Please help , if you can help me with a few more problems that would be lovelyyy just hit my dms my insta is ynlness.a :)
1 answer:
You might be interested in
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
#SPJ4
Domain, on an "even root" context, means, an even root cannot have a negative radicand, since say for example 
so... you end up with an "imaginary value"
so... for the case of 20-6x
"x", the domain, or INPUT
can afford to have any value, so long it doesn't make the radicand negative
to check for that, let us make the expression to 0, and see what is "x" then
now, if "x" is 10/3, let's see
now, 0 is not negative, so the radicand is golden
BUT, if "x" has a value higher than 10/3, the radicand turns negative
for example
so.. that's not a good value for "x"
thus, the domain, or values "x" can safely take on, are all real numbers from 10/3 onwards, or to infinity if you wish
so... you end up with an "imaginary value"
so... for the case of 20-6x
"x", the domain, or INPUT
can afford to have any value, so long it doesn't make the radicand negative
to check for that, let us make the expression to 0, and see what is "x" then
now, if "x" is 10/3, let's see
now, 0 is not negative, so the radicand is golden
BUT, if "x" has a value higher than 10/3, the radicand turns negative
for example
so.. that's not a good value for "x"
thus, the domain, or values "x" can safely take on, are all real numbers from 10/3 onwards, or to infinity if you wish