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HanoaGHColAlg11 7.o08_ pomts 474 Qno Inteteat nnu Annan InvostedpartJom much didahe Intest at 5057antl€ Kortl...

Question

HanoaGHColAlg11 7.o08_ pomts 474 Qno Inteteat nnu Annan InvostedpartJom much didahe Intest at 5057antl€ Kortl

Hanoa GHColAlg11 7.o08_ pomts 474 Qno Inteteat nnu Annan Invostedpart Jom much didahe Intest at 5057 antl€ Kortl



Answers

Subtract $252,631-47,882$ A. $204,749$ B. $215,859$ C. $204,947$ D. $215,895$

In this problem, I'm writing the reaction so just look at it carefully. S 04 two minus plus B A two plus. We will react to form B and so forth. This is white PPT and the second reaction happened. Something like this. S artery two minus plus, Be a two plus will react to form B. A. Absolutely. And this is also white PPT. Therefore, according to the option option C. H. Correct answer for this problem.

And this exercise, we're being given a table of data. The input is years from 1998, and two years increments. The output is the value of 3247: 5836. And we're being asked to find a model equation that would fit that data in quadratic form using translations, A H and K. Right? I have a graft all of that data here in Desmond's that we can look at it. I can see that it starts here at 1997. I'm sorry. at 1998 with the value of 3247 between 3000 and 3500 and the rate of increase gradually gets faster so that it looks like the values are getting higher more quickly. It's curving up, it looks like it might be a good candidate for a quadratic model. We could just start by using this initial point 1998 as the vertex of that. For Avalon I'm going to do that. That means that my starting year is shifting from 0 to 1998 to change that starting point and subtracting 1998 from the input and then To get the Vertex all the way up to 3247. I need to add that. That was my K value. If my stretch factor is just one, if it's a regular shape, I can see that. Yes, the problem goes through that. That's a fact vertex. It's very shallow right now because this scale is going in increments of 500. Every measure Interval, every major grid line represents 500. So it's being compressed a lot by the scale over the graph. My next task is to figure out what is the stretch factor A. That will make that problem fit the rest of data. So let's switch over to this whiteboard. I have already said that I am using the first point for my H and K values to get that vertex. So I'm going to rewrite my equation with this 0.0.1998 and 3247 ordered pair plugged in as my hk I'm shifting my starting year 1998. Yeah mm I am shifting my starting value up to 3247 247 So that's the beginning of my equation. Let me a great that seven. Better. I need to find this stretch factor. What I'm going to do is use another point of data. I've got this 0.0.2006 gives me 5,836 as an output. Yeah, That point has to be part of this equation. So if I put that X. Value 2006 in I need to get that 5,836 out. And that means I have to have an a value a stretch value that will make that work. So I'm going to do exactly that. I'm going to plug that point in with the input in the X. And the output in the UAE and then I'm going to solve to find out what a value makes that true so. Mhm. My y value is going to be 5836 if I have the right stretch factor. Mhm. When my input is 2006. Okay from this parable that's been shifted To a new starting year of 1998 and shifted up to give me a starting value Of 3247. Yeah. Mhm. Now I've got the equation I can solve for a, I'm going to start with parentheses. E. Is figure out what this is. So 5836 equals whatever my stretch value is, my stretch factor, 2006 19-1998 is eight years. That's still going to be squared plus 3,247. I can subtract 3247 from both sides and When I do that 5836 -3247 is 2589 equals a times eight squared eight squared to 64. So that gives me 64 Times my stretch factor. If I divide both sides by 64, 2589 Divided by 64 will give me my stretch factor. And when I put that into a calculator I get that that stretch factor a equals About 40.4. So value that I've calculated from this gives me 40.4. I'm going to go back to my graph And just check and see if that actually makes sense. Let's go back to Dez most if my a value is about 40, it fits This endpoint. If I put in 40.4 it does go through this end point but it's a little low. It doesn't include all of the other points which is never going to happen with real data. You're never going to get the curve that goes through all of the points exactly. But all of these points are above the curve which means that my model is a little love. I want to model that kind of includes all of the data. It doesn't have to go through all of the data so much as it goes through the middle of all of the data. So I'm going to try just a slightly larger value. 44. I've got one point that's too low and I've got three points that are above 45 is a little closer to the points that are above 46. Dance around 2048. It's kind of like 48 each. That's too big. Just testing to see what looks like balances the points the most. I think maybe like 48 is a good middle of the road. I've got three points that are a little above the curve. I've got one point that's quite a bit below the curve. So I would say a is anything between 44 and 48 is going to be a good model. I personally like 48. So I'm going to say about my model for this data. Yes. My health put value. Why Is equal to 48- nine years? My input year X- My starting year, 1998. Yeah. All squared plus. You make that squared better plus a shift up of 5,836. And that is my quadratic model for this data. Yeah.

Okay, This problem is asking us which of these molecules between all four of these is my m ing going to naturally to tomorrow's into. So let's go ahead and define a couple terms here. First off to Tom Rice or to Tamar Ization is essentially I saw memorization, right? Because whatever we have right here in this case, we have an I mean, the Tomur, the Tanamor is going to be in equilibrium. It exists in equilibrium with my I mean so in order to exist in equilibrium, that means that we're going to have basically the same atoms as we do in this. I mean, but just in a different order, right, we're gonna have different connections. Basically another item, er which is going to be called the totem. Or in this case, so we're looking among all these molecules, Which one has the same atoms just in different connections. So over here we have an oxide. Does this have the same number of atoms in the same type of atoms? The answer is no, because we have an oxygen here that is simply not a cook cook component of Miami. So this is not going to be an option. Next step. We have a hydro zone. Is this considered to be the same atoms as we do in my me? The answer is no. Because we have another nitrogen, right? We have the extra nitrogen. That is not a option. Next up we have this semi carpet zone. This is obviously not the right answer because this one has way too many nitrogen, right? We have one too many. This is not gonna be an option. The last option is my enemy. So does this have the same number of atoms? Same type of atoms and just in different connections? The answer is yes. So it's going to be my enemy. And what is that? Because if we're to simply use a base, for example let's go. You had a user base. My bass is able to Deep protein ate the aesthetic proton associative right here. So this is analogous to my alfa hydrogen usually associated with carbon nails. So I'm gonna take off the hydrogen. Moving the electrons onto this carbon or another way I can show is to move these electrons onto the single bond to create a double bond. As I do that, I'll have to move these electrons up to this nitrogen to produce this intermediate in which I have a nitrogen with a single bond because it has a negative charge. My method group right there and then my double bond right there. Right. So this is very, very similar to my enemy, which I have drawn right here. The only other thing missing is the presence of the hydrogen, which can simply be used over here. We have HB used from the protein nation of be in the previous step that can simply get pregnant by HB to eventually form my enemy shown right here. So this is a Tanamor because this exists in equilibrium, right? It just depends on the reaction conditions. But normally we're going to have my maybe the most stable form that is going to be the one that is present in the greatest excess, right? So enemy and I mean those are taught Immers, and they're going to exist naturally.

In this problem, I'm writing the reaction. Just look at it carefully. Any energy for at be your ford four H two will react to form any and at four at people for plus four as to and and ME energy for add bill for will react to form and the beauty plus an STD plus as to So according to the option. In this problem option. See each correct. Was there not that we have any Beauty which reacts with metallic oxide to give colored or to post fails.


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