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licre the ycllow and orangc precipirares arc, rcspectively (a) $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}, \mathrm{~K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{-}$ (b) $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}$ (c) $\mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{~K}_{2} \mathrm{CrO}_{4}$ (d) $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}, \mathrm{~K}_{2} \mathrm{CrO}_{4}$

Really asking us to do is to calculate the universe transform of the rational function escape plus full of the s squared plus X minus six s lead plus X Now the first step to performing the transformers to obtain the partial fraction to composition which will start by breaking the denominator in too linear terms. Since all the linear terms of distinct we can actually write this as some number over s minus two plus some number of s plus three plus some number over S plus five plus some number over s. To obtain each of these numbers, we plug in the value which would cause this to be equal to zero, um into this equation, ignoring the problematic terms where you will get an infinity in the denominator. For instance, to get this constant, we plug in X equal to two into this equation, ignoring this term here. If we do that, we'll get the particle a fraction decomposition forever. This Dean native 13 over 30 negative. Two of the three and five of the six. If we now take the inverse transform of this, then we'll get for over 15 times e to the two t minus 13 over 30 times E to the negative three T minus two over three times E to the negative five t plus five of the six times a constant.

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.

For this question. The first thing we're gonna do is to conducting the partial fraction the conversation. We have 33 Esquire Mons Extras two on the top and, uh, as minus one times as squared plus one on the bottom. So we're conducting the partial fraction. The conversation I'm gonna have s O. S were close one minus two os minus one and the function we're looking for. So let's call it out of the ass is equals Two Africa's times eat through the naked ass. So the strategy for this one this to apply the inverse will transform the first to admit ass and then shifted by one. So by shifting by one, I mean times the function you have t minus one, and, uh, we can apply to the charity of immersed up those transform. So the inverse of transform gonna be decomposed by the sum of two terms so well, I mean by that is too. Apply the investments to transform to as over as squared, plus one of T minus one. So, shit that one. Because we're from here. It's like the US pull us l of next two restaurants. One of two plus two minus one. Together times you have to minus one silver. Which part is pretty easy to grab the ourselves to transform. And they were gonna shred it down for the first time. We're gonna haveto deep two times e to the T minutes for one in the second term. We have co sign of T minus one. I'm sorry. Yeah, this is for that's Lee. So, for I just without the the order so far for this term is actually co sign of tea months one because his ass over as sort plus work. And for this firm is two times it to t minus one. And together we're gonna mount. Led by You have two months one and that's it.

Problem wants me to find the product of two training meals. So are plus two has minus three t times to our minus two s. Rusty. Okay, I said I do this. I'm just gonna distribute across edition. So first my gun belt by everything in my second polynomial about our than by two s And by negative three t kr Thames To our is to our square our terms to s is negative. Teoh r s are times t is just plus rt. Next time I'll use multiple about two s. So I'm gonna have four R s uh, minus four s square plus two s t. And finally I'm gonna have negative three tee times to our, which is negative. Rt negative six rt negative. Three tee times native to Asa's positive six s t the negative. Three tee times positive t is negative. Three and t squared. Okay, so adding up each one of these new pollen Emile's, we give a star equivalent simplified polynomial. So you two are square plus two Rs minus five rt minus four square plus aids s T minus three t squared. Okay, thank you very much.