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Question

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Answers

Match the inequality in exercise in Column I with its equivalent interval notarion in Column II. $\mathbf{I} \mathbf{I}$ $x<-6$
$\mathbf{I} \mathbf{I}$ A. $\quad(-2,6]$ B. $[-2,6)$ C. $(-\infty,-6]$ D. $[6, \infty)$ E. $(-\infty,-3) \cup(3, \infty)$ F. $\quad(-\infty,-6)$ G. $(0,8)$ H. $[-3,3]$ I. $[-6, \infty)$ J. $\quad(-\infty, 6]$

Okay, So for part a, we want why isn't good enough? Well, why or a number is considered to be negative if it's less than zero? So that is why it is less than though the report be we wouldn't see is greater than one. See, we have B. Is that most eight? So that means our maximum point would be eight. So that's going to be less than are able to eight on for D. We have W. It's positive, unless than or equal to 17. Okay, so that would be Darryl is less than W, which is just unable to 17 for E. We have. Why is at least two units from hi that is we have pai minus y. The distance between these two is at most two, so that's less than or equal to two.

Okay, so report A We want to find X is positive. So X is positive when were to the right of zero. So that's X is greater than zero Now for part B, we want t is less than four and for seeing we have a is greater than or equal to pi and for D. We have excess, Espen. 1/3 taxes us, then 1/3, and it's greater than native five. So this would be like this. And then we have to get a five here and for E. We have the distance between P and three, so that's the up Valuable. P minus three is at most five, so that's I'm going to be less than or equal to five.

Alright guys, So for this problem we kind of have a matching type of question. And so what we are given is the solution set to a linear inequality. And then I have to identify I'm a through J which interval notation of that solution actually matches up with what we were given. So um I know that the problem I'm working with is X is greater than -2, but it is less than or equal to positive six. Alright. So I've done a number line because I would like to go over this and kind of see it visually so that I can write the interval notation maybe a little easier or kind of, you know, hopefully will just help and as a support here. So, my number line goes from negative 10 to positive 10. Um going with intervals of one. If you needed to draw a number line, you could, you know, draw it too. However, works for you. All right. So, to go through this again, the goal is to write this solution set into an interval notation, So I know X has to be greater than -2. So my lower bound is -2. The solution does not include two. So this is going to be a parentheses on negative too, and it contains everything Greater than that until you hit your upper bound, which is six. Okay. Now you need to sit here and anything about, okay, what's happening at six? So X has to be less than or equal to six Because it's less than or equal to. This is a bracket because the solution set does contain the answer six and then you'd be doing everything that is left Of six until you hit your lower bound which was negative too. Alright, so to write this in interval notation, you're simply going to have a parentheses because this does not include negative too. Your lower bound is -2, comma your upper bound which is six. And ask yourself does this include six? Yes, because it's less than or equal to six. And then I'm going to use a bracket. All right. The answer that matches what I have written would be, hey.

Hey there guys. So for this type of problem we are going to have to kind of we are reviewing our knowledge of linear inequalities and um we are given the solution to an inequality and we have to be able to read that and be able to transcribe it into different forms of writing solutions. So we do need to know and understand how to write and inequality solution into set builder notation as well as interval notation. And for this problem specifically, we're really just going to be focusing on the interval notation. Yeah. All right. So, um I didn't want to go through this problem and I wanted to kind of um go through it visually to kind of help see why our solution is going to be written into the interval notation, specifically in the way that we find it. And so I'm just gonna get started for our example here we have X. Is less than negative six. And so um I've drawn a number line. My number line has zero in the middle, goes up to 10, positive 10 and goes down to negative 10. Um That's just so I cover my bases and so we're going to start by if X is everything that is less than negative six and does not include negative six, I'm going to put a my starting point is going to be at negative six and there's a parentheses there. And just to recap again, the parentheses on -6 means here's your starting point but your solution does not contain negative six as an answer. So -6 is not a true answer for this problem. It kind of just starts here. Now if this was a bracket on negative six then yes it would include negative six in the solution. But again this example it does not all right. And so because all of our X values have to be less than negative six, that's going to be everything going to the left or again decreasing from negative six. Which means essentially you'd kind of be going all the way out up until a negative infinity. So to rewrite this into interval notation, you're gonna start with your lower bound. Your lower bound means that like this number is going to contain we go all the way to negative infinity comma go up to your upper bound. My upper bound is negative six. Again, we do not include negative six in the solution. So I'm going to put in parentheses All right. And this is how you would write the interval notation. So for this problem um you would line this up and you would find that the answer to the interval notation of X wing less than negative six is f Negative Infinity to -6.


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