<p>For some reason I can't understand how to arrive at the answer of (pg 408, problem 5) without writing all the numbers out. I know there is an easier way to do this...can anyone explain how?</p>
<p>Theres a sticked thread with links to a bunch of blue book problems and solutions including this one for 408/5
<a href="http://talk.collegeconfidential.com/showthread.php?t=67458%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?t=67458</a></p>
<p>Thanks a lot!</p>
<p>Sorry one more one. How about 8 on pg. 473.</p>
<p>barrelbowl,</p>
<p>This is a GREAT example of how the SAT makes incredibly easy problems seem hard. Being able to see through the facade of complexity is key on lots of these problems. On a problem like this, it will be tempting to break out your pencil and starting doing algebra and recalling old Geometry laws about remote interior angles and blah blah. But that would be a waste time that wouldn't even bring you to the right answer. This question can be answered in 1 second flat by noticing that:</p>
<p>All of the a, b and c terms are inside of a quadrilateral. (The diagram is broken up into multiple triangles to obfuscate this fact, but if you look at the picture "macro" instead of "micro" you will see that they are all in one quadrilateral.)</p>
<p>A quadrilateral of course has 360 degrees in it. To get C, subtract the other variables, the 2 a's, and the 3'bs. 360 - 2a - 3b. Answer choice E. But you already knew that once I said it was a quadrilateral. That's why you have to see through the complexity. </p>
<p>Keep posting questions!</p>
<p>Andre</p>
<p>Andre Thanks a bunch! I always seem to not acknowledge these things. But seriously, thanks for offering a novel and fast approach to the problem.</p>
<p>Could you help me with pg. 490 Problem number 12? I got the answer right by estimating, but is there a correct mathematical approach?</p>
<p>On this problem, the key is to change the diagram according to the information that the problem gives you, or to see that Q is like a "half-way point" where both sides have to meet on your line. </p>
<p>We are given that BC is equal to 4. So write 4 on that side.</p>
<p>Additionally, we know that P and Q are "symmetric" about line AB, so that means they have equal length on both sides of line AB. So what you do now is you draw those old fashion congruency marks (the small verticle dashes). Since they are equal, give them each a number less than 4, since Q is inside the rectangle ABCD, whose width is 4. So give them each a length of 2, or any other number < 4. </p>
<p>Can you guess what the next step is?</p>
<p>They also tell you Q and R are symmetric about line CD. So put two congruency marks on both sides of CD. Now give them each a length that is 4 minus the number you gave the first two equal parts. If you picked 2, you would now also pick 2 (4-2 = 2). </p>
<p>Now you have 4 congruent parts: each length 2. The total width is 8, answer choice B.</p>
<p>Now you might be saying to yourself: How do I know the rectangle is bisected? You don't. But the answer holds true regardless what numbers you picked the symmetric parts to be. For example, if the first two parts had been 3.5 each, and the other two .5 each, PR would have still come out to a total length of 8. (3.5<em>2 + .5</em>2=8)</p>
<p>The most important insight to make here is to realize that you can create a line according to what the problem is giving you. When the SAT gives you a diagram, they want to confuse with it, but at the same time they want you to use what's given either in the text or the diagram to solve the problem. Since the diagram is not drawn to scale, they want you to use the text to solve the problem. The diagram is really only useful for you to make the connecton that both sides of QR and PQ meet to make a line. Knowing that, the text's directions allow you to solve the problem.</p>
<p>Andre</p>
<p>Thanks once again!</p>
<p>What if you just said that (1/2)PQ + (1/2)QR = 4
(1/2)(PQ + QR) = 4
PQ + QR = 8 ?</p>
<p>hey setzwxman, you failed to utilize AB = 6. ;)</p>
<p>Why do you need to use it? lol.</p>
<p>Does anyone know how to do #8 on p.369 or #8 on p. 375?
Thanks.</p>
<p>8, p. 369 looks really awkward/strange at first (at least to me), but after some thought and "connections" drawn w/ trig functions, it is JUST a periodic function. What other type of function would make the following equality true f(x + 5) = f(x)? Only a periodic one will. That means that you can add 5 to all of the x-values for the x-intercepts shown on the graph until as long as you are less than 12.</p>
<p>By estimating, there appear to be x intercepts at x = 1.5, 2.5, 3 and 4.5. Adding 5 to each of these values yields 6.5, 7.5, 8 and 9.5. Adding another 5 to those values yields 11.5, 12.5, etc... We can stop now because 11.5 is the last valid value for the specified domain. Let's count them up, and we arrive at a total of 9 x-intercepts on the domain (0, 12).</p>
<p>8, p. 374-375: They actually were very nice to us, as the slope is already found. We just have to move the x-intercept over two units to the left (changing it from -5 to -7) and then using that point in the point-slope formula to arrive at the value of k.</p>
<p>y - 0 = (4/5)(x + 7)
y = (4/5)x + (28/5)
So, k = 28/5.</p>
<p>ohhhhhh.. i get it now. thanks!!! :-)</p>
<p>
[quote]
Why do you need to use it? lol.
[/quote]
setzwxman -
you rather need to not use it. Your explanation just was way brief - could not you dilute it with musings on how the malicious ETS throws unnecessary data in to throw everybody off?</p>
<p>p.369 #8.
Since f(x) period equals 5, graph of f(x) consists of identical graph fragments connected together, each 5 units "long" along the x axis. Drawing the graph would really help.</p>
<p>x=0 to x=5 graph fragment repeats itself between x=5 and x=10, which gives us 4 extra f(x) zeroes (x-intercepts, in other words) on top of 4 zeroes for x between 0 and 5.</p>
<p>Between 10 and 12 we have a copy of the graph fragment between 0 and 2 (or between 5 and 7), which gives us one more f(x) zero.</p>
<h1>4+4+1=9.</h1>
<p>p.374 #8.
For line m y=(4/5)x+k
k is the y-intercept,
-7 is the x-intercept.</p>
<p>m is parallel to l, so there are two similar triangles (make a drawing)
(-5,0), (0,0), (0,4),
and
(-7,0), (0,0), (0,k).
k/4 = 7/5,
k=28/5.</p>
<p>Can someone help me on number 6 on pg 368 in the second edition of the Blue Book?</p>