<p>16 is correct.</p>
<p>Yesss! I got 16 =)</p>
<p>That makes me feel a bit better. I think that I did terribly on math... bleh.</p>
<p>
<p>it was like this:</p>
<p>(x-8)(x-k) = x^2 -5kx +m</p>
<p>and solve for m</p>
<p>i skipped that question cuz i never saw it on any practice test. can someone explain how to do it?
Just plug in any number for x and k. Then solve for m.</p>
<p>^Wow. That was simple. My approach was unnecessarily long, I guess =P</p>
<p>No, you cannot simply input any positive interger for k.</p>
<p>For example, let x = 2, k = 4</p>
<p>(2 - 8)(2 - 4) = (2)^2 + [(-5)(2)(4)] + m</p>
<p>(-6)(-2) = 4 + (-40) + m</p>
<p>12 = -36 + m</p>
<p>48 = m?</p>
<p>I think not.</p>
<p>I see everywhere a grin in that was 13- I don't remember this- could any of you refresh my memory what that question was?</p>
<p>PAHREEN is right, but I believe they gave us boundaries, right? x>m or something? I remember I got 16 but I did it again using different numbers and I got owned</p>
<p>(x-8)(x-k) = x^2 -5kx +m</p>
<p>This is how i did it</p>
<p>(x-8)(x-k) = x^2 -8x-kx+8k = x^2 - (8+k)x +8k
so x^2 - (8+k)x +8k = x^2 -5kx +m
and so i put 8+k = 5k so 8 = 4k and k=2
8k = m so 8(2) = m so m = 16</p>
<p>i hope that's right</p>
<p>i did without foiling.
for x^2 + bx +c =(x-x1)(x-x2) = (x-8)(x-k)
b=-(x1+x2)=-(8+k), c=(x1)(x2) = 8k.
the rest was the same.</p>
<p>damn i got all the way to 8k=m....then gave up cuz it looked too complicated.</p>
<p>
it said (x-8)(x-k) = x^2 -5kx +m for ANY x.
if you plug random numbers for x and k the equation might be true for those numbers only.
x=0, k=1 ---> 8=m.
can you make (x-8)(x-k) = x^2 -5kx + 8 for some other x and k?
no.</p>
<p>so...is 16 still the right answer?</p>
<p>i think i got 2 for k..but someohow i keep thinking i got 13 for the answer...0.0..</p>
<p>Does anyone remember what they got for the question that was like a triangle and then an equilateral triangle was in that triangle and the angles were 3k and 5k then you had to find xk but xk was a supplementary angle of the equil.</p>
<p>I'm guessing that was an experimental section because I don't think I got it</p>
<p>Okay good :)</p>
<p>Around what score do you think I'd get if I got like 4 wrong and left like 3 blank?</p>
<p>for (x-8)(x-k) = x^2 -5kx +m</p>
<p>was 8 one of the answer choices? Or was it only 16? Bc I can't remember what I got in that case.</p>
<p>did anyone get 13 for a question where it was like </p>
<p>2x + 3y = C</p>
<p>I got 13 for that u just simply plug it back in</p>
<p>For the questions with 11 integers and how many positive integers</p>
<p>It had to be 1 not 0 because the question said if the sum of the 11 integers was positive what is the least amount of positive integers that could be in the 11 integers?</p>
<p>If the integers were all 0, than the sum would be 0, which is not positive...
Yeah so word.</p>