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Roman55 [17]
3 years ago
6

A poster is shown below: A rectangle is shown. The length of the rectangle is labeled 12 feet. The width of the rectangle is lab

eled 3 feet.
 What are the dimensions if the poster is enlarged by a factor of five over two ?

7.5 ft by 30 ft
15 ft by 60 ft
1.5 ft by 6 ft
15 ft by 6 ft
please be right if i get this wrong i'll fail ;(
Mathematics
2 answers:
Vikki [24]3 years ago
8 0

Answer:

5.7 by 30

Step-by-step explanation:

i did the test broski and sub to pewds

klemol [59]3 years ago
6 0
It would be 7.5ft by 30ft
Because I just multiplied 12 by 5/2 which is 30
Then I multiplied 3 by 5/2 which is 7.5
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kiruha [24]

For the given function f(t) = (2t + 1) using definition of Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

As given in the question,

Given function is equal to :

f(t) = 2t + 1

Simplify the given function using definition of Laplace transform we have,

L(f(t))s = \int\limits^\infty_0 {f(t)e^{-st} } \, dt

          =  \int\limits^\infty_0[2t +1] e^{-st} dt

          = 2\int\limits^\infty_0 te^{-st} + \int\limits^\infty_0e^{-st} dt

         = 2 L(t) + L(1)

L(1) = \int\limits^\infty_0e^{-st} dt

     = (-1/s) ( 0 -1 )

     = 1/s , ( s >  0)

2L ( t ) = 2\int\limits^\infty_0 te^{-st}

        =  2[t\int\limits^\infty_0 e^{-st} - \int\limits^\infty_0 ({(d/dt)(t) \int\limits^\infty_0e^{-st} \, dt )dt]

        =  2/ s²

Now ,

L(f(t))s = 2 L(t) + L(1)

          = 2/ s² + 1/s

Therefore, the solution of the given function using Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

Learn more about Laplace transform here

brainly.com/question/14487937

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grandymaker [24]
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