<p>24.8 is correct... </p>
<p>I think that the answer is x = 1.
Both the z coordinates are the same.. so setting the z as a plane... u get it in 2D with slope = infinity... x = 1 has a slope equal to infinity (tan 90)</p>
<p>Both the x's were the same, too, so the set of all points equidistance could have had any x and z and still been equidistant in those dimensions.</p>
<p>Therefore, the solution set was a line. Only the y's were different, and the average was 3/2, so the answer was y = 3/2.</p>
<p>i figured it out the same way as theoneo</p>
<p>possible...yet if you normalise the plane..
x or z... u get a straight line with value 1</p>
<p>What? An xz plane? Then you wouldn't control the y's.</p>
<p>Oh and the one about the period.. something like lsin2xl.</p>
<p>was it pi/2? or pi/4 ? I can't remember exactly. can someone put the question?</p>
<p>2 pi. The period changes for a phase of pi... so x/2.. would have 2pi</p>
<p>I got pi/2 for the area problem and 2pi for the period problem.</p>
<p>which area problem?</p>
<p>The area of a triangle with two vertices at the x-coordinates of a function (I think that was |sin(x/2)|) and one vertex at the local max of the same function.</p>
<p>It ended up being A = bh/2 = pi/2 * 2 * 1/2 = pi/2 (choice A). I even calculated the integral, which was 2, and pi/2 was the only choice less than 2 (the triangle was inscribed, so its area was less than the integral).</p>
<p>can u pls state the full question? I forgot.. what was the function?</p>
<p>OO the period function was on my calculator haha. It was f(x) = |cos(x/2)|+3. The answer was 2pi.</p>
<p>I worked it out myself... thot so... whats the area questionplease?</p>
<p>How about the one that was values from [-4,4] and m^n had to equal n^m or something along those lines? How many values (m,n) make this work??</p>
<p>was it 2? hmm..my meomory is fading..</p>
<p>Ok the area question was f(x) = |2sin(2x)|. There was a triangle with vertices at the x-intercepts (0 and pi/2) and at the local maximum (pi/4) The base was pi/2 and the height was f(pi/4)=2, so the area was pi/2<em>2</em>1/2 = pi/2.</p>
<p>I got 5 for the 2^m = 4^n. The values of (m,n) that worked were (-4,-2), (-2,-1), (0,0), (2,1), and (4,2)</p>
<p>how do u guys remember so well? like the exact numbers, answer choices etc.</p>
<p>One of the questions was to figure out how many pairs of m and n there are between -4 and 4 when 2^m = 4^n. There are five.</p>
<p>2^4 = 4^2
2^2 = 4^1
2^0 = 4^0
2^(-2) = 4^(-1)
2^(-4) = 4^(-2)</p>
<p>oh yeah that was 5. my bad</p>