<p>can someone explain to me how to do #19 on pg. 523 in the blue book? thanks</p>
<p>Since the triangle is equilateral, that means all three sides have the equal length of 2. Also, XW and WZ are equal to 1 since they are split in half by WY.</p>
<p>Use pythagorean theorem:</p>
<p>(WZ)^2 + (WY)^2 = (YZ)^2
(1)^2 + (WY)^2 = (2)^2
WY = square root of 3</p>
<p>The problem tells you that WY is the diameter and area of a circle is pi*radius^2. So half of WY is square root of 3 divided by two.</p>
<p>Now, just plug the numbers in the area forumla:</p>
<p>(square root of 3 divided by two) ^2 * pi = 3pi/4. That is the answer (C).</p>
<p>thanks a lot, how about #'s 6 and 13 on the same section</p>
<h1>6:</h1>
<p>If Y is the midpoint, then XY and YZ must be equal to each other. I’ll plug in some numbers to make it easier to understand. Say that line XZ is equal to 10… that makes XY and YZ both equal to 5.</p>
<p>(I) has to be correct because half of the line XZ is 5 and YZ is 5… so they are equal.
(II) Isn’t correct because half of XZ is 5, and 2 <em>XY would be 5</em>2=10.
(III) Is correct because 2 * XY is 10 and line XZ is 10.</p>
<p>So the answer is (E).</p>
<h1>13:</h1>
<p>Plug in the x value and c value they gave you into the equation.
640 = [(600*20 - 200)/20] + k
Now simplfy:
640 = 590 + k
640 - 590 = k
50 = k</p>
<p>Answer is (B).</p>
<p>ah crap i meant 6 and 13 on section 4 lol. pgs. 532 & 534</p>
<p>No problem 
I like helping others because it helps me reinforce the concepts even better.</p>
<h1>6:</h1>
<p>The easiest way to do this is to plug in your own number for m. I chose m = 1. </p>
<p>10(1^2) k^-1 = 100 (1)
10*k^/1 =100
k^-1 = 10
k = .1 </p>
<p>Now that you know what m and k are, you can answer the question which is m^-1. Plug in 1 for m. It equal 1 ^ (-1) = 1.</p>
<p>Look at the answer choices, plug in .1 for everywhere you see k. Which ever one gives you m = 1 will be the answer. (D) will be the answer because 1/(10 * .1) = 1.</p>
<h1>13:</h1>
<p>2f(p) = 20… divide each side by 2 to get f(p) = 10. </p>
<p>P is a number that is substitued into the equation f(x) = x+1… put in p instead of x to get f(p) = p +1. You know that f(p) = 10. So subsitute 10 for f(p).</p>
<p>10 = p + 1
9 = P</p>
<p>Now, plug in p = 9 for f(3p)… to get f(27). Plug in that number into f(p) = p + 1</p>
<p>f(27) = 27 +1 = 28 <— this is the answer.</p>
<p>thanks alot, how about #15 pg. 551</p>
<h1>15:</h1>
<p>This is basically logic. If you look at the graph, notice how time isn’t affected by the # of practices a hamster is given. In fact, the numbers all seem constant since they are in the range of 40 - 50 seconds, regardless of practice. So the answer would be (A) since it’s a constant.</p>
<p>If you can’t think of it that way, try process of elimination. (B) can’t be the answer because t(p) does not equal p, it equals seconds (C) can’t be the answer because say you plugged in 4 practices… 4 * 44 gives you a # far higher than the seconds on the chart. (D) can’t be the answer because plugging in a number like 4 for p gives you 4/44… and that is no where close to the # of seconds. (E) can’t be correct because adding the # of practices + 44 seconds gives you a graph that is steadily rising (meaning that hamsters with 1 practice should take 45 seconds, 2 practices mean 46 seconds and so on) but the graph shown doesn’t have a steadily rising pattern. It’s scattered.</p>
<p>So therefore, you can narrow the answer to (A).</p>