math quest. has me stumped...

<p>to some, this will be an easy question but its not for me. i put down E (no i did not guess). i used the triangle inequality method and well i got 5. the answer is A.</p>

<p>If x is an integer and 2<x<7, how many different triangles are there with sides of lengths 2,7, and x?</p>

<p>A. 1
B. 2
C. 3
D. 4
E. 5</p>

<p>How did you get E?
The inequality rule is that the sum of two sides has to be greater than the other side. For example, sides A + B + C, A + B > C , A+C > B , B + C > A.</p>

<p>So looking at this. You can start from the integer 3 and go up to see which one works. There's no way you can get E, because there are only 4 possible lengths for X, since only the integers 3, 4, 5, 6 are available.</p>

<p>To help yourself, draw a triangle and label 2 of the sides as 2 and 7.
Obviously 3 does not work because 3 + 2 = 5 and that is less than the side with the length of 7. Only 6 works.</p>

<p>thanks for the reply. like u can see, im not a math person. i see it better and understand wat i messed up on.</p>

<p>You'll get it, and you're a sophomore, so you're fine...
I used to really suck at SAT math, and I thought I was just bad at it. But the math is really simple and it really just takes getting used to the types of problems asked. Keep going over your missed problems and make a mental note or physical note of that problem..</p>

<p>aha about being a sophomore ahaha.
haha no im barely starting freshman year....</p>

<p>i dont think i tricked anyone haha
im a junior so i need to improve more on math. i am seeing improvement in me. i just make silly mistakes so i gotta fix that.</p>

<p>Been many years since we did this... Took me some time to remember. :D</p>

<p>Haha this inequility rule is sth that i always forget :}
And come on people ,what does ''i am not a math person means '' ?! Everyone can be 'a math person' if practise enoguh</p>

<p>True, Ivan :)</p>

<p>Actually, all you have to remember are these two steps in finding the "missing side" of a triangle.</p>

<p>A-B+1 < X < A+B-1; X is the third side.</p>

<p>7+2-1=8, 7-2+1=6; X has to be 6, 7, or 8.
Since the restriction is 2<X<7, only 6 works.</p>

<p>so its A</p>