<h1>1) A pyramid with a triangle base is intersected by a plan. Which of the following could be the intersection?</h1>
<p>I. A line segment
II. A triangle similar to the base
III. A triangle not similar to the base
Y= 4.31x^2 + kx +2.5</p>
<h1>2) A real value for k is chosen so that the quadratic equation above has no real roots.Which of the following describes method to change the chosen value of k in order to guarantee that the equation has two different real roots?</h1>
<p>a) Decrease k by 10
b) Decrease k by 15
c) Increase k by 7
d) Increase k by 10
e) Increase k by 13</p>
<h1>3) A polygon is inscribed in a circle. If some diameter of the circle divides the polygon into 2 regions of equal area, which of the following must be true?</h1>
<p>I. The polygon is regular
II. The center of the circle is inside the polygon
III. The polygon is symmetrical about some diameter.</p>
<p>A)None
B) I only
c) II only
d)III only
e) II and III only</p>
<p>Glad I’m not taking the SAT… Yikes…</p>
<p>I have never seen questions this difficult on any SAT practice test. These weren’t from another test prep company’s test, were they?</p>
<p>As for the answers, the first one is all of them (I, II, and III)</p>
<p>The second one is a question who might see on the math level 2 test but the answer is decrease k by 15 (B). (Use quadratic formula and find what k must be to produce a positive value under radical)</p>
<p>the third one is none (A). (It is possible to have a non regular polygon that has sides that avoid the center and is not symmetric)</p>
<p>Were these from Princeotn Review? They specifically put harder questions in their practice tests.</p>
<h1>1</h1>
<p>I would say I, II, and III, however the question is ambiguous because it doesn’t specify whether the diagonal faces are similar to the base or not. If the diagonal faces ARE actually similar to the base, then the answer would only be I and II as far as I can tell.</p>
<h1>2</h1>
<p>I would say B. Let’s say that originally, k was -6.5 (and would therefore have no real solutions based on the discriminant [b^2 -4ac]). Adding 7, 10, or 13 wouldn’t give you numbers with real solutions because the absolute value of k needs to be greater than ~6.56 for there to be two real solutions. Similarly, if k was positive 6.5, subtracting 10 wouldn’t yield a value of k with 2 real solutions. For any value of k that doesn’t have two real solutions, subtracting 15 should work.</p>
<h1>3</h1>
<p>I would say C. This one would be hard to explain without sitting next to you with some paper to explain my reasoning.</p>
<p>I feel pretty sure about my answers, except on the first problem because it seems a bit ambiguous.</p>
<p>This is why I was a liberal arts major. I have no idea what y’all are talking about!!!</p>
<p>Actually I revise my answer to number three. It should actually be A.</p>
<p>Thanks!!! that was a lot of help!</p>
<p>so are you saying #2 is simply trial and error then?! (using the quadratic formula)</p>
<p>and #1 is pretty odd.</p>
<p>can it intersect it at only one point btw?!?!!?!</p>
<h1>2 is not trial and error but I guess you could figure it out that way. Just find the quadratic formula, set the stuff inside the square root as an equation. k^2 - 4(4.31)(2.5) > 0 and solve for k. And for question 3, there can be an intersection at one point (AKA any place where the edges of the pyramid meet).</h1>
<p>aha… yeah I get it thanks!</p>
<p>How about this one?</p>
<p>Which of the following could NOT be the intersection of a plane and solid right circular cylinder?
a) a circular region
b) a triangular region
c) a rectangular region
d) a point
e) a segment.</p>
<p>Those questions are just too stupid! srsly… I don’t get it o_0</p>
<p>Yes, it can intersect at only one point.</p>
<p>And ditto with phsyicsman, it’s not exactly trial and error. You have to understand what causes an equation in that format to have imaginary solutions. The cause is that when you plug the corresponding numbers of the equation into the discriminant part of the quadratic formula (b^2 - 4ac) you get a negative number. If you were to get 0, there would be one real solution. If you were to get a positive number, there are two real solutions. In this case, you have to realize that in order for the number to be positive and therefore have two roots, k^2 has to be greater than 43.1 (because 4<em>2.5</em>4.31 equals 43.1), and therefore the absolute value of k has to be greater than about 6.56 (because the square root of 43.1 is roughly +/- 6.56). Originally, k does not provide real solutions, so the absolute value of k is less than 6.56. No matter what value in that range it is, when you perform the correct operation, it has to end up so that the absolute value of k is greater than 6.56. When you use the values -6.5 and 6.5, you rule out every option except 15. Furthermore, if you really understand the problem, you should be able to tell that adding or subtracting any number greater than about 13.12 would provide two real number solutions because 6.56*2 = 13.12. So, it’s not really trial and error if you want to give yourself a good shot at getting the answer in a timely fashion.</p>
<p>B. a triangular region. You just need to imagine a plane and a circular cylinder in your mind and see the different shapes that result when a plane cuts the cylinder at different agles and locations.</p>
<p>As for your new problem, I get B by process of elimination. I can visualize ways in which the intersection points could by any of the others. Tell me if you need specific explanations on certain ones.</p>
<p>Edit: Damnit physicsman stop answering right before I do</p>
<p>that’s not from CB… just tough questions… I’m not sure they’re not even in the SAT… but I dunno… that feeling when you can’t solve a question. it’ so provocative!!! lol
I’m just looking for someone who’s rlly good at math i guess :P</p>
<p>OMG thanks guys really.</p>
<p>@physics, your explanation is amazing thanks… Now I know I’m stupid… it’s actually pretty easy the whole solid thing!</p>
<p>@Dumbandlethal, I ACTUALLY GET IT… haha… ohhhkay… people… I’m not usually THAT dumb… but yeah your explanation was so comprehensive… thanks! ^_^</p>
<p>Now I know who to use for math problems…<br>
I have some more… can i keep them coming? :D</p>
<p>Okay… this one is an SAT one for sure:</p>
<p>Which of the following points lies outside the circle with center (0,0) and radius 5?
A) (4.1,1)
B) (4,2)
C) (3.8,3)
D) (3,4)
E) (2,5)</p>
<p>I never knew how to solve those :/</p>
<p>@ 2200andbeyondXD</p>
<p>Just by glancing at the answer choices, I think the answer is E. If you can imagine the circle in your head, you’ll see that the point (2,5) lies outside of it.</p>
<p>If you want to be sure, just use the distance formula on each point and see which one has a distance from the origin greater than five.</p>
<p>Okay… i just embarrassed myself big time did’t i?
I actually put e, read the answer key wrongly, got frustrated, cuz it’s not the first time i get a question like this one wrong, posted and then got embarrassed… lol</p>
<p>What about this one:</p>
<p>Each of the 5 teams in a league plays each of the other teams exactly 4 times during a season. What is the total number of games played by these teams during the season?</p>
<p>a) 10 b) 36 c) 40 d) 80 e) 96</p>
<p>I wanna know a fast way to solve it. :/</p>
<p>Ok so there is team a b c d and e.</p>
<p>there are 10 matchups (write them quickly to be sure) you can also use calculator to find this by doing 5 C 2 = 10 (probability).
1ab
2ac
3ad
4ae
5bc
6bd
7be
8cd
9ce
10de
so 10 x4 = 40 is that the answer?</p>