<p>hint: you have a point on the line as well as the slope of the line, find the equation and set one of the variables to zero</p>
<p>answer for the last one is D</p>
<p>a slope of -2 means that for every 1 increase on x there is a 2 decrease on y, or you can think it the other way round, for every 1 decrease on x there a 2 increase on y.</p>
<p>So you know that in x=3 y=4. In x=2 y=6, in x=1 y=8 and in x=0 y=10. You are being asked the Y axis (x=0).</p>
<p>That's why the answer is D, 10.</p>
<p>Thanks gentlement, it shows that I am very close to the answer.</p>
<p>Since there is a line, so its form must be y=mx+b
m = -2
y = -2+b</p>
<p>But then I replaced it by the two given points, which makes no sense.</p>
<p>Thanks snipez90 and c0calait</p>
<p>no you have to use the point slope formula given by y-y1 = m(x-x1). this equation is explained by c0calait since if (x1,y1) is definitely on the line, then any other point (x,y) helps determine the slope if it is also on the line which is graphically rise over run or (y-y1)/(x-x1) = m. We're given the point (3,4) and the slope so x1 = 3, and y1 = 4. We also know that the y-intercept occurs when x = 0 so we can solve for y using point slope.</p>
<p>c0calait gives a more reasoned out solution, hopefully this equation makes sense because it often comes in handy.</p>
<p>
</a></p>
<p>I've worked out: 32, correct? gentlemen</p>
<p>Next soldier, </p>
<p>
</a></p>
<p>This is quite odd to me, first time I have ever dealt with. Although I love geometric and function questions of the sort.</p>
<p>Thanks in advance, gentlemen</p>
<p>Hey these are the last two I can help you with tonight since I need to write one essay and revise another before I go to sleep.</p>
<p>The correct answer to the first is 8, triangle APC is 3-4-5 (lengths) so AP is 4. triangle ABP is 30-60-90 (degrees) so AB = 2*AP = 8</p>
<p>The correct answer to the second one is D. The graph of absolute value of f(x) simply takes all the output values of the original function and makes then positive. Thus the graph of abs value of f(x) is simply the graph of f(x) except the portion of the graph from x = -1 to x = 3 is reflected across the x-axis because the resulting values are now positive. It should make sense why D is the correct answer now.</p>
<p>OK, thank you very much sir, good luck with the essays.
I also have one for US History need to be done.</p>
<p>A nice way of looking at it is realizing that since the answer is written as a fraction to begin with, you can see that r + t = 5 and that r - t = 2. Once you realize this, it is very easy to add the system and cancel the t's, leaving you with r = 3.5. Substitute that value into any of those equations and you get t = 1.5. 3.5 / 1.5 is the same as 7 / 3 or E.</p>
<p>You can take your time and test all values and realize that f(1) = -1, and |-1| is 1, thus making f(x) < |f(x)| true, but a good way to approach many of the problems where you are forced to test values is by looking for exceptions. Take a look at the values offered - the only one for which f(x) yields a negative number is x = 1, so you should start by testing this point.</p>
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</a></p>
<p>Sorry, I have to reproduce this by paint, looks a bit scratchy</p>
<p>I get 3pi^2, I don't know to get rid of that pi to get 3pi</p>
<p>Area of big circle = (3pi)^2 = 9pi^2
Area of medium circle = (2pi)^2 = 4pi^2
Area of the smallest circle = pi^2</p>
<p>Shaded area = 9pi^2/2 - 4pi^2/2 + pi^2/2 = 3pi^2</p>
<p>Did I do somthing wrong with the procedure</p>
<p>Well the area of any circle is pi(r^2), not (pi r)^2. Other than that, you're golden.</p>
<p>Can you resolve it again, so that I can learn the technique</p>
<p>Thanks :)</p>
<p>Did it wrong (I got confused, thought that the diameter was the radius)</p>
<p>So from the question, you know that the diameter of the large circle is 6. Since area=pi(r^2), the area is pi(3^2)=9pi. Since the majority of the shaded region is in HALF the large circle, cut that sucker into half= (9/2)pi. This number is assuming HALF of the large circle is completely shaded</p>
<p>Now, see which small circles are messing up the perfectly shaded half-circled region. You see there is half the medium circle coming in (substract) and you see half the shaded small circle is budging out (added ONLY half since you already included the other half in the HALF of large circle). Now it's all down hill.</p>
<p>You know the diamter of medium circle is 4 so the area is (2^2)pi. Divide that into halves and substract from (9/2)pi. You know the dimater of small circle is 2 so area is (1^2)pi. Divide that by half and add.</p>
<p>(9/2)pi - (4/2)pi + (1/2)pi = 3pi</p>
<p>EDIT: c0calait: you are assuming the diameter is the radius or the area of a circle is (d^2)pi.</p>
<p>You are right, I am editing it to avoid any confusion.</p>
<p>Thanks</p>
<p>Oh, I see, I almost grasp it.
c0calait gets that very frequent mistake just like me. I think it is very dangerous to mistake between diameter and radius. Comrades, I shall open my eyes as large as possible :)</p>
<ol>
<li>If the length of LM is 7 and the length of MN is 8, which of the following could be the length of LN?</li>
</ol>
<p>(A) 23
(B) 22
(C) 17
(D) 16
(E) 14</p>
<p>I bump this thread because I'm interested in the answer of the question on the previous post.</p>
<p>Are there any thing in this post that make you feel unsatisfied?
Please let me know if it is my fault, I gonna correct it.
I just intend to use this post as a regular basis to ask questions</p>