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OpenStudy (anonymous):
Is this right?
(question below)
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OpenStudy (anonymous):
\[\frac{ 1 }{2 } (x+10)>-(x+1)\]
OpenStudy (jdoe0001):
depends on what you mean by "right"
OpenStudy (anonymous):
This is what I came up with
\[\frac{ 1 }{ 2 }x+5>-1x\]
OpenStudy (anonymous):
I'm working on what I already have. Give me a few seconds
OpenStudy (anonymous):
\[\frac{ 1 }{ 2}>-6x\]
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OpenStudy (anonymous):
\[\frac{ 1 }{ 2 } /-6\]
OpenStudy (jdoe0001):
hmm, keep in mind that a/b * c => ac/b
OpenStudy (anonymous):
So I came up with x=-12
OpenStudy (anonymous):
Is that right?
OpenStudy (jdoe0001):
\(\bf \cfrac{ 1 }{2 } >-(x+1) \implies \cfrac{(x+10)}{2} > -(x+1)\\
(x+10)> 2\times-(x+1) \implies x+10> -2x-2\)
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OpenStudy (jdoe0001):
hmm, typos :( one sec
OpenStudy (anonymous):
No problem :)
OpenStudy (jdoe0001):
\(\bf \cfrac{ 1 }{2 }(x+10) >-(x+1) \implies \cfrac{(x+10)}{2} > -(x+1)\\
(x+10)> 2\times-(x+1) \implies x+10> -2x-2\)
OpenStudy (jdoe0001):
so, it's not 14 though
OpenStudy (jdoe0001):
I mean, -12
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OpenStudy (anonymous):
Thanks! I had a slight (ok more than slight) error! Thanks! \(\Huge\ddot\smile\)
OpenStudy (jdoe0001):
yw
OpenStudy (ybarrap):
$$\tt \frac{1}{2}(x+10)>-(x+1)\\
x+10>-2x-2\\
x+2x>-2 -10\\
3x>-12\\
\text{Answer:}~x>-4$$
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