Answers and the work for them please

Mathematics

1 answer:

Anon25 [30]2 years ago

3 0

Answer:

The screen is black? there is nothing there

Step-by-step explanation:

this is nonsense man

You might be interested in

How do I add and subtract radicals with no like terms

liq [111]

Most real number arithmetic is pretty vacuous

$\sqrt 2 + \sqrt 3 = \sqrt{2} + \sqrt{3}$

That's about as good a way as possible to write this particular real number. But it's far from the only way.

$(\sqrt 2 + \sqrt 3)^2 = \sqrt{2}^2 + 2\sqrt 2 \sqrt 3 + \sqrt{3}^2 = 5 + 2 \sqrt 6$

$\sqrt 2 + \sqrt 3 = \sqrt{5 + 2 \sqrt 6}$

Sometimes you can factor something out and you have a common radical:

$\sqrt{32} + \sqrt{128} = \sqrt{2^5} + \sqrt{2^7} = 4 \sqrt{2} + 8 \sqrt{2} = 12 \sqrt{2}$

But most of them are vacuous,

$3\sqrt{7} + 34 \sqrt{13} = 3\sqrt{7} + 34 \sqrt{13}$

8 0

3 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$