<p>il bandito, that is SO GOOD for Barron's! My highest was a 41, and I was ecstatic to get it!</p>
<p>pearfire - I don't think there's an entire practice test, but you can download this big file that includes quite a few math IIc questions here: <a href="http://collegeboard.com/prod_downloads/prof/counselors/tests/sat/2005-06-SAT-subject-tests-preparation-booklet.pdf%5B/url%5D">http://collegeboard.com/prod_downloads/prof/counselors/tests/sat/2005-06-SAT-subject-tests-preparation-booklet.pdf</a></p>
<p>Yoish. Someone want to explain the difference between answer choices A and C of the last question of that PDF file (math IIC section) to me? I'm not remembering matrix multiplication at all.</p>
<p>Label the two matrices in answer C A and B so that answer C is the product AB. Now, the first column in matrix B represents the number of cameras sold, with each row representing a different model. You multiply the entire first & only row of matrix A with the first column of matrix B, so that for Day 1, the entire amount of money is $99<em>20 cameras + $199</em>16 cameras + 299<em>19 cameras. Then the same thing is done for the second day, which is the second column of matrix B: $99</em>18 cameras + $199<em>5 cameras + $299</em>11 cameras. Do the same thing for the third day. Each of the three results represents the "total income received from the sale of the cameras for each of the three days."</p>
<p>Edit: By the way, Answer A is impossible because you can't multiply an entire row in the first matrix by an entire column in the second matrix.</p>
<p>Ooh, thanks much.</p>
<p>I don't think I encountered any matrix problems when I took the math subject test... I guess they're pretty rare. Personally, I focused more on problems 25, 26 and 31 while I was studying.</p>
<p>Yeah, I haven't really, either. The only question I've had involving matrices out of 13 practice tests I've taken was one this afternoon about determinants or something.</p>
<p>Answer B: no.</p>
<p>Look at answer B as being the product of two matrices: AB. The first row in matrix A is:</p>
<p>Day 1: 20; Day 2: 18; Day 3: 3.</p>
<p>These three figures are all for model X, day 1 but the column in matrix B has the prices for all three models! If you multiply this first row (matrix A) by the one and only column in matrix B, you are multiplying the same model by the costs of all three models:</p>
<p>(# sold, Day 1, Model X) * (Price of Model X)
+ (# sold, Day 2, Model X) * (Price of Model Y)
+ (# sold, Day 3, Model X) * (Price of Model Z)</p>
<p>Answer B doesn't work.</p>
<p>Does anyone know how to solve #31, 20 and 25 for Math IIC in that prep site listed above? </p>
<p>thanks so much</p>
<p>Answer B doesn't really give anything meaningful because it's pretty pointless to multiply the number sold of one model by the price of another model. It's like saying, "I bought 13 carrots, and beets are 60 cents each. How much did I spend?" 60 cents * 13 carrots gives an answer, but it's meaningless.</p>
<p>Edit: Answer B for Problem 32, that is.</p>
<ol>
<li>Draw a right triangle with vertices at (2,1), (3,5) - those two were given, and (3,1). The base is 1 and the height is 4. The tangent of the angle at vertex (2,1) is 4, so solve for the angle and subtract it from 180 to get the answer (opposite angles in a parallelogram add up to 180).</li>
</ol>
<p>Problem 31:</p>
<p>The number of daylight hours clearly has a maximum when sin(2<em>pi</em>t/365)=1. Therefore, Q, the number of daylight hours on the longest day, is:</p>
<p>Q = (35/3) + (7/3) * 1
Q = 42/3
Q = 14</p>
<p>Now, the answer is [Q - # daylight hours on May 1]. The # of daylight hours on May 1 can be calculated from the equation and with the other given information.</p>
<p>I found that the problems in this booklet were much harder than the average when I actually took the test.</p>
<p>Me, too. These took me a while to solve.</p>
<p>I guess I'll see in a couple days how effective it is to study for Math 2 out of <em>only</em> a free CB booklet.</p>
<p>I'm somewhat nervous...</p>
<p>haha, keep us posted :)</p>
<ol>
<li>Let's call the center of the circle O.
Draw in OB, which is a radius = 2. Angle OBC = 90 degrees because externally tangent lines form right angles with radii drawn to the point of intersection.
Draw in CO. This bisects angle ACB, so label OCB 25 degrees.
So we have a 25 degree angle and a 90 degree angle. Therefore we know angle COB is 65 degrees.
Now draw in AB. Call the intersection of AB and OC P. OC perpedendicularly bisects AB.
Since COB is 65 degrees and OB is 2, we can solve for PB. sin 65 = PB / 2.
Multiply PB by 2 for AB.</li>
</ol>
<p>29 is easy but it tries to trick you into thinking it's hard with lots of completely wrong answers.</p>
<p>x = 5 + t
y = 7 + t</p>
<p>To find y=f(x) and determine slope f'(x), simply solve x=5+t for t:</p>
<p>t = x - 5
y = 7 + t = 7 + ( x - 5 )
y = x + 2</p>
<p>Clearly, the slope is 1. The answer area rates this problem as being equally difficult to problem 32. hahaha</p>
<p>Parametrics are probably the most intimidating (but easiest in comparison) coordinate system.</p>
<p>Guys is there any way I could graph something like this:
Ax^2+Bxy+Cy^2+Dx+Ey+F=0, its the general quadratic equation and say I wanna graph a circle or an ellipse, is there a programme for which u only need to type in the equation with x and y and see the graph?
solving for y and then doing the square thing would take a lot of time.</p>