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H Alledd [OSetadmnlnneted ! nnd Quluinod zoro 1 1 Tal eaBttD...

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H Alledd [OSetadmnlnneted ! nnd Quluinod zoro 1 1 Tal eaBttD

H Alledd [OSetadmnlnneted ! nnd Quluinod zoro 1 1 Tal eaBttD



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(1) $\mathrm{SiO}_{2}+\mathrm{NaOH} \rightarrow ?$ (2) $\mathrm{SiO}_{2}+\mathrm{HF} \rightarrow ?$ The products of $(2) \&(1)$ respectively are (a) $\mathrm{SiF}_{4}, \mathrm{Na}_{2} \mathrm{SiO}_{3}$ (b) $\mathrm{H}_{2} \mathrm{SiF}_{6}, \mathrm{SiO}_{4}{ }^{4-}$ (c) $\mathrm{Na}_{2} \mathrm{SiO}_{4}, \mathrm{H}_{2} \mathrm{SiF}_{6}$ (d) $\mathrm{Na}_{2} \mathrm{SiO}_{3}, \mathrm{SiF}_{4}$

So this grammatical problem is about word choice, and in this case, we can't. So we're gonna cross out. First of all, answer choice, Jay, because they're the specific pronoun it's does not apply to. It doesn't make sense in this context. We're not referring to an object referring to the author narrators referring to herself. So a per the rules of grammar we have to use again. If she's referring to herself, then she has to use the, uh, the phrase that she is using a to moment. We have to keep the answer. We gotta keep the word the same.

In this problem, this component heritage development, No. In presence of and I alleged food will produce this component as the product. Here it is double one inherited four edge minus and I plus. So this will be A. And this in presence of AD plus will give this compound here in this side, ohh, in decided T o H or this component or edge or edge. So according to the option. In this problem, option B c D are correct.

So we have to evaluate the following limit and see if it equals one over E. So um the first thing to do is to plug in the value of the limit and see if it's something easy. So if I plug in the value H equals zero. I get one minus zero raised to the negative 1/0. And since you can't divide by zero, this becomes a problem. Now in order to solve this problem, we need to use one rule called low petals rules. So I just want to remind everyone what low petals rule is. So let's suppose that you have a function actually a quotation of two functions and you're taking the limit as X approaches some about you And you can write the expression as a ratio of two functions. Now if this evaluates to an indeterminate form that either looks like 0/0 or infinity over infinity and this is plus or minus infinity. Then you can rewrite the limit as follows. You can take the limit of the derivative of the numerator divided by the derivative of the denominator. So I just want to quickly interject and say that this is not the corset rule. Look up the quotient rule, remind yourself that is nothing that has nothing to do with this. This just says that you isolate F and take the derivative of that and then you take the derivative of G. And then you compare the slopes. Okay? So it essentially gives you an easier way of evaluating the limit because usually the derivative is a slightly simpler function, especially with polynomial. So that being said, let's evaluate this limit. The first thing I'm gonna do is I'm actually going to set the limit equal to evaluate why. You'll see why in just a second. So the limit h approaches zero of one minus H. To the negative one over H. I'm going to let that equal why. And the reason is this exponent is quite pesky in order to get rid of the expo and I'm actually gonna take the natural log of both sides. Now you can interchange a function and a limit as long as the function is continuous and in this case the natural log is continuous. So I'm gonna go ahead and say that the natural log of Y equals the limit as a jew approaches zero of the natural log of one minus H. To the negative one over H. Now, because the properties of logarithms, I can move this exponent down as a coefficient and then I get that the natural log of y is equal to the limit. As a jew approaches zero of negative one over H times the natural log one minus H. Just to remind you, we are trying to solve for why why is equal to the limit that we want? This is why. And here's the limit that we want in blue. Now to evaluate this limit. Um I can rewrite this as the limit As a church approaches zero of the Natural Log. Well negative the natural log of 1 -3. All over H. Okay. And if I plug in H equals zero at this point so I go and just plug that in, I will get negative natural log of 1 0 over zero. Well the natural log of 1 0 is the natural log of one And the natural log of one is 0. So I was able to write a ratio of two functions. And upon evaluating the limit, I get an indeterminate form. So now I can use low P. Tall and lo petals rule says that the same limit which was equal to the natural log of Y. Is the same limit. But now I separately take the derivative of the top and the bottom. So I'm actually just going to write that out. Just we don't confuse it with the caution rule. And I should specify that I actually made a mistake. Um It's not a big mistake here. We're taking the derivative with respect to H. I never use H. That's kind of line thrown up. I always say as X approaches something. So regardless, anytime you take a derivative, uh this variable in this variable better match. So that was my bad. Okay so this is just gonna be H. Now the derivative with respect to H. Of negative natural log of whatever. I can factor the negative out. And the derivative of the natural log is just one over whatever's inside. But I have to remember to multiply by the inside function using the chain rule. So I also have to take the derivative of um one minus H. And that's just negative one. So these negatives cancel and the derivative of um H. Is just one. So I have to remember that I can take the same limit but now I evaluated those derivatives. Okay, so just to rewrite what we've done the natural log of why we use loopholes rule in the numerator and denominator. And we said that now I'm going to take the limit, I'll be slightly simpler function. So these negative signs go away and I'm left with 1/1 -3 divided by one. Well that limit is super easy because now I can directly substitute the value at H equals zero And I get 1/1 0 which is one. I'm not done. The limit does not equal one. Because the limit that I'm interested in is why I set the limit equal to Y in the very beginning. So the natural log of Y is equal to one. So if I exponentially both sides, I get that. Either the natural log of Y cancels to why and the limit that I'm interested in why is equal to E. To the first power, which is just so this is the limit and it's actually not equal to um one over Eats. This is false.

In this problem we are going to use the properties of mattresses and they're in verses in order to simplify a given mattress expression. Now the given matrix expression is E C inverse whole inverse times E C inverse times E C inverse whole inverse times E. The envious Now in the question it is said that a B C and D. Are all in vertebral mattresses. Now let us begin to simplify this expression. Now, first of all, what we have is easy inverse inverse times, easy inverse. That means that the matrix easy inverse has its inverse multiplied with this original matrix. Now, since the product of any metrics with its inverse is equals identity matrix. Hence a C inverse inverse times a C. In verse will be the identity matrix. I. With this we multiply the remaining two terms A C inverse inverse times a. D. In verse. Since any matrix multiplied with the identity matrix, is that mattress itself? This will be equals to A C. Involves inverse times A. D. In Vegas. Now using the property of matrix inverse, which says that a B hole in verses equals to be inverse. The inverse E C inverse hole in verse will be equals to see inverse inverse times a inverse. And with this we multiply a. D. Universe which are the last two terms of that expression. Now see inverse inverse will be equals to see because the inverse of any inverse is the original matrix itself. Next we have inverse. E be endless. Now, since matrix multiplication is associating, we can group together the terms however we want without changing the order. So let us group together to terms a university. And then we use the property that the product of any matrix inverse is equal to the identity matrix. So inverse A. Will be equal to I. And we are left with C I. D. In most now since the product of any matrix of the identity matrix is the matrix itself and ci is equals to see. And with that we have at the end the in laws, so that means that our metrics expression is simplified to CB in Vals.


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