PLEASE HELP WILL MEDAL AND FAN I-4b - 8I+I -1 - b^2I + 2b^3 b = -2
@TheSmartOne
are those 1's?
l?
wait what
@TheSmartOne could probably help
or maybe @phi could
I can't make any sense of it.
I'm sure it's in the equation bar
Is it absolute value?
YEAh that
Sounds like you want to find the value of \[\Large \left|-4b-8\right| + \left|-1-b^2\right| + 2b^3\] when b = 12
b = 2 yes
I'm having trouble solving it
b = 2 or b = 12? Basically, you substitute whatever value of b in to it and simplify
B = -2 sorry
Replace every copy of 'b' with -2 to get \[\Large \left|-4b-8\right| + \left|-1-b^2\right| + 2b^3\] \[\Large \left|-4*({\color{red}{b}})-8\right| + \left|-1-({\color{red}{b}})^2\right| + 2({\color{red}{b}})^3\] \[\Large \left|-4*({\color{red}{-2}})-8\right| + \left|-1-({\color{red}{-2}})^2\right| + 2({\color{red}{-2}})^3\] Do you see how to finish?
b=-2 \[\left| -4 \times- 2-8 \right|+\left| -1-(-2)^2 \right|+2\left(- 2 \right)^3=?\]
I think so.. Do I multiply -4 and -2 then subtract eight etc..?
yes you would follow PEMDAS
Okay, so -4 * -2 = 8 - 8 = 0? then -1 - -2^2 = 4 then 2(-2)^3 = -16 idk if I did that right
The part `-1 - -2^2 = 4` is incorrect
the first and third parts are correct
Okay that's -5 then?
yes. Then you take the absolute value \[\Large \left|-1-(-2)^2\right| = \left|-1-(4)\right|\] \[\Large \left|-1-(-2)^2\right| = \left|-1-4\right|\] \[\Large \left|-1-(-2)^2\right| = \left|-5\right|\] \[\Large \left|-1-(-2)^2\right| = 5\]
So \[\Large {\color{red}{\left|-4*(-2)-8\right|}} + {\color{blue}{\left|-1-(-2)^2\right|}} + {\color{green}{2(-2)^3}}\] turns into \[\Large {\color{red}{0}} + {\color{blue}{5}} + ({\color{green}{-16}})\]
Do we solve the last part, and that's the solution?
you simplify it to get what?
-11
Correct. -11 is the final answer
Oh that makes sense. Thanks so much
no problem
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