Math for March SAT 2009

<p>

</p>

<p>Yea I have a feeling I got that wrong. Been a while since I had geometry and forgot that you have to square the sides.</p>

<p>I just did a test problem right now used a 3,4,5 triangle and a similar 9,12,15 triangle. </p>

<p>Area of 3,4,5 = 6</p>

<p>Area of 9,12,15 = 54</p>

<p>According to everyone who did 40/3, when you do 6/54 = 3/x. You should get 9 (the corresponding similar side). But you only get 9 if you do 6/54 = 3^2 / x^2</p>

<p>Was the math and science student one 104?</p>

<p>for math and science i got 90
the rope cut into pieces: I got 4</p>

<p>boddah, the fraction behind the answer your saying is right wont even fit in the grid, i don’t think they’d ever do that</p>

<p>same for both strat</p>

<p>What about the one with a and a+10 and the inequality?</p>

<p>hey guys, reason why i no longer think it’s 40/3:</p>

<p>it said the ratio of the AREAS of the triangles was 3/4.
therefore, you cannot set a proportion with 10/x = 3/4
10 is the length of a side.</p>

<p>therefore, you have to square root 3/4 to get the similarity ratio, and then set 10/x equal to the similarity ratio (square root of 3/4)</p>

<p>The math and science one was 90.
YOu have to subtract the students who took both the classes from the total.</p>

<p>i got 104 also</p>

<p>As I said previously, this proves the answer is 40/3</p>

<p><a href=“ImageShack - Best place for all of your image hosting and image sharing needs”>ImageShack - Best place for all of your image hosting and image sharing needs;

<p>

</p>

<p>Nope. 90.</p>

<p>There were… So and so kids in both classes. You had to subtract by 14 from each because 14 kids were in both. Add those together (76) and then add the 14 kids you took out.</p>

<p>master if 14 are in both, that means you subtract 14 because you know 14 of the total are already included in that count and that would be adding them twice</p>

<p>For the triangle one:
If you put in a random number for the base and calculate the proportion of the area, you get .75, no?</p>

<p>new fish: yeah, that was the only reason i doubted it.</p>

<p>but you definitely cannot set the ratio of areas to a proportion with a length of a side.</p>

<p>monoclide, do you remember what you got in that problem? 5/12 or 2.4?</p>

<p>I didn’t mean the one with the line where f(x)<3. It was the one where it said there is a number line with a and a+10 and x is in between a and a+10.</p>

<p>Master:
Yeah, but those kids are counted twice in the classes. That’s why you have to subtract.</p>

<p>

</p>

<p>I don’t remember even doing that question. Was it the experimental one?</p>

<p>fox, it asked for the total number of people in the group</p>

<p>Ohhh gotcha monkey that makes sense.
Fish: I was confused on it too, I put 8 then changed it to 12.</p>

<p>For the a and the a+10 one, I got
abs(x-a-5)<5
Did plugged in answers for that one, and only that answer worked.</p>