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$\mathrm{K}_{2} \mathrm{CrO}_{1}$ givcs a ycllow precipitare on rcaction with: (a) $C_{4}^{2}$ (b) $\mathrm{Fe}^{3}$ (c) $\mathrm{Ba}^{2}$ (d) $\mathrm{Pb}^{2}$

Okay. What we want to do today is kind of step through a application problem with with respect to the Newton's method. And I've already drawn, um, somewhat of a quick sketch. And what this represents is this is a This point right here is a submarines closest point of approach to a sound detector. Oh, detect D tech. I don't know how to spell detector detects Torre. That's terrible. Okay. And so what we have is if this point right here approaches his closest point of approach to sound detector, and the sound detector is down here at two. Negative one half. And the submarines, um, path is represented by a quadratic white called X squared. Okay, so what? The first thing we want to do is we want to find the X value that minimizes that distance. Um, between that, um, closest point of approach and where the sound detector is, and we also want to want to determine, um, that it is a solution. So we want to do a couple things, minimize the distance, and, um, and determine or show it is a solution of X equal. One over X squared, plus one. Okay, so a couple of things in the first part. Um and so we're talking about distance. Okay, um and so I'm gonna try to try to make it a little bit easier. We typically know that distance squared is equal to the difference of the X value squared, plus the difference of the Y value squared. So I'm gonna let D of X actually equal the distance squared. Okay. Just to kind of simplify things through, um, you could use, um, the distance squared and take the derivative and so forth. Um, you'll get the same result as what I did, but I'm just kind of trying to do a shortcut. Um, and so I'm gonna let my distance equal to X minus two because this value is XY. This value right here is X comma wife squared, plus the difference of the Y value squared. Okay, But I do know that each of the Y values on this parabola is determined by X squared. So that's the next thing I'm gonna do is replaced this. Why? With X squared plus one half. Okay, um, we know that minimizing or maximizing a function is where the derivative is equal to zero. So that's the next thing is, we take the derivative, and, um, this is implicit differentiation. So two times X minus to to the first times, the derivative of what's inside was just the one plus two times x squared plus one half times the derivative the once inside, which is times two x. Okay. And then we're going to simplify this guy so kind of distribute and we get, um this would be for X cubed and then plus a two X So we end up having, um Oh, and this should be a four, not a two. I would learn how to multiply. Okay, so if I rearrange, I get four x cubed, Um, plus four x minus four. And that's going to equal zero, right? Because that's where it's going to be. Um, yeah, minimal or maximal. Um, okay. And so I want to determine if this right here is actually a solution of that. And so what I'm gonna do is I'm actually instead of solving for X, I don't want to have to factor. A cubic is I'm going to take this X equal to one half x squared. I'm a multiply both sides by X squared plus one, and I get X cubed plus X equal to one. And then I'm going to have execute plus X minus one equal to zero. It's looking very close. So if I divide if I take up here and divide everything by four, I get X cubed plus X minus one equal to zero. So these two actually equal each other. So now I have shown that this equation is actually a solution to that one. Okay, so there's kind of a little way, um, to do that. And so now what we want to do for part B for Part B, What we want to do is in part B is actually to use Newton's method to solve the equation X equal to one over X squared plus one, or this equation right here, X cubed plus X minus one. Okay, that's what we want to try to solve. Okay? And so we know Newton's method is based off of that interest iterative process where each x value, um, each next value X value is related to the previous minus the function evaluated at that previous X value divided by the derivative as long as my derivative evaluated at that X value design equals zero, right, Because I can't divide by zero. And so what we're gonna do is one we know this is my function. Now, um, that's my function, not my derivative. And so now I'm gonna take the derivative. So I have three x squared plus X equal to my derivative. Okay, Now I need to I need to start. Um, I need to have a starting value, right, Because I'm going to fill out this table right here, and I'm going to know the first x value that I need to start with to get this to get this show on the road. Right. That's what I need to do. I need to find that first X value. Um, And so because we're looking to solve this, we want to know where it crosses the X axis, right? The solution where that equals zero. Okay, so we know that using the intermediate value theorem, if when I evaluate my function at one x value and it's positive and evaluate the next one is negative, Then somewhere in between those two x values, it crosses the X axis. Right? So that's what we're gonna do. And I automatically know that when I evaluate my function at zero, this function right here, it's gonna be a negative one. Okay, so now if I just quickly in my head, evaluated at one, I get one plus one minus one. So that is going to be a positive one. So I know somewhere in between, zero and one is where I want to start. Okay? Now, if I go back to my graph, I'm really actually finding trying to find this x value right here, right to minimize that distance. And so it looks like it is at about 0.5 just off the graph. I know it's between zero and one, and my intermediate value theorem confirms that. So let's just do let's just 2.5 for my starting value. Okay, Um, you could use zero. You could use one if you want to. It doesn't matter. Um, and so many evaluate my function at 10.5, and I get negative. 0.375 Evaluate my derivative. I get 1.75 and then I do 0.5 minus a negative 0.375 divided by 1.75 And I get point 71 49. So, for my next ex value, I use my result of my last column. So 0.714 29 I'm going to evaluate my function at that X value. And I get 0.78 seven to a evaluate my derivative two point 53 063 My last column, then, is 0.68 318 Okay, so my next ex value is what I just calculated. 2.68 318 Evaluate my function at that X value I get point 00 2044 Evaluate my derivative 2.4 00 to 0. And then take that second column minus third, divided by fourth. I get point 68 233 and what you're looking for. You know you're finished. When the The last column you have two of the same numbers, you end up with the two of the same number. So now I'm gonna do it one more time or a couple more times. However many long it takes for it did to do it. Um And so this is 5.2636 times 10 to the negative four. And I know I'm getting pretty close when my, um, function evaluated at that X value gets really close to zero. So, um, my derivative is 2.39 672 and my last column ends up being 6720.68 233 So that tells me X is equal. 2.68233 So where? Where? This submarine closest point of approach to the sound detector that is down here at two. Negative two. It's going to be at the X value of 20.68 to 33 to this value. Right here is 0.6268233 Hope this helps.

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Question is which of the following is not a crime try and receive is represented by that molecules that have actually get so we can observe that he'll start strong feeling and so it will be good to have and industrial. So first option is BDCL four who minus. So it will be, it always Bs people have palladium and platinum is always so people have read so correct on duty.

Hello everyone. The time average of the function after function is to find us His capital one upon teeth integration of function jude S. T. For the limited minus key way to keep this T. V. Two average value of sinus square of omega T. one way to 1- sign up omega T upon omega T into cause off. Who are you? No has been assigned a square theater plus causes square theater is cut to one. So using Sandy Square the topless causes square theaters called to one. The average value of cross the square of omega T Build Me 1- Average my work Sinus square omega T. Substituting the value. You will get one minus half one minus sine of omega T upon omega T. Cause of two negative. So finally you will get the answer that is half one plus sign off omega t upon omega T. Cause of to omega T. That so thanks for watching it.