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Question 6 LetT2 k_7 k--[A Atd 6- [{al+en K 3k2Find the value(s) of k for which A7 = b has:Mique solution110 solutioniil ) infinite solutions8packard bell...

Question

Question 6 LetT2 k_7 k--[A Atd 6- [{al+en K 3k2Find the value(s) of k for which A7 = b has:Mique solution110 solutioniil ) infinite solutions8packard bell

Question 6 Let T2 k_7 k--[A Atd 6- [{al+en K 3k2 Find the value(s) of k for which A7 = b has: Mique solution 110 solution iil ) infinite solutions 8packard bell



Answers

In the following exercises, determine whether each number is a solution of the given equation.
$$\begin{array}{l}{\text
k+\frac{2}{5}=\frac{5}{6}} \\ {\text { (a) } k=1 \quad \text { (b) } k=\frac{13}{30}} \\ {\text { (c) } k=-\frac{13}{30}}\end{array}$$

The question acts us to determine the constant value off que when the system has a no solution, be infinite number off solution and the sea unique solution. First, we need to change a system off the equation into a semantic metrics form. So take the core vision out. Yes, remember, we need to find the reducer or I should inform to find a final answer. So for row to we can do to times roll one might as a road to and for rosary we can do roll one by the Saros three Copy Roll one to the elementary row operation for row two and the rosary for row to we can do Maybe one times wrote to copyright. Wanna rosary? Okay, For rosary, we can do row to minus zero. Cherie copy robot either wrote you part. They asked us if the system has no solution. So if there's a battle rule in The Matrix, which means the system is inconsistent, Thea other words, there's no solution for the system. So for this case for minus K square should equal zero and negative two minus K should not equal zero. So for this one K one equal to que two you call negative too. And for this one, Okay. Equal. Negative chew. He ordered to satisfy both condition Que equal negative two should be rejected. So one k equal chew. The system has no solution. Part B asked if the system has infinite number off solution if there's all zeros in wire oath, which means there's a free variable. So for this case four minus case square should you call zero and negative two minus K should also equals zero. So for this one, Okay, one equal chew okay to equal in negative too. And for this one k equal negative too in order to satisfy both condition. Okay, should equal negative too. So one k equal negative to the system has infinite number off solution. Parsi asked, if the system has a unique solution If this position does not equal zero, the system will have unique solution. So for this case, four minus K square should not equal zero So occasion not equal either positive or negative too one, Kate

Let's calculate the percent association in the following point to Moller assets. For a we have point to Moller nitric acid. Nitric acid is a strong acid, and all ass strong assets undergo 100% dissociation, so there's no calculation needed for A for B. We have 0.20 Moeller nitrous acid, and we'll have to calculate its percent Association as nitrous acid is a weak acid will regard equilibrium here for nitrous acid. We'll fill in our ice table initially 0.20 Moeller Rick Kay expression Looking up the value for K A for nitrous acid, we can find that it's equal to 4.0 times 10 to the negative. Four. Substitute our values in here. And since this is a weak acid, we consume that the denominator 0.20 minus X of X is very small. Sequel the 0.20 Therefore, we have 4.0 times 10 of the negative four x squared over 40.20 Solving for X gives us 8.9 times 10 to the negative three, which means that our each 30 plus equals 8.9 times 10 to the negative three Moeller in Open Calculator Percent Association. What should be the polarity of the x 30 plus dissociated? Divided by the initial Milady went to zero Moeller times 100% and this will yield a percent association that's equal to you. 4.5% for see were asked to calculate the percent association for vino Uh, 0.20 Moeller h o c. 65 Beano is a weak acid and we will set up in ice table for, you know, starting with the equilibrium Initially 0.20 Moeller que expression. Looking up the K, we find that it is equal to 1.6 times 10 to the minus 10. And once again we're going to assume that the denominator I went to ZRA minus X of X is very small. And this will essentially be just went 20 on the denominator and solving for X yields 5.7 times 10 to the negative. Six makes each 30 plus my 0.7 times 10 to the negative six Mohler and our percent association is equal to our 0.7 times omega six, smaller, divided by the initial polarity, put two Moeller times 100% and we get a percent dissociation here, prince, equal to 2.9 times 10 to the negative. 3% for D were asked, How is the percent association of an acid related to the key value assuming the equal in initial concentrations. So, uh, for the same initial concentrations, the percent dissociation increases as the strength of he acid, which is a k a increases.

16 3 dot to Problem number 23 were asked to take a look at this linear system and find all values of the constant K that would guarantee me a unique solution for this system. So unique solution means that the determinant is non zero. So the determinate of the system is going to be the determinant. So right, the majors a coefficient, So one okay. And then K four. So to have a unique solution, this cannot be equal to zero. So that means that when you take the determinant here, so one times for minus K times K cannot be zero. That means that four minus k squared not equal to zero. So the only time that that would ever be equal to zero as when K is equal to plus or minus two. So OK, cannot be plus or minus two. So that's the condition. So every other value of K would give me a unique solution. So all values all real number for K, with the exception of plus or minus two to get a unique solution for this system,

Already. Number 33 off. They're getting tricky, aren't they? It's OK. Over K minus four. My five e cools You're over. Came us for that we want to do is get rid of the fractions and we got this infractions earlier. But before we do that, we want to put this in princes and this princes to indicate that that's really one number. Okay, so they came on its course one value. And this has the denominator. Once the common denominator here, it's came on us four. So I want to multiply each term. Times came on us four so I can get rid of that denominator. And then what happens when I do that? Is this quantity and this quantity simplifying? I'm left with K gotta distributed here. So I have five k less 20 and this quantity and this quantity simplify on them, left with four. OK, and now let's combine some like terms because I always like to simplify before I solved. So when I add these together and get a negative four K plus 20 equals four and now I can dio minus 20 minus 22 those sides. And so I have a negative four k equals negative. 16. Gonna divide by a negative for and find by negative for And now I have hey, equals four. So let's see what happens with that when we check it. So four over four, minus four, minus by equals for not yeah, over. Okay, four minus four. Let's see. So this is for zero. This is 40 and it seems to me like we can't have values with zero on the denominator, So this would be in no solution, but you can't have four in the denominator. There you go. That was sneaky of him. No wonder they want you to check.


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