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artcher [175]
3 years ago
10

This is the question I don't understand​

Mathematics
1 answer:
Jet001 [13]3 years ago
6 0

Answer:

D

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=f%28x%29%3De%5E%7B3x%7D%20.sinx" id="TexFormula1" title="f(x)=e^{3x} .sinx" alt="f(x)=e^{3x} .
Maslowich

Answer:

6

Step-by-step explanation:

đạo hàm cấp 2 của f(x) rồi thế 0 vào

8 0
2 years ago
Help me please please please​
julia-pushkina [17]

Answer:

702.1

Step-by-step explanation:

Use the formula for the diagonal of a cuboid.

√(l^2+b^2+h^2)

√(33^2+56^2+33^2)

√5314

=  702.125345

3 0
3 years ago
Read 2 more answers
These figures are congruent. What is QR? <br><br> If you put a link I WILL be reporting you!
Westkost [7]
46
I believe this is correct, if not lmk!
5 0
3 years ago
Find the surface area of the right rectangular shown below hurry plss
laila [671]

Surface area of a cuboid

= [2(length)(width)] + [2(width)(height)] + [2(height)(length)]

= {[2(4)(2.5)] + [2(2.5)(2)] + [2(2)(4)]} units²

= (20 + 10 + 16) units²

= 46 units²

7 0
2 years ago
Read 2 more answers
(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul
Aloiza [94]

Answer

(a) F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

Step-by-step explanation:

(a) f(t) = 20.5 + 10t + t^2 + δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 f(s)e^{-st} \, dt

where a = ∞

=>  F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt

where d(t) = δ(t)

=> F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt

Integrating, we have:

=> F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3}  )\left \{ {{a} \atop {0}} \right.

Inputting the boundary conditions t = a = ∞, t = 0:

F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) f(t) = e^{-t} + 4e^{-4t} + te^{-3t}

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt

F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt

F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt

Integrating, we have:

F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.

Inputting the boundary condition, t = a = ∞, t = 0:

F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

3 0
3 years ago
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