<p>Blue Book Practice test 1, section 3, number 18</p>
<ol>
<li>The shaded region in the figure above is bounded by the x-axis, the line x=4, and the graph of y = f(x). If the point (a,b) lies in the shaded regions, which of the following muust be true?</li>
</ol>
<p>I. a <_ 4
II. b <_ a
III. b <_ f(a)</p>
<p>I realies that I. and that II. doesn't but why does III. work?? </p>
<p>Same book, same test, section 6, number 6.</p>
<ol>
<li>Which of the following tables shows a relationship in which w is directly proportional to x?</li>
</ol>
<p>It talks about DIRECTLY PROPORTIONAL. What does that mean? I had two choices left. One was B (which I chose), and D (which I didnt choose). I had to choose and chose B and got it wrong.
B is basically w^2 =x
D is 3w=x</p>
<p>so why isn't squared directly proportional but why is 3w proportional??? what does DIRECTLY PROPORTIONAL??</p>
<p>Same book, section 6, 17.</p>
<ol>
<li>IN the xy-coordinate plane, the graph of x = y^2 - 4 intersects line L at (0,p) and (5,t). What is the greatest possible value of the slope L?</li>
</ol>
<p>The answer is 1. IDK how to solve it.</p>
<p>same book, section 9, 14</p>
<ol>
<li>If (a+b)^1/2 = (a-b)^-1/2, which of the following must be true?</li>
</ol>
<p>f(a) is the same as f(x) with domain 0<=x<=4. So that squiggly line is made of points a, f(a).</p>
<p>b is a y coordinate of a point in the shaded region with an xcor of a… If you look at all the possible values of f(a), it will make up the squiggly line right? Well the point a,b cannot possibly be above a,f(a) because it lies in the shaded region which is below the squiggle.</p>
<p>And it specifically refers to a,f(a) vs a,b. They both line on the same x coordinate, but the y coordinate, b must be lower then f(a) at any specific point x location “a”.</p>
<p>
</p>
<p>Directly proportional means a\b = k. where k is some constant. This means that as “a” increases, “b” increases with it as long as when they divide it always equals the same value. In this case it says which w is directly proportional with x. so the direct proportionality is this:</p>
<p>x\w = k, where k is some constant. IT CANNOT CHANGE. Only D works because the division of x and w always equals 3. Everytime.</p>
<p>
So it gives some points 0,p and 5,t and the equation of the specific graph. So lets put in these points.</p>
<p>0 = y^2 -4, by plugging in the points (0,p), so you get 4 = y^2. this is when you notice something unique. y = ±sqrt(4), so that means the point can either be 0,2 or 0,-2. Let’s keep that in mind and plug in the other point. Using 5,t</p>
<p>5 = y^2 -4 , y^2 = 9, y = ±sqrt(9). So the other point can be either 5,-3 or 5,3. If you connect one of these points with one of the other, you make a line with some slope. It asks for the greatest slope between the 2 points. If you look at the points the most obvious is a positive slope which makes the steepest line, which is made up of the 2 points:</p>
<p>0,-2 and 5,3. Find the slope and it is equal to 1.</p>
<p>
</p>
<p>Well 1\2 is the same thing as a sqaure root right? Well to get rid of a square root, you square it. Just niftily, by squaring both sides, you get</p>
<p>((a+b)^1\2)^2 = ((a-b)^-1\2)^2) = </p>
<p>a+b = (a-b)^-1</p>
<p>Something to a power of negative one means here 1(a-b) right?</p>
<p>So mutiply a-b to the other side to get (a+b)(a-b) = 1. FOIL it and you get</p>
<p>a^2-b^2 = 1. Answer E</p>
<p>I hope this helps. Don’t feel bad about not knowing. The SAT doesn’t test mathmatical skill anyway. I used trial and error on all of those questions, which is fairrly meaningless anyway. :P</p>
<p>^Yea, the math curve sucks pretty bad. It’s so easy to make a simple mistake or not notice something obvious until after you look at it again because of the questions’ wordings. And even if you try the question you’ll suffer the 1\4th penalty if you don’t notice something, so it’s better to try less and just assume you can’t do it. Pretty dumb if you ask me…</p>