<p>Yeah should be x cubed…
So sad I got the cone problem wrong… Similar cones -_- 4 omit and one wrong should still equal an 800 right?</p>
<p>@juchipan yeah the range would increase if there was a smaller number</p>
<p>The 99 numbers one was quite easy.</p>
<p>Think about it.</p>
<p>Say the data set was 1,3,4,5,6
The range would be (6-1)=5</p>
<p>If we add in a 5th number, this number cannot decrease the range. It can only increase the range or keep the range the same.</p>
<p>For example, if the 5th # were 7 for example, the new range would be 7-1=6(so range has increased). If the 5th number were a lower number like 2 it would not change the range as the max and min of this data set is the same(so range would still be 6-1=5)</p>
<p>sure basically lets say the list had the numbers 3,5,7,12, 12 in it. (i relize theirs more it doesn’t matter though for this example). If you add 3 you would decrease the median, the mode, the mean and the standard deviation the range however can’t be decreased by adding any number only increased since 12-3 = 9 but if you added six 12-3= 9 still.</p>
<p>That was ridiculously easy…</p>
<p>Spheres intersect at either a circle or a point</p>
<p>can someone post the abs question so far i think I’ve only got one wrong though my memory is foggy with what i put down for some</p>
<p>Intersection of spheres question was definitely sphere and point. Circle is not possible because any vaguely circular shape that the intersection could make would have thickness.</p>
<p>It was 2x-2y and does anyone remember which number question the sphere/point/circle question was? Was it in the twenties/thirties?
And also what did you guys get for the x=3costheta and y=5sintheta(not sure if problem is accurate) but it was an ellipse right?</p>
<p>It was the surfaces of the sphere, each of different diameters</p>
<p>@globalwolf </p>
<p><a href=“https://en.wikipedia.org/wiki/Sphere–sphere_intersection[/url]”>https://en.wikipedia.org/wiki/Sphere–sphere_intersection</a></p>
<p>by the way, it isn’t physically possible for two surfaces to intersect and form another surface unless they’re the same sphere, which was ruled out because they specified that the spheres were of different radius</p>
<p>jeffisaboss, I don’t remember the problem, but the answer was 2x-2y. If you give me the problem I’ll walk you through it</p>
<p>walahoo, the problem was accurate. it was an ellipse and the coefficients were fractions</p>
<p>the question not the answer and yah it was an ellipse</p>
<p>just remembered the abs problem. it was:</p>
<p>if x-y>0, find | y-x| - (y-x)</p>
<p>well, because it’s absolute value, we have:</p>
<p>| y-x| = | -(y-x) | = | x-y |</p>
<p>since x-y>0, |x-y|=x-y</p>
<p>Thus, x-y - (y-x) = 2x - 2y</p>
<p>q.e.d.</p>
<p>@tetratic: The question was asking the shape of the border that the intersection would make then? Because I interpreted it as the three-dimensional overlapping part that the spheres would make, and in this case if the smaller sphere fit completely inside the larger sphere, the overlapping parts would be a sphere.</p>
<p>Was the abs value problem choice A?</p>
<p>The parametric one was E right. The one that looked like an ellipse.</p>
<p>X-y greater than zero, abs(y-x)-(y-x) I believe</p>
<p>i agree with globalwolf</p>
<p>@medicalboy yeah I got e</p>
<p>well if the smaller sphere was within the bigger one, that still wouldn’t create a circle…</p>