<p>What did you guys get for the question:</p>
<p>n is an integer on the interval one to ten inclusive
m is an integer on the interval one to twenty inclusive</p>
<p>What is the probability mn is odd</p>
<p>I got 1/4 anyone confirm?</p>
<p>And i do remember getting 64a6</p>
<p>gunster:</p>
<p>The actual question was, on the interval 0<x<2pi, there exists an integer x such that:</p>
<p>sin(2x)^2+cos(2x)^2=2 –> 1=2 –> no!
2sin(2x)^2+cos(2x)^2=1 –> sin(2x)+1=2 –> yes
sin(2x)^2+2cos(2x)^2=2 –> cos(2x)+1 –> yes</p>
<p>Using the Pythagorean identity sin(x)^2+cos(x)^2=1</p>
<p>Zapa: I put 1/4: You have a 1/2 chance of getting an odd integer from the first set, and a 1/2 chance in the second. The only way to get an odd integer from two integers is if both are odd. So 1/2*1/2=1/4.</p>
<p>1/4</p>
<p>In order to get an odd integer as a product it requires the multiplication of two odd integers.</p>
<p>So yes, it would be 1/2 * 1/2 which is 1/4</p>
<p>^ Yeah it was 50/200 which is 1/4</p>
<p>Yeah I also had 64a6, which I think was Choice C. Do you remember it being Choice C?</p>
<p>Ah ok, I didn’t read the “exists an integer x”, so ok it is definitely II and III.</p>
<p>Also, i put 64a^6 as well.</p>
<p>x^30y = (x^5y)^6</p>
<p>x^5y = 2a</p>
<p>therefore, (2a) ^ 6 = 64a^6?</p>
<p>I vaguely remember it being C</p>
<p>:), i left the mn problem till the end and solved it in the last minute.</p>
<p>for the four digit no digit repeated question. the possibilities was 24</p>
<p>since you could make 6 with one digit </p>
<p>and there are 4 digits 6*4=24</p>
<p>wouldn’t it be higher than 24 because that would be correct for an answer for a 4 digit number, so if it is a three digit number, wouldn’t there be more possibilities?</p>
<p>^ It was asking for a four digit number.</p>
<p>it was 4 digits and u make 3 digit numbers out of them:</p>
<p>so, assuming u have 1, 2, 3, 4
For 1 u could make:
123<br>
124
132
134
142
143</p>
<p>Which is 6 possible combinations. Each of the other 3 will also have 6 combinations if u start out with each digit. Therefore, it is 6 * 4 = 24</p>
<p>gunster…
the problem was “might” not an absolute answer
obviously if it was absolute, there is no answer
but they were asking any possibility, thus the answer is II and III…</p>
<p>^ yes i’ve realized my mistake, as you can see in my previous posts. I misread the question and didn’t see the “possible” part :)</p>
<p>Shame, I studied all those trig identities, saw a problem such as that, and immediately marked none of the above. Now, I find out that I was wrong on the only trig identity problem on the test. >.<</p>
<p>Was the random integer question really looking for an odd? not an even? Well that’s disappointing, I put 3/4ths thinking that it was looking for an even. =</p>
<p>for the trig function one:
2sin^2(2x) + cos^2(2x) = 2
sin^2(2x) + 2cos^2(2x) = 2</p>
<p>both should have been false right?
i just set 2x to y (to avoid complication) and use the pythagorean identities </p>
<p>2sin^2y + (1-sin^2y) = 2
sin^2y + 1 = 2
chose a random angle… say 30degrees, and got 0.25 + 1 = 2… which is false</p>
<p>(1-cos^2y) + 2cos^2y = 2
1 - cos^2x = 2
chose a random angle… say 30degrees again, and got 1 - 0.75 = 2… which is false</p>
<p>How about choosing 45 and 180 degrees, respectively?</p>
<p>again, it has to be valid for ANY angle between 0 and 360… if it’s not for one, then the “identity” is false</p>
<p>^Please read the question and previous posts on it.</p>
<p>quote from Presidont:
The actual question was, on the interval 0<x<2pi, there exists an integer x such that:</p>
<p>I put none too, but if it actually did say “there EXISTS” then yes it is II and III.</p>
<p>Sadly…I didn’t read the question at all now it seems because I don’t remember ever seeing that part. Lesson learned…hopefully 
lol</p>
<p>guys the problem that is asking about the roots of 2-i is wrong. The question was actually asking other roots when 3 and -2+i are given.</p>
<p>the answer would be -2-i, not just 2-i.</p>
<p>^ scimmy boy
thanks for the clarification</p>
<p>ya i was confused whehter the given root was 2+i or -2+i</p>
<p>if it was -2+i it is -2-i which i remember to be C not A</p>
<p>A was 2- i~!!</p>
<p>VERY TRICKY!</p>