Answer:
The first one!
Step-by-step explanation:
Answer:um you simplify it
Step-by-step explanation:
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I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform

and

, then

Given that

and

So you have

From Table of Laplace Transform, you have

and hence

So you have

.
Hope this helps...
Answer: d=-3
-7d+8<29
-8 -8
-7d<21
-7d/-7<21/-7
d<-3