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Question

J7i catenin Gtaeitintt lunznon noee_Enedut I44 6mote PIRa IneunhAmria ninanIILaMenan L74ruaEnekmdaGdu 'Eom eula k DEcukratE y mn corqWDAeomaLuTraeemItnm TLeEiLlHleaihi MinnnoatmentuhnanMATetaetieeaineteMdeealuenEarclcuuabetCecdnhdattTeak PkTro Ovon % # der0 Trojm (nr k hI VRTeradin mgontrotounelnddmnen0CotMonocunjuch Debo cubomt "rotoulaetEtFecA ImAeenMoTtt3 ChateMWAnn4oludUpu LH DIEElufuMlou 344ACAlAutDSCAEtTMitthr EJZAl 5 [Znp Luin #tall Tro prjubllat WIIo ! Motabllul Ju7i 00+60NAne

J7i catenin Gtaeitintt lunznon noee_ Enedut I44 6mote PIRa Ineunh Amria ninan IILa Menan L74rua Enekm daGdu 'Eom eula k DEcukratE y mn corqWDA eomaLu Traeem Itnm TLeEiLl Hleaihi Minnnoat mentuhnan MA Tetaetieeainete MdeealuenEarcl cuuabet Cecdnhdatt Teak Pk Tro Ovon % # der0 Trojm (nr k hI VRTeradin mgontrotounelnddmnen0CotMonocunjuch Debo cubomt "rotoulaet Et FecA ImAeen MoTtt 3 Chate MWAnn 4oludUpu LH DIEElufuMlou 344ACAlAut DSCAEt TMitthr EJZAl 5 [ Znp Luin #tall Tro prjubllat WIIo ! Motabllul Ju7i 0 0+60 NAnea nenanEnaahin wpuubbe #Aatelut [ tlente4m LeetmuMula alan



Answers

$\mathbf{X}(t)=\left[ \begin{array}{ccc}{e^{t}} & {e^{-t}} & {e^{2 t}} \\ {e^{t}} & {-e^{-t}} & {2 e^{2 t}} \\ {e^{t}} & {e^{-t}} & {4 e^{2 t}}\end{array}\right]$

Hello and welcome to this video solution of numerous. Here we are given the rules of silver and copper are constantly reducing their scalability in So we have got option A. S case in which is the correct option. Now using the in the Synod process, what we do is okay let's let's try to find out what in what is happening in case of signing process. So let us take the Argentina, which is a G two S. Plus. We have Casey in now for this occasion this gives us a complex of silver. This is okay, ain't you? Him two plus it works now. This compound that is formed is soluble in the solution. He's so little now this is thing for copper also. Thus we can have a quick and select option skc I hope this is clear to you and have a very good thank you.

Today we are evaluating fractions polynomial fractions into equivalent fractions when given alias common denominator in this case the first fraction were given to work with is three over five M squared minus three um minus two and the least common denominator we are given is five AM plus two am minus one and M plus three. No, the first step is to factor out the denominator in this fraction and when we factor it out we are given a five M plus two and it's going to be m minus one as well. No, in order to make the equivalent value over the least common denominator, we need to multiply the top and bottom by whatever is missing from the given least common denominator. In this case that is an M plus three. So we multiply the top by M plus three and the bottom by m plus three, distribute the three out, gives us three um plus nine over our common denominator of five M plus two am minus one and plus three. And that is going to be your first fraction equivalent over the least common denominator. Now the second fraction they give us is six M oh man five M squared plus 17 plus six. So again, we want to start by factoring this out. If we factor it out, we are left with five M. A plus two and M plus three. Now, same as above. We want to multiply by whatever is missing in the least common denominator. In this case we are missing the m minus one. So we multiply the top and bottom by m minus one, distribute the six ml gives us six um squared minus six M over our common denominator of five M plus two um plus three. And now we have the m minus one as well, which is what we want. And this will be your second fraction equivalent over the least common denominator, meaning your answer to this problem will be these two fractions.

Okay, So for part eight, this is going to be a two by two matrix or he is going to be a three by three matrix parts. He's like to be an undefined on defined because it's simply not possible harpies going to be a two by three Matrix party's going to be three by two Matrix Heart F is going to be a three by three matrix, part G. The party's also going to be undefined, but it's simply not possible. Undefined on party. I don't know howto do off its top party. Her H is going to be a three by two matrix.

You're going to be subtracting these fractions. Ah We're going to start by trying to find the number, The lowest number that can be divided by 8, 5, 2 and 10. And that number will be 40. So we will make all of our denominators 40. So we have to look at what we multiply by each of our denominators In order to get 40. And in the case of eight we take eight times five to get 40. So therefore we need to take five times 5 to give us our new numerator of 25. five times 8 is 40 two times a gives us 16, two times 20 is 40 three times 20, gives us 60 10 times four is 40 And 11 times four gives us 44. So um when we subtract a negative will really be adding. So we can think of us as having a plus here, the negative 25 plus 16. Well give us -9 over 40 And then 60 -44 will give us 16. So the negative 9 -16 Gives us -25 Over 40. Now we're not quite done yet, Because both 25 and 40 Can divide by five, so our answer further simplifies into negative 5/8.


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