Okay welcome back, we are now on problem 15 and I tried my best to draw this graph ahead of time. Probably, one of the more difficult, for me with my shaky hand so this is .0 for the origin right here, this is point P, this is point Q, this is the Y-axis, this the X-axis.
And they say, this graph right here is Y=A-X² and this curve right here is Y=X². So the questions says and the figure above, the graphs Y=X² and Y=A-X² for sum constant A, those are the graph chart. If the length of PQ=6, that length is equal to six, what is the value of A?
What is interesting -- so the first thing that should jumped out of - these are completely symmetric graph. They both are kind of mere images around the Y-axis so if this length is six then the length of each of these and here is three. So what do we know about where this graph intercept? Well we know they intersect at the points, .3 and we know that they intersect as well at a point, minus three. Those are these two graphs intersect - that is a lot of information, the way once again we came to this conclusion was that.
These are both symmetric around the Y-axis so if this intersection is 6Y then it’s three on each side. Right through the positive side through the negative side so if it intersects with X=3 and X=-3. And if we want to figure the intersection, we can also just set these two equation, Y=X², Y=A-X². So we know that X² must be equal to A- X² or at X² to both side of this equation you get 2X²=A. And this will be true at the two X points that where these two graphs intersect.
Well, what do the X points? We figure it out it’s 3 - 4. And if you put three in here, you get 2(3²=A), 2(9=A), A=18. And if you minus three and it would be the same thing for if we do squared, the negative goes away. So A is 18, not too bad, the key conclusion that you have to make is that. They intersect at the points, positive and negative three and then you just set the equations equal to each other and solve for A. Let’s do that in the next time.
Set X has X members, Set Y has Y members, Set Z it contains all the members that are either X or Y with the exception of the key common members. Remember of Venn Diagram but I will show you what one is. So let us say that this is Set X and let’s make and let’s say that is Set Y right here. So this is Set Y, this is Set X and it says, Set Z consist of all the members that are either in Set X or Set Y with the exception of the K common members.
So this is the kind of the K common members, these are the things that overlap with Set X and Set Y. So what is Z consists of? Well Z is going to be if I do the fill tool, it could be that plus that, without the K. Because it says Z does not include the overlap, so what are those two things?
So let see and we go back to the paint tool, or I am just going to be tacky with all the colors that I am using. So there is X members in all of X, so how many members are in this blue region? So if there is X in the whole region, it just going to be X-K in this blue region. I just subtracted out this K part, similarly Y has Y members, there are capital Y set has Y lower keys Y members.
But if we take out this K that overlap, this is going to be Y-K, I am just subtracting it out. And so Z is this blue area plus this blue are so it will be X-K+Y-K = X+Y-2K and that is a choice D. Not too bad, alright, I think we are all done with this practice to this. Hopefully you found vaguely useful, I will see you soon.
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