SAT 2 Math Level II Discussion May 2 2009

<p>awkturtle
i think the way i did the circle one was that I made an isosceles triangle out of the problem. two sides are 5 and the angle between the two sides is 105. then i did the law of cosines to find the chord length.</p>

<p>the polar coordinate one was E, the one with the ridiculously large fraction for the “y” part of the polar coordinate (the question asked which coordinate did NOT represent)</p>

<p>-inequalities with k, k max value: 1/3
-volume of cube 27x^3
-probability of choosing even integers in set: 16/81
-graph question with f(100): 1?
-cosx when given sin y: -4/5?
-cosxsinx max: .5</p>

<p>my answers
-inequalities with k, k max value: i got none… O.O
-volume of cube 27x^3 yes
-probability of choosing even integers in set: 1/6
-graph question with f(100): 1? yes
-cosx when given sin y: -3/5 (I think i got this one wrong… anyone want to explain this one to me?)
-cosxsinx max: .5 yes</p>

<p>I got all of those too, except for the probability one. there were 4 even out of 9 total, so the prob of picking two of them is 4/9 * 3/8 = 1/6 no?</p>

<p>@ldftalk, for the one w/ choosing even integers, was there replacement? I thought I read without replacement, so I had 1/6.</p>

<p>i got 1/6th as well, how about the k, k max value? i remember putting 1/3</p>

<p>i got 17 for the circle question with an angle of 105. i did .5absin(c)=.5(d1*d2), found d1 with law of sines and solved for d2…</p>

<p>also for the probability of choosing even integers in set, wasnt it just (4/9)^2=16/81? im pretty sure that was the answer…it never said one is removed after choosing the first one (not 3/8 for second selection)</p>

<p>but i thoguht the inequalities one was "|3k-4|is less than … right? so 1/3 would be wrong… i think?</p>

<p>1/6 for me.</p>

<p>|k-2| < 1, |3k -6| <3 for the inequalities one. For the choosing one, I’m pretty sure it said without replacement.</p>

<p>i thought the problem was |3k -6| <1???</p>

<p>|x-2| < k
|3x - 6| < 1</p>

<p>what value of k satisfies the inequality.</p>

<p>3(|x -2|) < 1
|x - 2| < 1/3</p>

<p>k = 1/3</p>

<p>my god i don’t even remember the law of sines! i think i skipped that one but i might have guessed in the end i don’t remember. that’s what happens when i took geometry like 4 years ago lol!</p>

<p>the probability one said nothing about replacements so i just assumed there wasn’t one…so 1/6
cos x when given sin y- i think i put positive 3/5…
why is the f(100)=1? i think i put 0…maybe i just counted strangely…</p>

<p>The graph of f(x) repeats every 7 units. 14 * 7 = 98, so f(100) = f(2) = 1</p>

<p>is cosine of the big angle the same as the cosine of the small angle?? the one that had a 3-4-5 looking triangle angle</p>

<p>I used my calc to do that one fast, but this is how to do it:</p>

<p>cos(x) = cos(180 - y)</p>

<p>= cos180cosy + sin180siny
= -1 * 4/5 + 0 * 4/5
= -4/5</p>

<p>how was the probability one 1/6…wasnt it a set of nine numbers with four evens in it- so the probability of choosing two even numbers would be (4/9)(4/9)=(4/9)^2=16/81</p>

<p>what about the very last question? how many seconds after the ball reached the maximum</p>

<p>for the polar question, was it (-1, -4(pi)/3) [was there a negative sign in front of 4]? The original was (1, pi/4)</p>

<p>no f(100) is 0
because 100/7 = 14 and remainder of 2
u don do f(2), u look at the second value in a period, which is 0
</p>