Question

whats the standard form for: 3|x-2|+12>9

Answer 1

Depends on how "standard form" is defined.

Answer 2

simplest form

Answer 3

Ah, ok. Well, for starters, you can divide everything by 3.

\[\frac{1}{3}(3|x-2|+12 > 9) = |x-2|+4 >3\]

Now just subtract four from both sides
\[|x-2|>-1\]

And since the absolute value of ANYTHING is always greater than or equal to zero, that means this inequality is true for ALL values of x. :)

Answer 4

Check out the graph:

Answer 5

simplest form

Answer 6

thanks