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The traffic flow rate (cars per hour) across intersection is r(t) = 300 + 700t 180t2 , where ` is in hours; and t0 is 6am. How many cars pass through the intersecti...

Question

The traffic flow rate (cars per hour) across intersection is r(t) = 300 + 700t 180t2 , where ` is in hours; and t0 is 6am. How many cars pass through the intersection between and am?PreviewcarsLicensePoints possible: This is attemptSubmit

The traffic flow rate (cars per hour) across intersection is r(t) = 300 + 700t 180t2 , where ` is in hours; and t0 is 6am. How many cars pass through the intersection between and am? Preview cars License Points possible: This is attempt Submit



Answers

The traffic flow rate past a certain point on a highway is $q(t)=$
$3000+2000 t-300 t^{2}(t$ in hours), where $t=0$ is 8 AM. How many
cars pass by in the time interval from 8 to 10 AM?

For this problem. The first thing we need to do is identify what happens to the traffic and 8 a.m. 12 noon and 5 p.m. So we need to have these within the context of that given equation. So you know that T equals zero is 6 a.m. So then 8 a.m. is t equals two, right, cause then teak was one would be seven than eight. This allows us to do some synthetic division to solve for what happens at 8 a.m. For the traffic, right? So then, too, is it divisor? And then we have are coefficients Negative one 25 negative 1 92 for 32 and zero. And this is through the remainder theorem saying that if our T value is to then whatever this value is is what happens at that t value. Okay, It's the first time drop straight down, multiplied by two negative two plus 25 is 23 again multiplied by two 46. We'll continue this pattern, right? Seven Oops. When X equals two or why if you will t value says the T values to the traffic is to 80. But not only is it to 80? It's to 80 and notice that it's positive. So it's greater than zero and zero is the average. So it's to 80. A. Both average. Okay, so then we follow the same pattern with 12. New if eight is too thin. 12 is six. All right, so we do the same thing. No surprises. Pretty straightforward and follow the same procedure since six times 19 is 1 40 14 right, So then our value here is negative to 16 and no negative. So it's below average, and then we don't actually need that negative sign here because we're too. Noting that it's negative by saying that it's below all right, and then make a little room for myself here for our last one, which is 5 p.m. Translating to 11 t equals 11. So once again, same pattern as before. Make sure you keep track of your negative signs so that you don't make any careless mistakes. It's a very that. Then again, that's my bed. It says 14 1 54 and then that equals negative. 38 got ahead of myself. 18. It's than 1 54 right? So then transiting that back to our problem here. It's positive. So it's 1 54 above average, and that answers part ache. Part B, then is to use rational zero serum to solve for the zeros of this equation. Now the rational zeros says that we have to take 4 32 and our leading coefficient, which here to negative one. And it's saying that all of the factors of 4 32 let's call those p and all of the factors of negative one que Then that's saying, with all those options of the form P divided by Q, are possible rational Sears into function so there won't be any zeros outside of this list, But it doesn't guarantee that every zero in this list is a zero up this function. Okay, so beginning, positive or negative one it's possible for 32 is even so positive or negative, too. Rights that I'm going through all of the factors of 4 32 actually has a lot. But when she did notice that I'm not going because like 4 32 actually has huge factors like 1 92 we're not gonna deal with those because think about time. If his T equals zero is 6 a.m. Then t equals 24 is also 6 a.m. There's no reason for us to go that high, so I'm just going up to 23 which, actually 18 is the highest one possible. But then again, we're dealing with time, so we're only gonna deal with positive values because we time goes up, out, down. Okay, so then, considering with the work we just did in part, a two is not a zero and six is not a zero in case we took care of those options. Now we have to do a little bit of tedious work to check these other ones. Okay, so we have one check, if one is zero, and this is the same process we use synthetic division to check. And then for this 1st 1 is gonna become pretty clear that it is not a zero. He's not worried about that one. All right, so then let's try three based on that 66. Okay, so that is also not a zero. All right. Next we have four, and this is, unfortunately a very tedious process. Think of it as guests and check your making an educated guess, and you're just check and see if it's right. So far, we have not been right, but they are still good. Check. Good guesses. Just because they haven't been right doesn't mean they were bad guesses. Necessarily. It just means they weren't quite right. And then here we go. We have a zero. So four is a zero. Okay, So then that means our job became a little easier because we can now use this new polynomial to continue checking, or it's the next I'm gonna try eight, but then you want to just make sure. I mean, I'm sure you do by now, but you definitely want a calculator for this process. Or else you will spin your head in circles around these numbers, and it's a good chance you'll make a careless mistake. All right, so that is not what we wanted. So eight is not one. All right. So, nine. And then again, I'm taking this resulting polynomial instead of starting from scratch. And we came out with the winner nine as a zero. And then this allows us to say OK, the resulting polynomial then is X negative. X squared? No, sorry. Just x to the first plus 12 set that equal to zero and 12 to both sides. I sleep the variable may get X equals 12. So positive 12 with our last serum. And then here's where I'm going to stop because I have three identified here, and t equals zero is also zero The function. The degree is four telling me I have four zeros, so I've identified them all. Okay, so now I'm gonna clear out my workspace cause I've taken care of what I need, and then it can continue moving on to the last step of this problem, which is that we're going to utilize the zeros that we just solved for. Okay, so the zeros we found X equals zero X equals four X equals nine and X equals positive 12. Okay, so then this frees us up to grab Yes. And then this in mind one that get my graph here in order so that it's useful for us. Okay. And then with this, we're also identifying the medics and the men minimum sorry of this function, but we're only making estimates, which is huge. Okay, so then drawing in my two axes and I'm counting by twos here just to allow myself some space, and then it will be a little more readable this way. Okay, so that are zeros. So the origin Adam it go ahead and count these as two. So this is gonna be to I mean, sorry. That's gonna be four. And then so for eight. So this is gonna be nine, and this will be 12. Okay? And then we have our values up here. Where we found to to 80 11 is 1 54 and six is negative. 2 16 Okay, so then, for my y axis, I'm gonna count by 50 since me 5100 1 50 200. So if to is to 80 then it's about right here. So I'm slipped up right there. So 102 100 to 50. So about right here, Okay. And then six is negative. 2 16 So 102 100 right there and then 11 is 1 54 So Ok, so then this allows me to connect and sketch my graph, and that's what we end up with. Okay, so then you want to use your calculator to cut, calculate thes maximum minimums and you'll find you're made. Clear this out. So I have room to write. So you see that at 8 a.m. Were at 2 50 So it's smart to check around that area, putting in pot potential t values. And we'll get about 300 above average again above, just noting that were above zero. And that's gonna be at 7:30 a.m. Which would be a T value of 1.5 and then our minimum we see at six, which is about 2 20 below average. Right? So that's why we're not writing the negative sign because of that below. So it 6 p.m. Oh, that is my bed. That should be 5 p.m. And I've written it wrong in multiple spots. I'm so sorry. The numbers are all right. Do you really? All right. That is six. So excuse me. All mix up. Those are good. Supposed to be six at 5 p.m. Which is a T value. Okay, T value of six at new. And I'm sorry that I got all mixed up, but that does make sense, right, because it's not in rush hour. So basically, this is a big problem of analyzing rush hour versus non rush hour, and we have all the information we need, and we're all done

In problem 32. We want to show that the analysis of this intersection doesn't give us any information about X, Y and Z. It is insufficient information here. The intersection means at each node the total enters equals the total exists. Then, for notes, A BNC we can get three equations at E. We have 80 enters equals 300 out plus Why Out plus XL. And the analysis of new would be save that 700 out equals X. N plus dead in. And the analysis of let's see gives us 400 in plus Why in equals 200. I would plus that. Let's rearrange this system of equations. Get the variables in the left hand side and the absolute values in the right hand side from the first one we have X plus why? We don't have that equals 500. The second equation we can we have X plus, it equals 700. And the third equation we have, we don't have X. We have wide minus Z equals minus too hard. Let's try to solve this system of equations from equation two and three. We can eliminate that. Then we add the question two plus, equation three. We have X plus Y equals 500. Oops. The new equation is the same as the first equation. Then we have a redundant equation. If we added the equation for or subtracted equation for from equation one we can get zero equals zero. This means we have infinite number of solutions for X and Y. We can put X and Y any values. Then there is insufficient information here to have a unique values for X, Y and Z.

In this question, we're going to consider the applications of systems of equation If we have Theatre section of 31 way streets. Andi want to keep it? That the traffic always flows? We must set up a system of equations governing the traffic flow such that the cars entering is equal to the cars leaving. So, for example, in Section one, we have 10 cars entering an ex cars entering. So have temples. X on one side on leaving we have 14. And why so why? Plus 14? They're similar equations governing in sections two and three. Yeah, a family of solutions can be found then by a gas in gas Jorden elimination or may treating version. And this will give a family of solutions if we pretend, or if by construction, we have said is equal to four. What will be the flow between the other? The other intersections. So given that it is equal to four, um, this sent houses that t is equal Tofel so that X is equal to six. Plus T is equal to 10 on y is equal to two plus t is in quarter six and this gives us our traffic flows in between each intersection. Thank you

India. And and they did provider that before I went in the sex annotate. The question is X plus 10 is a quarto y plus 14 and we have to remind 40 question I too, and I think that is helpless there. Here, 12 plus digger equals 26 plus six, and it's explicitly 43 is the cool so eight Plus it and we have solved. Disagree isn't and great over dessert as yeah, see, Lex did away. This is their works. The upturn from the guards. Integration? No, we have to find the value for Z equals 24 for their this going for here here for their little before. So why minus four will be toe live with six and I would like Is going toe here Any off the question, then 01 x that is one x minus y equals 24 So x will be four plus six that distance So X equals two. Then why equals 26 and 30 close to four, whichever the Kerr solution


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