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dantipina2012
@dantipina2012
August 2021
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Пожалуйста, помогите решить систему. И распишите как решали.[tex] \left \{ {{4x-15 \leq 0} \atop {x+7 \geq 3}} \right. [/tex]
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ShirokovP
Verified answer
4x - 15 ≤ 0
x + 7 ≥ 3
4x ≤ 15
x ≥ - 4
x ≤ 3,75
x ≥ - 4
/////////////////
---------- ( - 4) --------------- ( 3, 75) ----------> x
Ответ
x ∈ [ - 4; 3,75 ]
2 votes
Thanks 1
isom
4х-15≤0 4х≤15 х≤15/4 х≤3,75
х+7≥3 х≥3-7 х≥-4 х≥-4
х∈[-4; 3,75]
2 votes
Thanks 1
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Answers & Comments
Verified answer
4x - 15 ≤ 0x + 7 ≥ 3
4x ≤ 15
x ≥ - 4
x ≤ 3,75
x ≥ - 4
/////////////////
---------- ( - 4) --------------- ( 3, 75) ----------> x
Ответ
x ∈ [ - 4; 3,75 ]
х+7≥3 х≥3-7 х≥-4 х≥-4
х∈[-4; 3,75]