Mathematics
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OpenStudy (anonymous):
Step by step to answering |8*x−1| ≤2*x+11.
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OpenStudy (anonymous):
\[|x| < a \rightarrow -a < x < a\] Use it and solve
OpenStudy (kc_kennylau):
Entonces, tenemos:
\[-(2x+11)\le8x-1\le2x+11\]
OpenStudy (kc_kennylau):
Porque el "x" en su ecuacion es "8x-1", y el "a" en su ecuacion es "2x+11"
OpenStudy (anonymous):
????
OpenStudy (kc_kennylau):
@sourwing ?
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OpenStudy (anonymous):
creo que si entiendo un poco
OpenStudy (anonymous):
is this spanish?
OpenStudy (anonymous):
yes
OpenStudy (kc_kennylau):
@AntonCosme si aun no entiendes, puede lo pensar como asi:
\[|y|<a\rightarrow -a<y<a\]
OpenStudy (kc_kennylau):
Si no usas "x", lo puedes entender
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OpenStudy (kc_kennylau):
entonces, "y" significa 8x-1, y "a" significa 2x+11
OpenStudy (anonymous):
Si lo entiendo mas mejor.
OpenStudy (kc_kennylau):
no lo entiendes?
OpenStudy (anonymous):
Si entindi como ponerlo en la formula que usted escribio pero no se como completar el problema
OpenStudy (kc_kennylau):
tenemos \(−(2x+11)≤8x−1≤2x+11\)
Los puedes separar a: \(-2x-11\le8x-1\) y \(8x-1\le2x+11\)
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OpenStudy (anonymous):
Ok
OpenStudy (kc_kennylau):
Solve las dos ecuaciones separadamente
Disculpe, tengo que comer ahorita
OpenStudy (kc_kennylau):
Cual huso horario eres en?
OpenStudy (kc_kennylau):
GMT+8 para mi
OpenStudy (anonymous):
esta bien tengo que hirme a dormir es tarme 12:44AM Tampa Fl
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OpenStudy (kc_kennylau):
Entonces es GMT-5?
OpenStudy (kc_kennylau):
Hasta manana :D
OpenStudy (anonymous):
si gracias por la ayuda.
OpenStudy (kc_kennylau):
(u hoy)
OpenStudy (kc_kennylau):
adios
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OpenStudy (anonymous):
bye
OpenStudy (kc_kennylau):
de nada :)