<p>@ban8431 & rdward96 i thought you had to use the opposite parts of the graph when its 1/cosx</p>
<p>@rdward96 I got 45 and 315</p>
<p>@Ban8431 it was a phase shift. Translation of -1, 2</p>
<p>@swagkinggg i think since it was cos, and had to be positive it would fall in the first and fourth quadrant. the reference angle of 45 degrees is already in the first quadrant, so then to find the angle in the fourth quadrant you subtract 45 from 360 and get 315. if that is the question you’re referring to</p>
<p>any other questions that people had trouble on? </p>
<p>@ban8431 yes i got the reference angle to be 45 (luckily from guessing and checking) but i thought since it was 1/cos i thought you would use the opposite ones. Either way i probably got some credit and I’m proud of pulling those points from out of nowhere </p>
<p>Last one we all got 58 and 39, right?</p>
<p>@swagkinggg hahah i’m not too sure thats just the way i did it, but hey props to you</p>
<p>for the log~base (5x-1)~ 4= 1/3 , did you guys get x=13?</p>
<p>@Ban8431 yes, and it checked as well</p>
<p>@Ban8431 yep, that’s what I got</p>
<p>@ban8431 yup you could have also checked it </p>
<p>was it 3 log 2 + 3 log x</p>
<p>awesome thanks guys. this is a great thread my anxiety levels have decreased so much </p>
<p>@galaxyexplorer you could have plugged it in with a number for x I’m pretty sure i got the right answer i just don’t remember what i put </p>
<p>@ban8431 yes for me too! anyone in here taking the english tomorrow?</p>
<p>Do you guys know the answer for that secx - radical(2) are 60’and 300</p>
<p>@galaxyexplorer it was actually log2+3logx. The whole thing wasn’t cubed in the origional, it was only the x. </p>
<p>Do you know how exact we had on the function we had to graph? I moved the x-intercepts of the first one 1 to the left and 2 up, plotted those points, and kinda sketched from there</p>
<p>@swagkinggg yes for english! </p>