This problem has two answer to it, Help!
1 answer:
as you should already know, an absolute value expression, is in effect a piece-wise expression, since it has a ± versions.
![\bf \cfrac{|-2-2r|}{3}=4\implies \begin{cases} \cfrac{+(-2-2r)}{3}=4\\\\ \cfrac{-(-2-2r)}{3}=4 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ \cfrac{-2-2r}{3}=4\implies -2-2r=12\implies -2-12=2r\implies -14=2r \\\\\\ \cfrac{-14}{2}=r\implies \boxed{-7=r}\\\\[-0.35em] ~\dotfill\\\\ \cfrac{-(-2-2r)}{3}=4\implies \cfrac{-2-2r}{3}=-4\implies -2-2r=-12 \\\\\\ -2+12=2r\implies 10=2r\implies \cfrac{10}{2}=r\implies \boxed{5=r}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ccfrac%7B%7C-2-2r%7C%7D%7B3%7D%3D4%5Cimplies%20%20%5Cbegin%7Bcases%7D%20%5Ccfrac%7B%2B%28-2-2r%29%7D%7B3%7D%3D4%5C%5C%5C%5C%20%5Ccfrac%7B-%28-2-2r%29%7D%7B3%7D%3D4%20%5Cend%7Bcases%7D%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B-2-2r%7D%7B3%7D%3D4%5Cimplies%20-2-2r%3D12%5Cimplies%20-2-12%3D2r%5Cimplies%20-14%3D2r%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B-14%7D%7B2%7D%3Dr%5Cimplies%20%5Cboxed%7B-7%3Dr%7D%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B-%28-2-2r%29%7D%7B3%7D%3D4%5Cimplies%20%5Ccfrac%7B-2-2r%7D%7B3%7D%3D-4%5Cimplies%20-2-2r%3D-12%20%5C%5C%5C%5C%5C%5C%20-2%2B12%3D2r%5Cimplies%2010%3D2r%5Cimplies%20%5Ccfrac%7B10%7D%7B2%7D%3Dr%5Cimplies%20%5Cboxed%7B5%3Dr%7D%20)
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Answer:
x=4
Step-by-step explanation: 64/16=4
It is in all 0.0098934551 in all in total
True. The domain of the function gives all real numbers.
Answer:
For 17 it's 5: 7.5 i hope it helps?
Step-by-step explanation:
#5 is 1845 and #6 should b 4
