Find the Laplace transform F(s)=L{f(t)} of the function f(t)=sin2(wt), defined on the interval t≥0. F(s)=L{sin2(wt)}= help (formulas) Hint: Use a double-angle trigonometric identity. For what values of s does the Laplace transform exist? help (inequalities)

a. b.  F(s) exists when s ≥ w + wi√3

Step-by-step explanation:

a. Laplace Transform of f(t) = sin²(wt)

f(t) = sin²(wt)

Using the double angle identity

sin²(wt) = {1 - cos(2wt)}/2

f(t) = [1 - cos(2wt)]/2

Now Laplace of f(t) = F(s) = L{f(t)}

= L{[1 - cos(2wt)]/2}

= 1/2L{1} -1/2L{cos(2wt)}

=

b. Range of values of s for which F(s) exists

The range of values of s for which F(s) exists is when F(s) ≥ 0

Using the quadratic formula,

So F(s) exists when s ≥ w + wi√3 or s ≥ w - wi√3

So F(s) exists when s ≥ w + wi√3

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10

83-45=38

38 was the other number

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