Solve l 3x-7 l = 2
and solve l 2x-5 l< or equal to 8
the lines (l) signifies absolute values therefore l 3x-7 l is the absolute value of that and same thing for 2x-5

1) Solutions are x = 3 and x = 5/3

2) Solution are x ≤ 13/2 and  x ≤ -3/2

Step-by-step explanation:

1) Given absolute inequality,

|3x-7| = 2

⇒ 3x - 7 = ± 2

⇒ 3x = 7 ± 2

2) l 2x-5 l ≤ 8

⇒  2x-5  ≤ ±8

⇒   2x ≤ 5 ± 8

1) Solutions are x = 3 and x = 5/3

2) Solution are x ≤ 13/2 and  x ≤ -3/2

Step-by-step explanation:

1) Given absolute inequality,

|3x-7| = 2

⇒ 3x - 7 = ± 2

⇒ 3x = 7 ± 2

2) l 2x-5 l ≤ 8

⇒  2x-5  ≤ ±8

⇒   2x ≤ 5 ± 8

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐