5

1 =l 2 N T- 325....

Question

1 =l 2 N T- 325.

1 =l 2 N T- 3 25.



Answers

Solve each equation.
$$\frac{1}{25}(20-t)=\frac{4}{25} t-\frac{3}{5}$$

Right too. They were final actress Transform of three T squared minus E to the two T. Now using a table and linearity is quite an easy problem. The last transport T squared is equal to to Over s cute and the class transformed. Each of the two t is one over X minus two is a founded the table in your book. Uh, simplifying. We find the live past Transformers six over s cubes minus one. Ever es my message.

Today we are subtracting fractions with unlike basis specifically those involving Paula, no meals that needs to be factored. So these are the two fractions we're dealing with here. And I want to start bye simplifying this. Oh I forgot the negative 30. My apologies. I'm Gonna Simplify it by taking out the 2 1st. This might not go anywhere, but it's a simple thing we can do. So I'm going to do it and then I want factor the bottom Which it can be factored out into plus nine & T -3. Okay And then now we see something cool up top here if we distribute this negative one make this a positive and distribute now you want into the denominator we get negative three and plus T. Which is the same as t minus three which is what we have down here. I'm gonna rewrite that. Just that we don't have any confusion. We have T -3 now up here which means we are only missing this ti plus nine for a common denominator. So we'll do that. We'll multiply both sides by G. Plus nine of the top and bottom rather. Mhm. That now gives me two T. Plus nine plus 18. Sorry over our common denominator O. T. -3 & T. Plus the line. And it does actually look like I did this for no reason. So I'm gonna go ahead and redo it over here. Just really quickly Leave it at two T -30, Adding them together. We get 40 minus 12 over tm minus three. T plus nine. Now we aren't quite done. We're going to take out that four that we see in the numerator. It leaves us with T -3 and the numerator as well as the denominator. Which is great for us because that means we can cancel them out and we're left with four over T. Plus nine. As our answer for this problem.

Hi. We're finding the last class. Transform a three T to the full minus 22 squared, plus warm. Now using linearity in a table. This is quite easy. It's, you know, the three times the last left transform of tea to the full minus two times the last blast Transform off T square plus the lack last transform of tea, which is a glass has transom over constant, which is one ever es simple flying Be fine for the lap as transform its 72 over s to the power of five minus four over s cubes plus one over s.

Okay, So I can write, tend to the fifth and expanded form as 10 times itself another four times, or for a total of 10 times. And then I could simplify. Left rights of 10 times 10 is 100 times 10 times 10 is 100 times 10 ah, 100 times 100 is 10,000 times 10. This will give us 100,000 as our final answer.


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