|-4b-8|+|-1-b^2|+2b^3;b=-2
Help ?
its actually pretty easy
Can you explain it to me? @AllTheLonelyKillers
im pretty sure it is -23 but if you can check and make sure im not 100%
use pemdas just plug in the (b) and then solve the parentheses use what you have left in order then you can solve the full equation
I don't think I can use PEMDAS because it has absolute value brackets.. that is throwing me off @AllTheLonelyKillers
@funinabox Help?
oh he is good lol
Lol he is ! Hopefully he cn help me lol @AllTheLonelyKillers
what answer or how far have you gotten otonogold?
you can use PEMDAS, just think of the absolute value brackets like parentheses
i knew it
So it doesnt make a difference ?
would you combine like terms too ?
well, i mean it does, the absolute value brackets are not parentheses, but they are treated as such when you are using something like PEMDAS in other words, do the things inside the brackets first, always. THEN apply the absolute value
|-4b-8|+|-1-b^2|+2b^3 b=-2 so plug b in, as usual |-4(-2) - 8| + |-1 - (-2)^2| + 2(-2)^3
can you simplify each part down?
That makes soooo much more sense, I get it. thank you :)
what answer do you get? this problem is actually fairly tricky with the order of operations, its easy to make a mistake
saved the day again
-32
not quite, remember with absolute values, |-x| = x
so 32
so positive ?
|-4(-2) - 8| + |-1 - (-2)^2| + 2(-2)^3 first step, simplify it allllll down |8 - 8| + |-1 - 4| + 2(-8)
do you see how i got from the top to bottom?
no, the answer isn't -32 or 32 :)
how did you get the -8 at the end?
because when you plug -2 in for b, you have (-2)^3, which is (-2)*(-2)*(-2), which is -8
i put the parentheses in there to make it clearer what is happening
oh ok i see
alright well, can you solve |8 - 8| + |-1 - 4| + 2(-8) then?
-11?
hold up i might be wrong
yes!!! good job
-11
im right? lol
yep both of you are correct
do you see why the answer is different from -32 now?
Thank you so much :) & Yes I do, we had to break everything down
yeaa told u he was awesome
to recap - treat absolute value signs like parentheses first, solving everything inside of them, and then actually do the absolute value part to get rid of the brackets entirely :)
ok
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