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  • 03-09-2017
  • Mathematics
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Solve the inequality for b
10 ≥ - 2/3 (9+12B)

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Аноним Аноним
  • 03-09-2017
 [tex]10 \geq - \frac{2}{3} (9+12B)[/tex]

[tex]10 \geq \frac{-2}{3}*9 +\frac{-2}{3}*12B [/tex]

[tex]10 \geq (-2*3) +(-2*4B)[/tex]

[tex]10 \geq -6 -8B[/tex]

[tex]-8B \leq 10 + 6[/tex]

[tex]-8B \leq 16[/tex]

[tex]B \leq \frac{16}{-8} = -2[/tex]

B ≤ -2

Hence, B is any number less than or equal to -2.
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