SAT May 2009 Math

<p>I did it twice and I still got 1319.</p>

<p>Wait, what’s the answer to the tin/copper problem? I know I put 19, but was that what it was asking for or did I misread the question?</p>

<p>^it was 250, but I FORGOT TO ADD 10 and put 150 ;________________;</p>

<p>I reread the question so many times and I could have SWORN it was 19:1 T:C.</p>

<ol>
<li>10 char</li>
</ol>

<p>ratio of copper to tin is 19:1</p>

<p>alloy is 380 g. 380/20 = 19 </p>

<p>tin = 1/20 = 19</p>

<p>^ Someone confirm the tin one, I’m nervous.</p>

<p>Can someone explain the rectangle problem? I got 252 =/ made the triangle a 6-8-10 then divided by two to make it 24/square. 300 - 48 = 252.</p>

<p>God it if I read that question wrong, but I’m pretty sure I put 361.</p>

<p>how about the parallegoram question? it was liek 16 degrees and 85 degrees with a line and an exterior angle? i got 101?</p>

<p>yeah i got 101</p>

<p>I got an answer of 101 for a problem.</p>

<p>101 was what I got.</p>

<p>i got 101 too!</p>

<p>Tin one was 19, you just had to read the question right, meh.</p>

<p>The Rectangle/2 Square problem was 250. It was a 45 45 90 triangle cut in half. Square one leg of the 45 45 90 and you get the area that isnt shaded. 300-50= 250.</p>

<p>One leg of the 45 45 90 was 7.something</p>

<p>-2/3 for slope question?
and y/(x+y+z)%</p>

<p>significa-</p>

<p>you can divide the squares in half down the diagonal to make two triangles (x, x, xroot2). xroot2=10, so x=10/root2
1/2 b * h so .5<em>10/root2</em>10/root2 = 25
there were two triangles so 25*2= 50 </p>

<p>subtract that from 300</p>

<p>Ask some other questions, we are debating the same problems over and over.</p>

<p>Quadratic- 12
Vertices- 1/5
Triangles with shaded part – 3pi
b-a = 9
**1319 ** Please confirm this
Logic- calc/wed
1<x<2 area=“” of=“” figure=“” with=“” triangles=“” on=“” inside=“” rectangle-=“” 250=“” f(a)=“f(b):” line=“” slope=“” 0=“” pq=“”>PR
Wire- 6
Beads- blue
Tin/copper- 19
Scissors- 1600</x<2></p>

<p>I seem to remember getting 101. What was the math to get that, again?</p>