<p>what calculator do you have? a lot of draw in the vertical asymptotes like ti 83, 83+</p>
<p>nah it was def III too ( i checked) and i wasnt sure but when u do f(x)=g(x) arent u setting the y's equal to each other? so....shouldnt u be solving for x?....so i put II and III....or I and III...whichever one that was.</p>
<p>the thing about the f and g functions...I put roman numeral I and III were true...that was the one about finding the intersections, right?</p>
<p>what was the answer to the area of the trapezoid #49</p>
<p>omg...it drew in the asymptotes.......LAMEE...i shouldve known that too....asymptotes...never thought of it like that...</p>
<p>These are wut im pretty sure are right answers
A+B+C was 5 for the 10 digit number (100% sure)
7.92 for trapezoid
f(x) = 1- x^2 for the thing with the sin
.5 for probability
2 lines for the asymptotes
y = x for inverse</p>
<p>so thats 2 omits and 2 wrong so far for me....gah...</p>
<p>how did you do the a+b+c one?</p>
<p>I got all those answers as well</p>
<p>The only one that seemed weird to me...domain was {1,2}, what could range not be? I said {1,2,3} but i couldn't remember what the syntax meant...</p>
<p>i had time so i did it guess and test....but i guess i shouldnt have rushed through problems like the asymptote one</p>
<p>ur right arsenal. function = every input must have only 1 output. if u have 2 inputs, u cant have 3 outputs :p</p>
<p>what is 2 omit, 5 wrong</p>
<p>yeah it was {1,2,3} b/c you cant have 3 y values for 2 x values in a function. the brackets just meant those points. ex: x could be 1 or 2, y could be 1, 2, or 3.</p>
<p>oh yeah...that range question...I put (1,2,3) couldn't be it.</p>
<p>answers i remember:</p>
<p>15.6 (side of triangle, involved sin laws)
38.4 degrees
n^2 + 2, the geometric sequence problem
lim = 5
3.87, x is not defined
.27, lnx intersect e^(-x)
{1,2,3} range not possible if f(x) is function at {1,2}
period = 2pi, it was absolute value of sumthing
tan2x period = pi/2
the circle one was r^2 = 4
2-3i goes with 2 + 3i
the height of the tree was like tan(sum degree) = x/100</p>
<p>the height of the tree was tan(something)=x/100, but you had to add the guy's height to that too</p>
<p>for the 2+3i one couldnt it have been -2+3i also? b/c (-2+3i)(2+3i)=-13 so i put not enough info</p>
<p>****, i hate missing easy problems</p>
<p>but if you think about the quad form its a +- bi... so i think that it was the inverse</p>
<p>I put a-bi...why isn't there enough info.?</p>
<p>im not sure about the complex # one....i always learned the conjugate was neg the imaginary part...but it does work out to be -13.......man...if i got another one wrong....&%$*</p>