Answer:
L(f(t)) =
Step-by-step explanation:
let f be a function defined for t ≥ 0
we can write the function f(t) in terms of unit function as follows
f(t) = 2 u,(t) - 1 where
0≤ t < 1
f(t) = (2 * 0) -1 = -1
when t ≥ 1
f(t) = (2*1 )- 1 = 1
Now the Laplace transform L(F(T)) = 2L( u, (t) ) - L(1) --------equation 1
this is because L(u,(t)) =
c = 1 hence L(1) = 1/s
back to equation 1
L(f(t)) = 2 - 1/s laplace transform
also L(u(t) ) =
Answer:
Impossible
Step-by-step explanation:
In 5x^2-4x+3=0, standard form, substitute these values in the quadratic formula:
a = 5; b = -4; c = 3
The quadratic formula is
(ignore the weird capital A)
Substitute a b and c:
Simplify:
Because is the square root of a negative number, the answer would be imaginary.
Therefore, there are not solutions to this equation.
A solution is the same as the roots or zeroes, where the graph would cross the x-axis when graphed.
The graph never meets the x-axis. It looks like this:
