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J MR 2 J MA...

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J MR 2 J MA

J MR 2 J MA



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What does $a_{i j}$ mean?

The symbol A I J. Represents an element that belongs to the eye. It rule Andi j it call him off a matrix.

This problem asks us to find the loss in verse of E to the minus two s over s minus. What? And a good rule of thumb when you're trying to find a plus in verse is to split up your term. So that's a little bit more recognisable when you go to your tables. So here we're gonna pull out e to the negative to s. And we're going to say that is multiplied by one over s my s one. You know, the easiest term here to find a plus in verse. Is this here? And we know that e to the minus a s Neill Applause Inverse of that is our unit. Step a little trinket over here. The plus in verse of E to the minus eight s equals our unit. Step turning on at T equals a So looking at each of the minus two s, you'll applause inverse of that need to the minus two s. We know that's going to equal you to the T minus two. Okay, so for the second term here, we're gonna look at our tables and we know that, um, one over s plus a equals no excuse beetle plus inverse equals goes to will say e to the minus anti. So when we look at these two terms, we can say that our A is equal to negative one here. So when we take the applause in verse of one over s minus one, there you go Equals E now. Ah, minus minus. Since we have a minus A t and R az minus one. When we take it, a plot inverse of this, we're just gonna end up with e to the T. So now that we have Philip lost fingers off both of those terms up there, we are going to remember that what we want to dio is we want to put this together into a form that looks like this. So f t minus a U T minus a. So to do that, we know that when we took the inverse slope loss of f of s, we got each of the tea so we can write off to the side. We know that f of t is equal to e to the T, but what we want is we want to make sure we get f A T into half of T minus a. So if of t minus a turns out to be e and we know that a here equals two. So we're gonna have t minus two. This is our a term here, So T minus two. So only put this all together. Asked a team of T minus a multiplied by U of T minus A. We will have e to the T minus two multiplied by you. T minus two. This is the inverse LaPlace.

Problem. We're being asked to simplify the given expression so we have to raise to the empower. And that entire quantity is getting raised to the K. Power. Well, this is an example of a power to a power rule. So let's review how that rule works. So that says if we have a base X getting raised to the P. Power and this entire expression is then getting raised to another power. Let's call it cute. Well, according to our rule, all we need to do is multiply our two exponents and keep the same base. So in other words, we can rewrite this as X. Getting raised to the power of peak times. Cute. So notice the base stays the same and all we do is multiply the exponents. So we're going to apply that rule to our problem. So when we take two to the empower and raise it to K. Power. We can rewrite this as to raise to the and then we're gonna multiply or two exponents. So we'll have the MK power. And now it's simplified, it's to raise to the M. Times K. Power.

I guess so. We want to simplify the following well seven times Fine. That's a 63 and then have a nice word. So that's a negative 63 times a negative one, and that's equal to a positive 63.


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