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(I)Write each of the following in inequality notation and graph on a real numberline:Uii JYlAi Je WJlsAilisll ~i Jyij_ LS(D [-8,7](2) (-6.6]() [-3,3)(4) (-4.8)(5) [...

Question

(I)Write each of the following in inequality notation and graph on a real numberline:Uii JYlAi Je WJlsAilisll ~i Jyij_ LS(D [-8,7](2) (-6.6]() [-3,3)(4) (-4.8)(5) [-6,=)(6) (~o,5)(II)Write each of the following in interval notation and graph on real number Iine:Uu_lJsjl #yjJ _si()-3-<5(2)-6<*$4(3) *2|(3)* <3(III) Solve and graph: ~Jbh(I) 7r-8 < 4x+7(2) 26*-31+5<r-[(3) 3-*2503-*)r-[s6- 4 (5) - 20 < <3(6)5-3< 44(7) -4 <5+3*517(8) 1026-2r> 0(9) 12->2 2(9- 2y)43

(I) Write each of the following in inequality notation and graph on a real number line: Uii JYlAi Je WJlsAilisll ~i Jyij_ LS (D [-8,7] (2) (-6.6] () [-3,3) (4) (-4.8) (5) [-6,=) (6) (~o,5) (II) Write each of the following in interval notation and graph on real number Iine: Uu_lJsjl #yjJ _si ()-3-<5 (2)-6<*$4 (3) *2| (3)* <3 (III) Solve and graph: ~Jbh (I) 7r-8 < 4x+7 (2) 26*-31+5<r-[ (3) 3-*2503-*) r-[ s6- 4 (5) - 20 < <3 (6)5-3< 44 (7) -4 <5+3*517 (8) 1026-2r> 0 (9) 12->2 2(9- 2y) 43



Answers

Write each inequality using interval notation, and graph each inequality on the real number line. $$ -2<x<0 $$

Foreign inequality in X, where accept such that X is less than be. They say that ex belongs toa the open intervals that is minus infinity toe be now. If the inequality given this excess such that X is less than minus four than we can say, that ex belongs to the open and double off minus infinity to minus four. Now let's plot it on a number line, marking the point at minus four on putting the bracket off open and trouble on dhe, then drawing a line on the left off minus force to reach negative infinity and finally drawing an addle. We get this as the solution off the given inequality.

For the lenient inequality off X, which is half open. We know that excess such that four is less than or equal to X is less than six. So since it's 1/2 open and double so we can see the decks belongs to the half open and double of four and six. Now let's plot the given inequality on a number like so, plotting the points four on six and marking with a closed bracket on an open bracket, respectively, on highlighting the area between four and six will get the solution for the given inequality.

Inspection based on inequality. Actually let's move forward Taking the night town. That is the creator and a with a 1/3 and constant on the other side. So level 5568 -1 by 40 Mana. Come here we descend by my nation with wealthy will be the name by now. So I think this Taking a common where the 55 or 6 -1 x four remind us giving back Well which is better than by three. Now move forward. Eight affected. So. Mm hmm. Um multiplying. Be worrying. But yeah. And when they were back video So to be electric to 10 minus tweet upon well -7 point where you learn to my truth. So this is okay. You're disappointed 8 7 by 12 -7 by well given by two this time Because it was zero To me it is better than two. So we are we are getting zeros. So this statement is false for anything, not the statement that's not the kind of company. And this statement cannot be true since zero is always less than puberty. And here the company serious everybody. So yeah, these and then when you four. Hey, so it does not have any venue. Since this statement can actually folks you trust are certainly should be organized. E belongs to. The symbol means belongs to but and he said, we're building like this number or you can like it belongs to and he starts doing this. You did. It does not have any either.

Now for a variable X, which belongs to an open in trouble. It's an equality will be given us access such that B is less than it is less than a So we'll say that experience. We open in trouble, baby. Now, if you're given an inequality, next X is such that minus one is less than X is less than five. So we can say that for this ex belongs to the open and terrible of minus one and five. Now let's plot the given solution honor number late now taking the tour ends at minus one and five on Dhe. Putting the brackets for the open and taller at these two points on cheating or highlighting region. Between these two points will be the solution for the given inequality.


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