<ol>
<li>1 ( possible value of x in</li>
</ol>
<p>^ I don’t remember this one. What was the question asking?</p>
<p>Guys, the circle question had a y-coordinate of 15.6.
Allow me to explain:</p>
<p>The image of the circle was NOT entirely on the 1st quadrant. Some of the circle fell into the 3rd quadrant. So…what does this mean? This means that 7 was NOT the radius of the circle. So, how do you get the radius? </p>
<ol>
<li>Create a right triangle towards to origin.</li>
<li>(a^2 + b^2 = c^2) ----> 7^2 + 5^2 = 74</li>
<li>sqrt(74)= approx. 8.6</li>
<li>Now that you have the radius, add it to the center’s y-coordinate and you will get 15.6</li>
</ol>
<p>I know…I thought it was 14 too UNTIL I went back to check it when I had extra time.</p>
<p>By the way, I am sorry for the double post but, which one was the experimental math section?</p>
<p>Was it the one with the star or the one with the surface area and volume?
(I would like to know because my grade HIGHLY depends on that.)</p>
<p>uhh why would u take approx? i just leave it with root 74 and find the distance between P and that centre because they need to be same! and yes,i got 14</p>
<p>it said integers from 1-10…</p>
<p>wolf you do realize that P wasn’t the top of the cirlce? If point P was at that top of the cirlce then 15.6 would be correct, but P was on the y axis and below the highest point of the cirlce. 14</p>
<p>That one had an answer of zero.</p>
<p>Anyway, what was the experimental math? (Sorry being redundant…I just can’t wait 18 days…I will implode)</p>
<p>By the way, how can you use the distance formula when you don’t have P’s y coordinate?
I am pretty sure it had to do with the radius.</p>
<p>It was definitely 14… the distance from the origin to the center, vertically, was 7. The distance from the origin to the center, vertically, was equal to the distance from the center to the y-intercept. 7+7=14. You overthought it.</p>
<p>Star was 24 (each side was 3, and there were 8 sides around shaded area). A/V one was A^2/3.</p>
<p>Oshiz, I think you are right about that one.</p>
<p>…And now I see what kikocupcake meant.</p>
<p>hey can you post the link for the CR section. cant seem to find it</p>
<p><a href=“http://talk.collegeconfidential.com/sat-preparation/1143343-may-sat-2011-cr-section.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/1143343-may-sat-2011-cr-section.html</a></p>
<p>another answer that I remember guys was like 110 million people had cell phones in 2000 or something and all you had to do was 110/280 and you got like .390293123 so the answer was 40% anybody know what im talking about?</p>
<p>Also there was one that asked for a possible value of x for 2x>x^2.
One was the simplest answer obviously.</p>
<p>do you guys recognize this answer? 7k=k^2 I saw it somewhere can someone re jog my memory?</p>
<p>LedbyExample…I got that too.</p>
<p>So basically this is how it is for me:
- If the star one is experimental then I get 2 FRQ wrong and 1 MC
- If the Area and volume was experimental I get 2 FRQ, 3 skips and 1 MC.</p>
<p>I am not in a good postion,</p>
<p>Ledbyexample, yes, that’s the correct answer.</p>
<p>I’m seeing some people answering x = 70 in the triangle question. However I think it’s wrong. Wasn’t the answer x = 10?
…90…
…/…|
…/…|x
…/…
…/…
…/…
…/…
…/50…50…y</p>
<p>Anyone remember this question, where y=40, isnt x = 10? :|</p>
<p>x=10
y=40</p>
<p>What is wrong with that? It makes sense.</p>
<p>50 + 130 = 180 (supp angles)
…90…
…/…|
…/…|x
…/…
…/…
…/…
…/…
…/50…50\130.y</p>
<p>180 - (50 +50) = 80 so, x is 10.
If x is 10 and we have an angle of 130,then y= 40 .</p>
<p>@tizil, im pretty sure that was part of the experimental.</p>
<p>Was that the section with the star and the perimeter? If so, college board has gone mad.</p>
<p>was the triangle question experimental? (x=10)</p>