SAT January 2012 - Math

<p>Michael, it asked for least posible integer of d which was 7</p>

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<p>That would imply that a = 0 b = 4 c = 6 d = 7. However, a cannot be 0, since the restriction was set that a, b, c, and d are positive integers.</p>

<p>I got 10, with intercepts at (0,8) and (6,0).</p>

<p>what was the proportionality one</p>

<p>No. I am 100% sure that D was 8.</p>

<p>Also, I caught myself making a stupid mistake. For the one about the proportion of camp counselors, the right answer was 81. I almost put 4.</p>

<p>For the parabola one, the vertex formula is -2a/b, I graphed it too. I forget if it was 4/3 or 2/3 but it was one of the two. From what other people are saying, I guess 4/3.</p>

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<p>Yes it is, as a quick Google search would show.</p>

<p>what about the second side of triangle?
Perimeter was 20</p>

<p><a href=“http://talk.collegeconfidential.com/sat-preparation/166335-0-integer.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/166335-0-integer.html&lt;/a&gt;&lt;/p&gt;

<p>What was the proportion one.
P is proportional to</p>

<p>1/t
1/t^2
t
or
t^2</p>

<p>I put t^2</p>

<p>@Roy</p>

<p>Seven</p>

<p>@njersey</p>

<p>I meant positive integer.</p>

<p>it was 3/2. the parabola was x2-3x-40.
the vertex formula -b/2a. -(-3)/2(1)=3/2</p>

<p>i also graphed it, its 1.5 or 3/2.</p>

<p>For the sequence of 12^10, how many do you need to get 12^12, I said 144. I originally had 24, but doing in the calculator, it worked for 144.</p>

<p>Yeah I put 3/2… Don’t know why I said 4/3. My short term memory is pretty bad…</p>

<p>the formula is actually -b/2a making the answer 3/2. b was 3 and a -1</p>

<p>I used my graphing calculator to calculate the minimum, and I got 3/2.</p>

<p>the parabola one was 3/2 for sure. and the number line has to be 8.
the order was a, b, c, and d in a number line. all of them had to be positive integers. Therefore, makes sense that A = 1 because it is the lowest positive integer. if b, c, and d had to be integers, the distances have to be integers as well, so CD is 1, BC is 2, and AB is 4. with this logic, A is at 1, B is at 5, C is at 7 and D is at 8. makes sense to everyone?</p>

<p>The vertex formula was not necessary for the parabola question. The equation was:
f(x) = x^2-3x-40;
f(x) = (x-8)(x+5);
The x intercepts are at x = 8 and x = -5.</p>

<p>You can find the midpoint of these two intersects, [8+(-5)]/2 = 1.5 or 3/2.</p>

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<p>No, the vertex formula is the reciprocal: -b/2a (see [Using</a> the Vertex Formula - Quadratic Functions- Lesson 2](<a href=“http://www.algebra-class.com/vertex-formula.html]Using”>Using the Vertex Formula-Quadratic Functions)). Thus, since the formula (unless I somehow misread the equation) was “x^2-3x-40”, b=-3 and the denominator would equal 2. Put together, the vertex would be a 3/2.</p>