<p>Yeah, I think a = 4 and b = 2, so 6. I don’t know if there’s an actual mathematical way to do it, but I kept plugging in numbers until I got it to equal 20,030.</p>
<p>lol why do you guys keep saying 20,030…don’t u mean 20,300?</p>
<p>ice cream? I don’t remember an ice cream question yet I did have four math sections</p>
<p>I think there is a problem because 2<em>10^4+3</em>10^2=20300 and not 20030?</p>
<p>the answer is 6 but i think they just mistyped 20,030 instead of 20,300</p>
<p>Was the section with the question 3^(x+1) = k (or something along those lines) an experimental section??? Thanks.
Was the section with the question 3^(x+1) = k (or something along those lines) an experimental section??? Thanks.Was the section with the question 3^(x+1) = k (or something along those lines) an experimental section??? Thanks.</p>
<p>It was just a typo. Either way I got the answer right. 2x10^4 and 3x1^2…4+2=6</p>
<p>The answer to the 3^(2x + 1) question was 3k^2.</p>
<p>3^x = k</p>
<p>3^(2x + 1)
= 3(3^2x)
= 3(3^x)^2
= 3k^2</p>
<p>Yeah, but was that question part of the EXPERIMENTAL MATH section?!</p>
<p>@chosen00, i feel like it isn’t. as far as I’ve seen, i think almost everyone had that question</p>
<p>All possible SAT curves: <a href=“http://www.erikthered.com/tutor/SAT-Released-Test-Curves.pdf[/url]”>http://www.erikthered.com/tutor/SAT-Released-Test-Curves.pdf</a></p>
<p>3k^2 was not experimental.</p>
<p>I don’t remember that 3^2x+1 one, but by now I’ve forgotten a lot of them. 
</p>
<p>And yeah, I meant to type 20,300 for that a+b one. Whatever the number was, it was still 6. LOL.</p>
<p>Are these experimental Math questions: </p>
<ol>
<li><p>Find the fraction of the volume of the smaller cylinder to the whole thing… It was like a cylinder within a larger cylinder or something like that… The larger cylinder had like side length 10, the other had side length 5.</p></li>
<li><p>There are n numbers in a list that consist of only 3s and 5s. The mean of the list is 4. What could be a possible value for n? ( I, II, or III)</p></li>
<li><p>Find perimeter of the triangle that is equilateral and has an area of 1… or something like that ( this was one of the last problems in the section #18,19,or 20) NOT the maximum perimeter of triangle question that was 33… I had that in another section.</p></li>
<li><p>Problem where you have to solve for (a) that was in f(a)… They said that a does not equal 5… So you had to factor it out and set equal to 0… So it was like (a+1)(a-5) … So the answer was like a=-1</p></li>
</ol>
<p>Math wizard here I get 800s consistently on math and these are some of my answers that I can remember</p>
<p>1600
(I II and III)
33
165
128
22/9
6.2(any number between 6 and 7)
2:1
(-1,3)
65 degrees
(-5)
50
130 degrees
330
48
6 (the power problem)
12 (perimeter of a triangle inside circles or something)
54
slope = 5
x=5
3ksquared
F(x) were equal at 4
0 (median—mean problem)
250 seniors
root 130
2xroot3
1/8 humback whales
x+y was the greatest
x+y+z < 270 (or something like this I can think I put E for this problem)
3 solutions for the 10 = 3x (something like this)</p>
<p>These are 30/54 answers I can remember. I am fairly certain they are right, but feel free to disagree.</p>
<p>^ What’s the question for (-5)?</p>
<p>Also, you can add the x = 70 degrees problem and the 2B8/11 has a remainder of 7 question.</p>
<p>Thanks for finally making a compiled list!</p>
<p>Also, I think you meant 9/22, not 22/9.</p>
<p>ahh yes, I got 7 for a remainder problem, and @shadow mist I did mean 9/22, and the (-5) one was like the first problem, thats the only reason I remember it.</p>
<p>and @studio I believe it was x = 65 degrees unless I am thinking of a different problem?</p>
<p>^ Pretty sure it was -3. 9 + (-12) = -3.</p>
<p>No, y = 65 degrees and x = 70 degrees (one with two isosceles triangles).</p>
<p>Also, you can add 998 (the number in the (3, 8, 13) sequence).</p>
<p>Yeah I remember that one too, I am talking about the one where it was like "which one is not between -4 and 3. and it was -5. And can you refresh me on the x = 70 degrees I am drawing a blank on that problem</p>