did anybody find the math section a little bit harder?

<p>i did too.. not sure which was exp.... anyone have any ideas?</p>

<p>im pretty sure the one with t^2rs as the last problem was the real one. i also ha d math experimental but had the other one section 7 and the trs one in section 2. they wouldnt do experimental in section 2 would they?</p>

<p>i dont know, i had the t^2rs in the 7th section. i definitely had an experimental section. i hope its the one with the f(x) and the chart, because i bugged out and guessed, and it wasnt even hard lol.</p>

<p>wasn't f(x) and the chart a part of the last math section (16 questions, 20 min)?</p>

<p>not for me, i had grid-ins last.</p>

<p>ok so if I had WXYZ problem, then it's not experimental? correct? because I thought those (section 7) grid-ins were really hard..I thought it was experimental.</p>

<p>i dont think grid-ins can be experimental, so that section can't be the one.</p>

<p>what was the section with the rectangular pyramid that had the variables m, h, and e? experimental?</p>

<p>what was the section with the rectangular pyramid that had the variables m, h, and e? experimental?</p>

<p>no</p>

<p>but what was the answer for that? im not sure i got it right. Im like only 80% sure</p>

<p>wait if youre sure its experimental why does it matter if you got it right?</p>

<p>I'm pretty sure the pyramid one wasn't experimental.</p>

<p>im sure it was NOT experimental...</p>

<p>for the question about the area of a square with opposite vertices at (-2,-2) and (2,2) did you all put 16 or 32? At first I thought that opposite meant diagonal but then I changed it so that the line formed one side of the square.</p>

<p>ok sweet</p>

<p>haha....sorry i aced that one but missed 1 or 2 on the trs one so i was a little ****ed because i read wrong</p>

<p>as for the pyramid one it was A im pretty sure....some thing like hsqrt(2)/2....dont remember exactly but i got A and im pretty sure that was right</p>

<p>i put 16...</p>

<p>I got the answer as msqrt(3)/2...if you put m as 10, making m/2=5, then you're just solving for the other leg of the triangle. I'm pretty sure it ended up being a 30-60-90 right triangle, so that would make it msqrt(3)/2.</p>

<p>what did people get for the pyramid? I got A for that one which i think was a root 2/m or something lie that. There was one choiuce that had a root 3 (i think it was C) that I heard some people put that I didnt get.</p>

<p>i got A for the pyramid as well</p>

<p>i put m/root(2)...i found the diagnol of the bottom square (mroot(2)) took half of that as the base and put it into the pythagorean theorum and got m^2-(mroot(2))/2)^2 = X^2</p>

<p>i got m/SQRT(2)</p>