Math Iic Discussion Thread

<p>i dont really remember lol</p>

<p>I put f(x)<0. I was having trouble thinking of an instance where it would still work if f(x) wasn't negative.</p>

<p>i think ur right...i just thought of that...funny though i dont remember it being a choice</p>

<p>It was C (f(x)>0 was B, yours was D, and symmetric over origin was E. A was wrong, don't remember it.) The thing is: I'm also having trouble thinking of an instance where it isn't symmetric over x-axis.</p>

<p>"I agree, cities one was 2.3 or something like that"</p>

<p>Thank you!</p>

<p>And for f(x) < |f(x)|, I also put that f(x) < 0.</p>

<p>if f(x) < 0 what wat about stuff like x^2- 2 would that be f(x)<0?</p>

<p>"An airplane is at an altitude of 8 km flying towards cities A and B. Angles 37,25... Distace between A and B?
I got 2.3 or 2.5... forgot what the exact decimal was, but it was the only choice close to my calculations.
I used tan to get the base to the left of the first city, then did law of sines to get the answer."</p>

<p>The degrees were 39 and 27. "8/(tan 27)- 8/(tan 39)= 5.81" I think</p>

<p>To blah: but in that case, the absolute value would touch it in most places (you'd have to confine the range to only negative values, but (unfortunately) then it would still be symmetric :S</p>

<p>^koax i think i did the same as u</p>

<p>isnt it multiply though?</p>

<p>Isn't what multiply?</p>

<p>oh nvm i did tan(63)8 - tan(51)(8)</p>

<p>Is it possible CB screwed this question up? For the life of me, I can't find anything wrong with either answer.</p>

<p>for the plane/cities problem, i did what blah123 did</p>

<p>koax: i think it's just two takes on the same problem, though?</p>

<p>I'm referring to the f(x) doesn't share a point with |f(x)| question. It seems like the answer is both that it's reflected over the x-axis and that f(x) is less than zero.</p>

<p>"It seems like the answer is both that it's reflected over the x-axis and that f(x) is less than zero."</p>

<p>The question specifically said that they don't share a point. Even if it is reflected across the x-axis, it's still possible for them to share a point. On the other hand, any negative value will be less than |f(x)|.</p>

<p>actually the answer to the f(x) and lf(x)l was that it was a constant.
ie if f(x) = -4, lf(x)l would be equal to 4, thus having no points in common</p>

<p>"if f(x) = -4, lf(x)l would be equal to 4, thus having no points in common"</p>

<p>That's the exact same thing as f(x) < 0. lol.</p>

<p>
[quote]
"if f(x) = -4, lf(x)l would be equal to 4, thus having no points in common"</p>

<p>That's the exact same thing as f(x) < 0. lol.

[/quote]
</p>

<p>Exactly, constant wouldn't work since if f(x) = 4, |f(x)| would also = 4, thus sharing points.</p>

<p>iamnervous is correct. Good for me! I had forgotten about 0.</p>

<p>what was the one that asked for the frequency and said f= 1/t (t=period)</p>

<p>i put 1/2pi. I plugged the function into my calc and found that the period was 2pi.</p>