<p>Yeah, but if the first coefficient was zero and there were two pairs of complex roots? Idk still…</p>
<p>@LoveArtForever–If the first term is taken out, then the polynomial wouldn’t be 5th degree anymore. IT would be quartic</p>
<p>The terribly worded cone question…</p>
<p>it wants a “similar” “shaped” cone that has half the volume.</p>
<p>If you actually work out the volume for the cone of depth 3.2 and the “similar” radius of 1.6 the volume is .2 off… 8.3 vs 8.5</p>
<p>The realy problem is the usage of the word “similar”, if we use the rigorous definition which implies a reasonable amount of congruency i.e. same angles and thus ratio of radius to height, NONE of the answers come out exactly (or within the rounding on the SAT).</p>
<p>After alot of debate during the test I determined that the SAT had decided “similar” meant something else and a cone of radius 2 and depth 2 does indeed have the volume that is wanted by the problem.</p>
<p>similar in the cone problem meant that the slant height and radius were scaled proportionately…</p>
<p>Yeah, but nothing ever said the coefficients must be non-zero… I’m sorry I’m being stubborn, haha, I know I’m wrong on this problem but just trying to justify it…</p>
<p>@tetratic:</p>
<p>Okay the cone problem said:</p>
<p>The radius is 2
The height is 4
If there is another similar cone with half the volume, what is the height of the smaller cone?</p>
<p>Actually - I’m not sure if the words smaller or similar were mentioned. It had to be one or the other… anyone remember?</p>
<p>If you do the problem with 3.2 and the radius of 1.6 (which is proportional) you do not get “half the volume”</p>
<p>@ tetratic:</p>
<p>I don’t think we had to consider the slant height in the problem…</p>
<p>This is because SAT’s answers are rounded, you get exactly half the volume before rounding. The exact height would be something like be 2 times the cube root of 4. There is a difference between rounding an answer and taking a rounded answer and plugging it back into the equation. </p>
<p>I can see a reasonable justification for both questions, but if they wanted the new cone to have a radius of 2 as well, they would have directly told us. I don’t remember the exact wording but it certainly did not tell us any of the dimensions of the new cone. Assuming similarity is only logical.</p>
<p>Volume of given cone of R=2 and H=4
16pi/3</p>
<p>Half of the Volume
8pi/3 = 8.377</p>
<p>[(3.2/2)^2]pi x 3.2 / 3 = 8.5786</p>
<p>Even with SAT rounding… 8.377 =/= 8.5786</p>
<p>I believe the wording was “similar shaped”</p>
<p>@alyssa193</p>
<p>Okay the cone problem solved:</p>
<p>initial volume = pi 2^2 x 4 /3 = 16 pi /3</p>
<p>half volume = 8 pi/3</p>
<p>to keep proportionality, height = 2 times radius (h=2r)</p>
<p>plugging stuff in:</p>
<p>8 pi/3 = pi r^2 h /3</p>
<p>8 = r^2 (2r)</p>
<p>r^3 = 4</p>
<p>r = 1.58</p>
<p>Yeah it was 2x-2y.</p>
<p>Anyone else solve the parametric one by subbing the t in y = 5sin( ()T ) term by isolating the t in the x =3cos( ()T ) equation and getting something like y=5sin (cos inverse( ))
on my calculator it looked like the upper half of a circle with its center at the origin, but did not exist below the x-axis. I got answer (C)… thoughts?</p>
<p>@tetratic:</p>
<p>what was the height then? the radius is roughly 2…</p>
<p>If you graph the parametrics using your calculator (set under PAR in MODE), you see that it is an ellipse.</p>
<p>tetratic gets it for the cone question, I’m confident he is correct. The height is two times the radius (as it is similar to the original cone) so you multiply 1.58 by 2, then round.</p>
<p>See if if you put in a filled sphere inside a filled sphere, the resultant points from the intersection forms a sphere.</p>
<p>If we assume that similar meant identical proportions, then the correct height, or value of “d”, is (32)^1/3. If it said that the radius stays constant, d=2.</p>
<p>Did anyone get that problem where the surveyor measured a 30 degree angle between the ground and the top of the cliff, and then he moved 30 more feet and measure 45 degrees between the ground and the top of the cliff?</p>
<p>how many can you miss or omit to get at least 750? is 750 considered a good score?</p>