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Wenment TVIDIOcodMhkh0' tJ=EannePrtScnhan...

Question

Wenment TVIDIOcodMhkh0' tJ=EannePrtScnhan

Wenment TVIDI Ocod Mhkh0' t J= Eanne PrtScn han



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$$\mathscr{L}\left\{t^{2}-3 t-2 e^{-t} \sin 3 t\right\}$$

We're gonna look at the T statistic and find the critical values that associate with each of the specific T guys that were given. So typically, this would be an appendix in the back. Your book, a charge or table or teacher professor might give you one, but all you need to do is look at that and match up with what were given per problems over the 1st 1 We have tea, which is going to be 18 and 180.9. So when you go on there, you would see that we actually need to look for negative t at 18 and 0.10 because in the chart, we're looking for the entire not just the one sided. So we have to find the complement of 10.9 to get toe one, which is 10.1. So that's going to give us negative 1.33 Okay, so that's a critical value for the 1st 1 The 2nd 1 we're gonna have tea at 9.99 So the same thing for this one, we want the compliment. So negative T at nine. And then 90.1 We're just gonna give us negative 2.82 Then the next is T when we're at 35.975 So again, the complement of this one just simply negative t We're gonna keep the 35 then find the compliments of one minus the 10.975 which would give us 0.25 And that's gonna give us negative 2.3 for the critical value. And then our last one that were asked to look at in this chart is T at 14 0.0, 4.98 0.98 So again, we want the complement of that. So we need to find so negative t keep the 14 but we're gonna have one minus 2.98 Which 0.2? It's Give me negative 2.62 Thanks. So again, just a reminder. We're finding the compliments of these, which are going to be one minus the values given on the chart on. Then you'll see the critical values or the native 1.3392 point eight to thinking of 2.3 and negative. 2.62

Okay, so here were given Ah, loss. Transform of t cute Minus t Uh, times e to the power of T plus e to the power of fourty times co Sign of tea says the first thing you can do is just break this up in the three little chunk. So we will evaluate the low cost transform of t cubed. So track that from the transformer tee times e to the power of T now that to the transform for E to the four tee Times Co sign t then to evaluate all of these. We just want to use that table 7.1. Um, that tell us what the u a pause transforms are. So the transform 42 the power end is going to be an factorial. In this case, it's three over s plus and plus one. So again, this case and his three So it's gonna give us two out of four. Okay, then from Atlas attract in this case for the he's out of tee times like t to the power of and we get n factorial of this case is gonna be one over Ah s minus. Whatever each of the team's multiplied by in this case, it's going to be once we get s minus one for the power of and plus once and is what tea is raised to in this case is one us That's just gonna be U to the power of two. And then finally, we'll add that to this last function, which is going to be s minus a being. What he's multiplied by in this e functions has been before over again. That's minus a squares the s minus for it's where 1st 1 um and one being sort of what is multiplied with what to use multiplied by in the coastline function. Um, so if you're looking at table 7.1, that will make sense. And when we violate this will three factorial is going to be a six, and then one factorial is just one. It's worth noting that this is gonna be for s straighter than four. Okay, and then that is our solution

So the given different jelly question is being B minus three. Why? Double dash less duty? Why? Single lash minus y equals to be square. Then the standard form off the question will be by double dash less. Don't be divided by okay, the mine STD my Nash minus y upon he being minus three equals two peace. Where upon be B minus three Here. Creepy big party, too. You would be divided by be the U minus three and GOP would be quickly minus one important be B minus three and GP with a big bagel be divided by T minus 30. So here, beauty all my you'll be Coleman GP our corn tenuous continuos in the interval zero you three and three don't. Infinitely. And the point being on close to one nd interval zero 23 Therefore, the differential equation has unique solution in the interval Zero Do you the thank you

To finding the last class transform of E to the negative to tee times Sign of to see well e to the three tee times t squared Now, using a table this is equal to the laughter show in some of the first, which is to I have a pass plus two square plus for the two Here comes from the negative due here before is the square of two. And this to also comes from this to the lab past Transmit The second term is too over s minus three cute.


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