<p>Anything on the consolidated list is 100% the right answer. Everything that is being debated in these recent pages has already been proven.</p>
<p>That means that k-4, XZ, 43 degrees, and 88 are right. Period. End of story.</p>
<p>Anything on the consolidated list is 100% the right answer. Everything that is being debated in these recent pages has already been proven.</p>
<p>That means that k-4, XZ, 43 degrees, and 88 are right. Period. End of story.</p>
<p>wait they weren’t perpendicular? ***</p>
<p>I thought they were perpendicular and I still got 43. Strange. Regardless, 43 is the answer.</p>
<p>Same, I think… :x</p>
<p>what was the question with the c=0 and k=3? </p>
<p>i dnt remember getting anything with tht!</p>
<p>can someone remember what the diagram looked like? what angle did they give?</p>
<p>They might have been perpendicular. I just remember a couple lines.</p>
<p>They were, it was 2 parallel lines bisecting one line, and yeah what was c=0 and k=3?</p>
<p>Gahhh. For the parabola question, I put down (5,0) then went back to change my answer at the last second! There goes my chance at an 800…</p>
<p>So far, everything else seems right though.</p>
<p>They were not perpendicular. CB tried to screw with you guys by making it look so, but there was no little box indication. If they were, it would have been 44 degrees, but it was 43.</p>
<p>Okay I guess they were perpendicular then? Regardless, 43 degrees.</p>
<p>what page are the consolidated answers on? Or can someone please repost?</p>
<p>can someone explain the semi bi root question. I thought it was sqr(2(128)) = 16, and then there is sqr (16/2) = root 8 as the answer?</p>
<p>treysongz its the other way around.</p>
<p>wait what do u mean. do u mind writing it out?</p>
<p>sqr((sqr(128/2))*2) = 4</p>
<p>"a) x^2
b) ??
c) y/z
d) y - z
e) xz "</p>
<p>Can anyone remember / confirm any other answer choices?</p>
<p>For the last question in the grid-ins, the answer was “2” right? Can anybody confirm?</p>
<p>The answer is xz.</p>
<p>Vincit- Do you remember the question?</p>