T minus 30 seconds.

T minus two minutes.

T minus three minutes.

So it would be e to the t minus 2 cosine of t minus 2.

So leaves at 6 o'clock minus t.

So Dirac delta of t minus c.

Delta t minus c and times dt.

So it must have been t minus 10 to go, and then t minus 10 to come back.

Well, if you 6 o'clock plus t minus 6 o'clock minus t, the difference in time is just 2t.

He leaves at 6 o'clock minus t, right?

Oh sorry, not f of t minus c.

Well, I would go to t minus 2.

Or we could just rewrite this as 2 times the unit step function starting at 2, where that's when it becomes non zero of t times f of t minus 2. f of t minus 2 is this with t being replaced by t minus 2.

The ship will detonate in T minus ten minutes.

This is f of t. e to the minus st times f of t dt.

So I t is equal to minus 5 plus or minus the square root.

So the Laplace transform of e to the t cosine of t became s minus 1 over s minus 1 squared plus 1.

Let's say that x is equal to t minus c.

The ship will automatically destruct in T minus five minutes.

The ship will automatically destruct in T minus one minute.

0 minus 2 is minus 2, 3, minus 1 is 2, for t is a member of the reals.

We could just take the integral from t is equal to c to t is equal to infinity of e to the minus st, the unit step function, uc of t times f of t minus c dt.

So the Laplace transform of our shifted delta function t minus c times some function f of t, it equals e to the minus c.

It's going to be zero everywhere until we get to minus t, and then at minus t, we're going to jump up to some level.

So this function right here is f of t minus c.

The President will reach Home Base in T minus two minutes.

And we are good to go in T minus 10 seconds.

So it's 3 s times the Laplace transform of t to the n minus 1, so t squared.

So e to the minus sc f of c times my delta function t minus c dt.

That is equal to c1e to the minus st, f of t, plus c2e to the minus st, g of t, and all of that times dt.

The option to override automatic detonation expires in T minus five minutes.

The option to override automatic detonation expires in T minus three minutes.

The option to override automatic detonation expires in T minus one minute.

So this is equal to minus t s, e to the minus st, evaluated from 0 to infinity.

So we can write that x is equal to minus 1 be careful with the colors minus 1, plus minus 1 times t.

I could've put P2 there, just as easily plus t times minus 1, minus 1, 3, where, or such that, t is a member of the real numbers.

This should simplify to t I'll just arbitrarily switch colors. t is equal to 300 va minus what's 100 6?

So you get x plus c there, and then times the function f of t minus c, but we said t minus c is the same thing as x.

And then above c, it's going to be e to the minus st times f of t minus c.

This right here is e to the minus st times f of t.

So if a minus s is less than 0, and this is a negative number, e to the a minus s times well t, where t approaches infinity will be 0.

We took the integral with respect to t. e to the minus sA, right?

So when t equals 0, it becomes e to the minus s times 0.

How would I represent my let's say I want to do t minus 3?

So our function in this case is the unit step function, u sub c of t times f of t minus c dt.