[" 31If "|x+3|geqslant10," then "],[[" a "x in(-13,7]," b) "x in(-13,7)],[" c) "x in(-oo,-13]uu[7,oo)," (- oo) "-13)uu[7,oo)]],[" d) "x in(-oo,0)]

If |x+3|>=10, then x in(-13,7] b.x in(-oo,-13)uu(7,oo) c.x in(-13,7) d.x in(-oo,-13]uu[7,oo)

If |x+3|geq10 , then x in (-13 ,\ 7] b. x in (-oo,-13)uu(7,oo) c. x in (-13 ,7) d. x in (-oo,\ -13]uu[7,oo)

If |x-1|>5, then a. x in (-4,6) b. x in [-4,6] c. x in (-oo,-4)uu(6,oo) d. x in (-oo,-4)uu[6,oo)

If |x-1|>5, then x in(-4,6) b.x in[-4,6] c.x in(-oo,-4)uu(6,oo) d.x in(-oo,-4)uu[6,oo)

If |x+2|lt=9 then x in (-7, 11) b. x in [-11 ,7] c. x in (-oo,7)uu(11 ,oo) d. x in (-oo,-7)uu[11 ,oo)

If |x+2|lt=9 then x in (-7, 11) b. x in [-11 ,7] c. x in (-oo,7)uu(11 ,oo) d. x in (-oo,-7)uu[11 ,oo)

The solution set of (x + 3)/(x - 2) le 2 is a) (-oo, oo) b) (-oo, 2] uu [7, oo) c) (-oo, 2) uu [7, oo) d) [7, oo)

Solution set of the inequation (x^2+4x+4)/(2x^2-x-1)>0, is x in (-oo,-2)uu(-2,1) x in (-oo,-2)uu(-2,-1/2)uu(1,oo) x in (-oo,-2)uu(-1/2,1)uu(1,oo) x in (-oo,1)

If -3x+17<-13, then x in(10,oo)bx in[10,oo)c.x in(-oo oo,10]d .x in[-10,10)

If -3x+17<-13 ,\ t h e n x in (10 ,oo) b. x in [10 ,oo) c. x in (-oo,\ 10] d. x in [-10 , 10)