Show that laplace transform StartSet e Superscript at Baseline t Superscript n EndSetℒeattn(s)equals=StartFraction n exclamation mark Over (s minus a )Superscript n plus 1 EndFraction n! (s−a)n+1 in two ways.

(a) Use the translation property for F(s).

(b) Use the formula laplace transform StartSet t Superscript n Baseline f (t ) EndSet (s )ℒtnf(t)(s)equals=(negative 1 )Superscript n Baseline StartFraction d Superscript n Over ds Superscript n EndFraction (laplace transform StartSet f EndSet (s ))(−1)n dn dsn(ℒ{f}(s)).

Step-by-step explanation:

a) The function is:

Is is necessary to demonstrate that:

you can use two identities of Laplace's transforms:

By using the previous transforms you obtain:

b) The other way is to use the following formula:

Then, you use the following identities:

then, by replacing you obtain:

answer: 370

step-by-step explanation:

remember use pemdas! there are no parentheses or exponents so lets move onto the first part of the equation: multiplication and division. first, lets break the first part of the equation up into something easier using the distributive property. 28 * 12= (20 *12) + (8*12) you know that 20*12 is 240 and 8*12 is 96. add them and you get 336. then, lets solve the last part of the equation (the last part of pemdas is addition and subtraction). 336 plus 34 is 370. hope this !

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