Evaluate the expression when x=-6. x² + 5x-5
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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 =
=
=
= 2 L(t) + L(1)
L(1) =
= (-1/s) ( 0 -1 )
= 1/s , ( s > 0)
2L ( t ) =
=
= 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
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Answer:
Step-by-step explanation:
The given inequality is 3x-2>4 or
We group similar terms to obtain: 3x>4+2 or
Simplify to get:
3x>6 or
x>2 or
This implies that the solution set is all real numbers.
The solution in interval notation is
