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QuescionFor the Braph belov. write ineduafor the variable x whose solution the graph-25,45 7221 -2 0r, 2 $ ~294 Xs-2 0r $ >...

Question

QuescionFor the Braph belov. write ineduafor the variable x whose solution the graph-25,45 7221 -2 0r, 2 $ ~294 Xs-2 0r $ >

Quescion For the Braph belov. write inedua for the variable x whose solution the graph -25,45 7221 -2 0r, 2 $ ~294 Xs-2 0r $ >



Answers

Solve the given problems. Draw a graph of the solution of the system $y<|x+2|$ and $y=x^{2}$

The question is, will provide squared minus X squared equals one terrible equation y squared by these four minus X squared by a square by one because the one so is equal to plus minus one and visible to plus minus one. Let's check the origin point. Another feasible is a non more so zero square minus zero square. Pleasant one. So zero lesson one. So that points, right? So piece of reason comes under Wasn't point comes with visible so Lester than other equation Excess far less white Square nine. So this is Circle equation Explorer plus y squared ableto are square So our isn't going to plus minus three this circle point let's take the origin point Whether comes into resemble isn't or not dealers Choir plus zero is quiet rather than nine. So it's, you know, great the night That is not true. So the physical isn't are sort of the circle This is the origin point doesn't come on. A feasible is um so I will find him. Come combined crab is this one You can see this one common reason is a feasible reason final feasible. Is there

Okay, so we'll start by writing, um, each part of our system of equations as if it was an equal of time instead of an inequality. So for the first time, you've X squared plus y squared equals one. And we can recognize this as the equation of a circle centred for him. Because we have the square right here and it's Senator about the origin because we have no modifications. That's of the why. And we know that this equals r squared so r equals square to pull in. Articles one are right. It's gonna be one. And we're gonna wall a dashed line here because we're not including where X squared plus y squared equals one. There's no equal right there and you are stuck it equation. We have X squared plus y squared. Both nine This is R squared are equal screwed. And nine, which is three and the gun were centered about the origin because you don't have any pluses or minuses are on the X or the y and again a dash line because we're not including the equal. But we will. 30. Bert Green bettered about the origin with radius equal to one and then we can touch the point to see where to shave. You can also just look back of the equation. It says X squared plus y squared When, um, the radius is greater than 10 that should be dashed. Sorry about that. Um, and because where the radius is greater than one we know he's just shooting. I'm not so good where the radius is over one and for our smoking equation and that we have a circle with radius equal. The three you can draw back. Oh, gonna It should be the Dutch by here side. And here we see that this is extracts. Wife's worried is less than mine. So anywhere we have a radius less than three. Which means we should cheat inside, have the circle. But because of the system of inequalities, we only want where the green and red sheeting overlaps, so will erase everything else, which would be everything outside of our dashed red circle. And anything inside are green circle. We'll just kick start growth. And this was the final solution

In this question, we have to graft inequalities X squared plus y squared greater than one and x squared plus y squared less than nine. For the 1st 1 we will write X Square plus of I square equals one V. C. This is a circle with the center zero comma zero on radius is equal to one. For the 2nd 1 we will write X Square. Place of I Square equals nine vz. This is a so called with the centre as a zero coma zero on the radius as equal to three now reveal. Upload these to SoCal's on the graph people. So for the 1st 1 will take the center at zero comma zero under. The idea says Equal toe one. So these will be deep, warm, soft reference. So it will be our dash the SoCal because the intake of ality does not include the equality symbol. The 2nd 1 will also be a dash the SoCal with the centre zero comma zero on the radius as equal toe three. So these will be deep Points off. Reference has marked in the graph. Now we will test the inequalities at the 30.0 Komen zero for the 1st 1 Richer's X squared plus of I square greater than one. We get zero greater than one that is falls. So appreciate the reason not containing the 10.0 coma zero. It will be outside off the circle like this. Now, for the 2nd 1 we will get X Square plus of I square less than nine, which gives a zero less than nine, which is true. So for this we appreciate the region containing the 90.0 comma zero. That means it will be inside off the SoCal like this. The solutions that will be the intersectional, these two shaded half planes, which is as shown in the graph paper.

Let's take the equation. X squared plus y squared for tonight. This is circle equation. So X squared plus y square Because to our square, so as far to nine are physical too plus minus three. So this is real circumlocution the sector origin So zero Squire, As you know, square lesson nine So you know less than nine. That is true this point come under feasible, is it? I think that I'm the question Why could relax? Why? So take this one point is one comma 001 So it is not true. But that point don't come under the usable So isn't well finally feasible Reason Chedid part our circle on the line.


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