<p>To tangent spheres technically intersect? I don't think so...</p>
<p>stop thinking so mathematically and just stop to realize that most of the cone's volume is in the top half. just from there and without doing any math the only possible answer would be 3.2. CLOSE THREAD, THANKS.</p>
<p>just kidding about last comment</p>
<p>visions of cones dance in my head</p>
<p>There's a reason you don't remember the word "similar"; the exact wording was that its shape was identical to the original cone.</p>
<p>Ah stupid mistake with the largest angle on the triangle (my way accidently gave me the supplement to the real angle of 112 degrees) and stupid mistake of substituting arcsine for cosecant on the inverse function graph thing in my mind, which subsequently made me skip over the graph of x to the third.</p>
<p>But that's only two wrong and no omitted so easy 800. Couple that with an 800 easy on bio, and i would say i had a pretty good haul. (other than US history....eich).</p>
<p>Does anyone remember the answer to the question with the diagonal going across the base of a cube with a 30 degree angle forming with the base? It asked for the volume of the cube...or something.</p>
<p>The diagonal/cube one... I skipped that one. It looked easy. 30-60-90 triangle, so the height of the cube should've been 5 sqrt(3), since the diagonal is 5 (length 3, width 4, 3-4-5 right triangle). But when I multiplied 5 sqrt(3), 4, and 3, I got an answer that was not one of the choices.</p>
<p>i think it asked for the perimeter. Does anyone remember the answer to that one??</p>
<p>Tangent spheres intersect at one point... period. The SAT Subject test on math is not a test on semantics.</p>
<p>The fact that the question specifically stated that the diameters of the spheres were not equal implies that the intersection would only be a sphere if both surfaces intersected entirely (aka the same sphere). Also the answer was to be all inclusive, so if the sphere inside a sphere solid intersection logic mandated a sphere as one answer, the same logic would produce many irregular and crazy shaped intersections.</p>
<p>
[quote]
About the problem concerning the number of real roots, it didn't say that a is not 0. If a were 0, then you could have 0 real roots. So is it 0 the answer? I hope not ;-) However, it didn't sound ok 'at least 0 roots' as the statement would have been, not that it does matter. So what do you guys think?
[/quote]
</p>
<p>I realize now, that for that question, I should've graphed it. But I put 0. :(</p>
<p>the volume of the cube was 34.64</p>
<p>3,4,5 triangle, so 5 is the length of the diagonal of the bottom face. tan(30)= height/5. tan(30)5(3)4= 34.64</p>
<p>lol not a cube but you know what i mean</p>
<p>what you did wrong with 30-60-90 triangle is that 5= height(sqrt 3). 5/(sqrt 3) *4(3)= 34.64</p>
<p>shucks. DARN IT
That was incredibly dumb of me.</p>
<p>what does about 4 wrong and 4 omitted give me?</p>
<p>lol i bombed this test so bad. i didn't study at all. if i pulled 600 out of my ass I'll be happy.</p>
<p>was one of the answers </p>
<p>1)/sqrt(k^2+1)</p>
<p>I got k/sqrt(1+k^2)</p>
<p>yea sumthing like that</p>
<p>yea that was proabably it skyway.</p>
<p>all i could remember was the denominator. The numerator i guessed that i put 1 but i most likely put k also. Was 1 even a choice?</p>
<p>1 in the numerator was a choice. The answer was k/sqrt(1+k^2) because the tangent was given such that the tangent was k/1, and the opposite side of the given angle was provided as k (with the adjacent side thus being 1). Hypotenuse, of course, was sqrt(1+k^2).</p>