Thursday, September 15, 2011

can someone tell me how to solve this:
pretend square brackets are absolute value signs:
[3x+1] +x greater than or equal to [2x-1]
squaring both sides doesn't work ): and i've never leant how to do the "plus x" bit ):

physics, massive fail. i promise myself never to study the night before. this will never again happen. my results will reflect my poor effort. i kept on telling my: key words, key words, key words and then i forgot about "Gravity" and red-shift and doppler effect. all those easy concepts to write about.
and i am bad at bs-ing so they are going to take marks of me for incorrect info ):
the test itself wasn't hard, and time wasn't an issue, if ONLY I MADE A BIGGER EFFORT APART FROM THE NIGHT BEFORE then i think i would have been fine, but obviously i'm not a cramming genius, like some people are ):
and i have a really bad memory ):
mc was hard ):
and i was talking to a girl and she said she's done 90% of the paper before D= those curry tutoring places *shakes fist
but i defs need phys tutoring next year.

and english, i also promise myself to have essays with less than 900 words so that i can add more stuff to it during the exam. i didn't even have time to write my last word: vulnerability, so i ended up writing a v and then a squiggle.
the question wasn't as hard as i thought it would be, but i think i didn't answer the question properly ):

i hope tomorrow is going to be productive, unlike today after phys which i was like ceebs, i have tomorrow....
and i was just watching the yr12 finale, variety night, and i have to agree, 3:54 is very amusing, well, i couldn't stop laughing.

3 comments:

  1. I also failed phys (again) because of night-before study but that's okay because I'm probably dropping phys.

    To solve equations with multiple absolute values the cleanest way is to take each absolute value as a conditional statement.

    From the definition of absolute value, |x| is:
    x for x>=0
    -x for x < 0

    so |3x+1| is the same as
    3x+1 for 3x+1>=0,
    -3x-1 for 3x+1<0

    So basically you have two critical points, -1/3 and 1/2 where the graphs change. Below -1/3 |3x+1| = -3x-1 because 3x+1 is less than zero, and 2x-1 = -2x+1 because 2x-1 is less than zero.

    Then solve each section of the equation (x<-1/3, -1/3<=x<=1/2, x>1/2) with the absolute value, apply the condition to the result you get, and then get all the conditions together.

    So for example:

    If -1/3 <= x <= 1/2:
    3x+1 + x >= -2x +1 (since |3x+1| = 3x+1 when 3x+1>0, and |2x-1| = -2x+1 when 2x-1<0)
    4x + 1 >= -2x + 1
    6x >= 0
    x >= 0 FOR -1/3 <= x <= 1/2
    0<= x <= 1/2

    Solve in the same way for other two conditions.

    Note that if you solve into equation and get no solution, you can just ignore the solution.

    ReplyDelete
  2. Squaring both sides *should* work...

    But failing that, there's a good, Icy method, right there.

    ReplyDelete

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