
<p>i don’t remember a circle question…which one was that one?</p>
<p>@satmath
The limit approached 2</p>
<p>For some reason i could not get one of the starting questions that just asked you to calculate mean. the question was…</p>
<p>“There are 25 students in a math class. The above graph shows the number of caluculators owned by a number of students. What is the average calculator ownage?”</p>
<p>Im guessing i threw a wrong number in somewhere cos my answer wasnt a choice. I went with 2.67~ or some number that was close to that i think. anyone happen to remember the correct answer?</p>
<p>(Also, i just wanna jump on the bandwagon on saying i think that was a hard test. i got an 800 in the BB with 12 minutes to spare. this time i skipped five and ran out of time to go back.)</p>
<p>^ I think the answer was 2.48?</p>
<p>Does anyone remember getting the answer 11 for question #11? I kind of got worried since my answer seemed weird since it was question #11 lol.</p>
<p>omgosh i’m so worried right now.</p>
<p>has anyone ever gotten a 42 raw score or heard of anyone who has gotten a 42…but still got an 800?</p>
<p>this sounds really nerdy but i really don’t want a 790. lol</p>
<p>@Banned: what was question number 11? did you input all the values into a list and compute the mean for the calculator question?</p>
<p>@sporty04: The lowest I heard about is a 43 raw score…but someone on an earlier post was talking about how one year the raw score 41 was still an 800. But, turns out that number was for an SAT II math back in 1980 or something. So, might want to consider canceling your score? That’s what I’m thinking of doing. But don’t do that until you have checked all the ones you’re unsure of on this forum.</p>