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juneham
16.09.2020 •
Mathematics
-3r+15≥4(r-2) please help me
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Ответ:
-3r + 15 ≥ 4(r) - 4(2)
-3r + 15 ≥ 4r - 8
+ 3r + 3r
15 ≥ 7r - 8
+ 8 + 8
23 ≥ 7r
7 7
3²/₇ ≥ r
r ≤ 3²/₇
Solution Set: x ∈ (-∞, 3²/₇]
Ответ:
Subtract 4r from both sides.
−3r+15−4r≥4r−8−4r
−7r+15≥−8
Subtract 15 from both sides.
−7r+15−15≥−8−15
−7r≥−23
Divide both sides by -7.
−7r/−7 ≥ −23/−7
r ≤ 23/7
Ответ: