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anyanavicka [17]
2 years ago
15

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

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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

so

\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}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

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

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

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

3 0
3 years ago
Y- 4/5=1<br> what does y equal to?
navik [9.2K]

Answer:

y=1.8

Step-by-step explanation:

first change the fraction into a decimal.

4/5 =0.8

plus the 0.8 to both sides

y=1.8

6 0
3 years ago
Read 2 more answers
If Lisa gets $31.42 an hour, what is the minimum number of hours does she need to work to get $3392 a month.
Leto [7]

Answer:

108

Step-by-step explanation:

All you need to do is take 3392 divided by 31.42 and you'll get 107.9567...

Since Lisa needs AT LEAST 3392 a month, she would have to work 108 hours in order to meet that requirement.

8 0
3 years ago
Read 2 more answers
Solis have 30 cupcake.and she have 12 more of the cupcakes.how many in all?how many in total do solis have
prohojiy [21]
42 you just add 30 and 12
7 0
3 years ago
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