<p>What about the circle O r=1 with a tangent line, where you had to find the length of a chord that extended to the tangent line?</p>
<p>What school do you go to in Oakland county, ead? I go to detroit country day.</p>
<p>i <3 my TI it does all the work on this test
it was easier thani thought cuz i totally did not study, crammin for aps and all that fun</p>
<p>For the x^3 + x^2, was it all except the one where the range of f(x) is positive?</p>
<p>y = tan x was the only odd function, right?</p>
<p>^Yeah 10 char</p>
<p>I'm back, had a nice long nap :-), not as many people posted as i thought :(. Was the ellipse thing centered around (-2,5)?</p>
<p>thats what i put. i can't believe i remembered conics</p>
<p>TI-89 shouldn't be allowed.</p>
<p>Define f(x)=when(0<=x and x<1,x, when(x>=1,f(x-1)))
-> Done
f(4.7)
-> 7/10</p>
<p>or</p>
<p>tan(-x)
-> -tan(x)</p>
<p>or </p>
<p>solve(sin(42)/4 = sin(x)/6,x)
-> false</p>
<p>or</p>
<p>sum(4/3<em>pi</em>r^3) | r={3,4,5}
-> 904.7784234
solve(Ans = 4/3 pi r^3,r)
-> r = 6.</p>
<p>or</p>
<p>(1-i)^2
-> -2i</p>
<p>or</p>
<p>e^x+x+k|x={1.27,1.28}
-> {-0.19, 0.27}</p>
<p>or</p>
<p>solve(0=ax^3+bx^2,x)
-> x = -b/a or x = 0</p>
<p>or</p>
<p>Define f(x,y) = x*(x+y)
-> Done
solve(f(x,y)=f(y,x),x)
-> x = -y or x = y</p>
<p>Waaaait hang on</p>
<p>4x^2 + 9y^2 + 16x + 90y - 245
4(x^2 + 4x) + 9(y^2 + 10y) = 245
~(x+2)^2 + ~(y+5)^2 = ~</p>
<p>(-2, -5)</p>
<p>How is this (-2, 5) ??:|</p>
<p>I thought it was easy (especially after flipping through Barron's), BUT, I also took this test cold turkey, and after a year into Calc.. which we all know rots our brains in terms of non-Calc math. I also just bought an 89-Ti after losing a Ti-83. I miss its responsiveness, seriously, I kind of regret getting the Titanium. Anyhow, 9 blank, 1 unsure, mostly due to time constraints and not enough practice.</p>
<p>It was -9x^2 not +</p>
<p>I believe the answer was >99 for the percentile, as its z-score was like 5. While, you know, just a z-score of 2 is 95th percentile..</p>
<p>Oh wait, it was! Good thing my calculator froze up during the first execution of the program, because I had typo'd it earlier.</p>
<p>vvvvvvvvvvvvv
A plane</p>
<p>Solution (A)
Imagine a plane perpendicular to the xy plane at x = 3; that's your solution set.</p>
<p>Solution (B)
Since our only requirement is x = 3, the points:
(3,1,0)
(3,0,0)
(3,1,1)</p>
<p>all work.</p>
<p>We can define a line with the first two points:
(x,y,z) = (3,0,0) + t<0,1,0></p>
<p>(3,1,1) isn't on this line (0 + 0t can't possibly yield 1), so we've determined a plane: by definition, a plane can be determined with a line and a point not on that line. (Well, it can be determined by two parallel lines, so by Playfair's axiom, we can construct one)
vvvvvvvvvvvvv</p>
<p>For the x, y, z plane problem does anyone remember the choices? One of them was half plane. What was the answer?</p>
<p>pike, it was plane. Answer choices included line, point, plane, half-plane, and.. I don't remember the last one.</p>
<p>the last choice was half-space whatever that means</p>
<p>
</p>
<p>The locus of points (x,y,z) such that x > 0 is a half-space.</p>
<p>It's just like a half-plane: the locus of points (x,y) such that x > 0.</p>
<p>what kind of math is that?</p>
<p>Geometry?</p>
<p>--
The following errors occurred when this message was submitted:</p>
<ol>
<li>The message you have entered is too short. Please lengthen your message to at least 10 characters.</li>
</ol>