Hard Math Question

<p>Given three segments of length x, 11-x, and x-4, respectively. Which of the following indicates the set of all numbers x such that the 3 segments could be the lengths of the sides of a triangle? </p>

<p>A. x>4
B. x<11
C. 0<x<7
D. 5<x<15
E. 5<x<7</p>

<p>
[quote]
Given three segments of length x, 11-x, and x-4, respectively. Which of the following indicates the set of all numbers x such that the 3 segments could be the lengths of the sides of a triangle? </p>

<p>A. x>4
B. x<11
C. 0<x<7
D. 5<x<15
E. 5<x<7</p>

[/quote]
</p>

<p>Well, you have to understand that in a triangle, one side may never be larger then the sum of the two other sides. With that in mind, we see that only E is correct.</p>

<p>The sum of 2 sides of a triangle is greater than the 3rd side. So,
x<11-x+x-4
x<7....(1)
Also, x-4 has to be 1 or greater.
x-4>1
x<5....(2)</p>

<p>Combining 1 and 2,
5<x<7</p>

<p>Omg I wish I got these type of questions on my SAT -_-
If you understand rules and properties of triangles, this should be one of the easier questions on the test.</p>

<p>I'm not sure I understand Fantasy's solution - it has some errors in it. Here's one way to do it:</p>

<p>First recall the triangle rule: The third side of a triangle is between the difference and sum of the other 2 sides.</p>

<p>Now, using x as the third side, we add the other 2 to get (11-x) + (x-4) = 11 - 4 = 7. So by the triangle rule, x < 7. So we can eliminate (A), (B) and (D).</p>

<p>Can x = 5? Well, if x = 5, then the 3 sides are 5, 6, and 1. But 5+1=6 violating the triangle rule. So x cannot be 5, and we can eliminate choice (C). </p>

<p>mmm... well I still have one thing on my mind:</p>

<p>If we try x - 4 < x + (11-x)</p>

<p>then x < 15 , why don't we take the 15 and take the 7 from the other inequality instead?</p>

<p>Yes, x does have to be less than 15. But it also has to be less than 7. If a number is less than 15 and it's less than 7, then it's less than 7.</p>

<p>In other words you always have to take the intersection of all the restrictions - generally the condition that is most restrictive.</p>

<p>As a simple example, if 2<x<10 and 0<x<6, then 2<x<6</p>

<p>Furthermore, if you didn't see that x had to be less than 7, you can simply plug in more numbers to eliminate answer choices as I did in the second part of my argument.</p>

<p>I take back the second part of my answer.... I don't know what the crap I was thinking when I wrote it... it looks like rubbish to me now.... thanks DrSteve!</p>

<p>I understand it now! Thx a ton DrSteve! :)</p>

<p>Here's another hard math question:</p>

<p>y = -2(x-2)^2 + 3</p>

<p>In the xy-plane, line l passes through the point (4, -5) and the vertex of the parabola with the equation above. What is the slope of line l?</p>

<p>(A) -4
(B) -1/4
(C) 0
(D) 1/4
(E) 4</p>

<p>Answer is (A). Please tell me the shortest way to do this problem. I was able to do it, but after time ran out! I first found the intersection points of the parabola with the x-axis, then added them and found their average, which is the x-coordinate of the vertex of the parabola. Then I used the equation to find the y- axis of the vertex, and thus found the slope. Is there a faster way?</p>

<p>Vertex is (2,3)</p>

<p>Now you have two points to find the slope of line l..</p>

<p>(-5-3)/(4-2) = -4</p>

<p>The original function is in vertex form. You can look this up... Or, you could expand the equation and find the vertex using the formula, -b/2a.</p>

<p>Thanks JefferyJung! Just discovered the vertex formula now!</p>