SAT 2 Math Level II Discussion May 2 2009

<p>i believe the area of the triangle was 6.</p>

<p>I put r+s+1 for the prime number. it worked out for a few numbers I tried.</p>

<p>Area was 6.</p>

<p>didn’t prime question say “have to be” prime?</p>

<p>it said COULD be prime</p>

<p>It said some had to be prime (not all of them) I think.</p>

<p>yeah, it said could be prime. I think r + s + 1 has to be prime and can be proven by fermat’s little theorem [maybe], but I’m not sure.</p>

<p>-1!. (prays he didn’t miss another question or by by good score)</p>

<p>no it does not. 13+7+1=21 which is not prime</p>

<p>It was “could” be prime.</p>

<p>last time… i hope some new person will remember:
for the angle with the 3-4-5 triangle, what was on the x axis? the 3 or the 4?? i thought it was 3. can anyone back me up on this?</p>

<p>Never mind then :[</p>

<p>To consolidate, I think the consensus is:
Dog Age: I and II correct
Primes: r + s + 1
f(100) = 1
Last answer: 1 second between max and 20
f(x) = nx –> -1 and 1
k max value = 1/3
Prob. of even integers = 12/72 = 1/6
max of cosxsinx = .5
cosx when given siny = -4/5
Area of triangle using 1/2absin(t) = 6</p>

<p>I think I got these right, but omitted two.
Anyone remember what q48 was? I omitted that I think.</p>

<p>the 4 was on the x axis. siny = 4/5 = opposite /hypotenuse. y was the top angle. 3 is on the y axis.</p>

<p>The answer is cosx = -4/5, which has been posted several times by a variety of means.</p>

<p>Oh and area of the log graph, wasn’t it 90 + 45 = 135?</p>

<p>i think i already missed 3 and omitted 2 or 3(i don’t remember). :(</p>

<p>i think i remember omitted or guessing on 48.</p>

<p>i want to say it was the one with f(x)=e^5x what is inverse of …?
i think i put c…which i think was .34. not sure though?</p>

<p>question 48 was a simple application of vieta’s formula. It asked that if s and t were the roots of a function f(x), which of the following has roots s + 1 and t + 1. The answer was x^2 - 9x + 10.</p>

<p>Yes it was 135. Took me a while to do that one properly.</p>

<p>wasnt the probability of choosing an even number (4/9)^2=16/81 b/c there were 4 evens in the set of 9 numbers and you had to choose twice…</p>

<p>for the inverse, it was .34 definitely. for the area of the log graph, its 135. The reason being height = 100 - 10, b1 being log10 = 1, b2 being log100 = 2.</p>

<p>A of trapezoid = 1/2 (3)(90) as the area, so 135</p>

<p>nvm. 135 is the right answer.</p>