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Havc bolcon. Ihis balecn Jos 13 Rs 6f Nz sof Ndvrx & 2.2LL 1 renove 0.59 rales c Nz Gcar Ki blasn 7 ~hol ik Yolunc;...

Question

Havc bolcon. Ihis balecn Jos 13 Rs 6f Nz sof Ndvrx & 2.2LL 1 renove 0.59 rales c Nz Gcar Ki blasn 7 ~hol ik Yolunc;

havc bolcon. Ihis balecn Jos 13 Rs 6f Nz sof Ndvrx & 2.2LL 1 renove 0.59 rales c Nz Gcar Ki blasn 7 ~hol ik Yolunc;



Answers

The orcs of $A g$ and Cu arc concentrarct using rhcir solubiliry in (a) $\mathbf{K C N}$ (b) $11 \mathrm{NO}_{3}$ (c) $11_{2} \mathrm{SO}_{1}$ (d) $11 C l$

Statement who is correct? Because n. c. -1. Let's n. c. r. is actually same as endless one. c. r. It's a very important for me actually it's very easy to prove. So what we do is we take the left hand side. The plant is in factorial by n minus R plus one factorial into ar minus one factorial. And this is in fact a well by end minus our factory into R. Factorial. So what will take common? So let's take common first the right length. In fact well by end minus R. Plus one in two and minus R. Factorial into our minus on factory in plus N. Factorial by end minus our factory into our into our minus one fatto. So then let's take common. We can take common in factorial by n minus R. Factorial into our -1 factorial in Bracket. What we get is one by n minus R. Plus one plus one bite. Uh huh. So let's take calcium in the bracket. So it becomes in factorial by N minus R. Factoring into ar minus two factorial. So when you take and see him you need to be in place one do not need to be in minus R. Plus one into art. But what is in fact a leading to N plus one? It is N plus one factorial divided by what is N minus R. Factory. And doing minus or plus one is n minus R. Plus one factory. In our into our minus one factories are factories. So which is nothing But and plus one C. R. So statement to is correct. Now let's use this formula and find the value of 15 c. four plus 15 C. Five plus 16 C six. 15 C four plus 15 to 5 will be 16 C. Five. Because in this and his 15 R. S. Endless one cr 16 five. Now that is getting added to 16 6 now here in in 60 RS six, so it should be 17 C six, but right hand side is actually given us 17 67 17 C six is not same as 17 C seven. In fact it is same as 1711. Alright, so statement one is wrong and statement tweets Okay?

In this video, we're gonna go through the answer to question number 17 from chapter 9.4. So we asked to find where these vectors x one x two x three Ah, linearly dependent on where they are linearly independence, which FYI is teeth independent much varsity that linearly dependent. So okay, so for them to be linearly dependent would need values off. See one c two c three it such that c one times x one c two times x two plus C three times Next three equal to zero association into values for these x one x two x three then we write This is a system of three equations. So the 1st 1 is just gonna be Ah, well, it's the the first element of each of the equations Time each of the vectors X one experience to text three times by the corresponding um constant C one C to C three. So we're gonna have Well, there's no, uh, ex threes are no component, Maxime, because the top of the next three is zero. So we're going off, uh, ex one next Tuesday, at both of which contain eats the to tease it. So take a common factor out. Get either to t C one plus C two. Ah equals there. Next up, we're gonna have eats the two tea. Close. It was C one that's actually to eat the TT plus C three he to the three tea equal Syria. And finally five. Easter to tee times five C one minus C two equals Sarah. Okay, so comparing the first of these equations on the last of these equations we find that C one don't see too must be equal to zero as you see that because, well, first look in the first equation e to the two tea for any tea can never be zero. So we basically just cancel by its beauty and saying with the bottom equation s So therefore, we have this bit is equal to zero this busy zero for those two to both equal to zero. Then C one and C t o. You can show that by rearranging one of those sub student in the other. Um and you're find this. You wanna see t birthday frequent zero. So therefore I see two is the rocks that this is zero. So then we have to see three times Three tier sequence There we can we can divide by eats and three team because that's never zero for any tea on DDE. All we're left with is C three sequences So far, all t ah the C one c to the T three. Must could only be for this for this equation. Thio be satisfied. Uh, this equation to be satisfied then Theo Dissolution of the Trivial Solution. Therefore the vectors x one x two x three are linearly independent for that tea in any value between minus infinity to infinity.

Today we are adding and subtracting fractions with unlike denominators in this case, specifically around denominators that have polynomial is in them in this case these are the two fractions that we are subtracting. Yeah. Now I want to start by factoring this polynomial here, your weekend. Simplify this by writing it as p minus seven P plus three. And then you'll notice something here. If we distribute this negative one into the bottom, we get AP -7. So I want to rewrite that just for clarity's sake we get a p -7 which means the only thing we're missing is a people street. So we can just multiply top and bottom by P plus three. That gives us a common denominator. And it also gives us a piece square A plus four P. That's four p.m. Plus three here. Which is all over the common denominator of P -7 People ST get great. Now we can add the top. Since we have that common denominator, we're left with a p square four,- -5 p. and negative. 13 17 plus three is eight of 14 over p by seven people's three. Now we can simplify the top even more by factoring it out. It factors out into p minus seven people's too And that's gonna be over. That common denominator, P -7 people stream. What do you know? We can cancel these out and we're left with P plus to overpay plus three as our final answer?

In this problem, I'm writing the reaction. Just look at it carefully. 903 Plus two i 2. It is dry. We will react to give the productive I food oh nine Plus nine or 2. Therefore, Here I four or 9 is produced when odeon react with dry i Odeen along with oxygen gas. Therefore, according to the given option, option A each correct and said for this problem, I hope you understand the solution of this problem.


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