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Nutka1998 [239]
3 years ago
15

Seperate real and imaginary parts tan(2x+i3y)

Engineering
1 answer:
ahrayia [7]3 years ago
3 0

Answer:

First you have to separate real and imaginary parts of Tan(x+iy)=Tan(z)=sin(z)/cos(z)

sinz=sin(x+iy)=sinxcos(iy)+cosxsin(iy)=sinxcoshy-icosx sinhy

cosz=cos(x+iy)=cosxcos(iy)-sinxsin(iy)=cosxcoshy−isinxsinhy

Now if you plug in Tan(z) and simplify (it is easy!) you get

Tan(z)=(sin(2x)+isinh(2y))/(cos(2x)+cosh(2y))= A+iB.

This means that

A=sin(2x)/(cos(2x)+cosh(2y)) and B= sinh(2y)/(cos(2x)+cosh(2y))

Now,

A/B=sin(2x)/sinh(2y)

If any questions, let me know.

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Experimental Design Application Production engineers wish to find the optimal process for etching circuit boards quickly. They c
Veseljchak [2.6K]

Answer:

Hello your question is incomplete attached below is the missing part and answer

options :

Effect A

Effect B

Effect C

Effect D

Effect AB

Effect AC

Effect AD

Effect BC

Effect BD

Effect CD

Answer :

A  = significant

 B  = significant

C  = Non-significant

D  = Non-significant

AB  = Non-significant

AC  = significant

AD  = Non-significant

BC  = Non-significant

 BD  = Non-significant

 CD = Non-significant

Explanation:

The dependent variable here is Time

Effect of A  = significant

Effect of B  = significant

Effect of C  = Non-significant

Effect of D  = Non-significant

Effect of AB  = Non-significant

Effect of AC  = significant

Effect of AD  = Non-significant

Effect of BC  = Non-significant

Effect of BD  = Non-significant

Effect of CD = Non-significant

8 0
3 years ago
Select the correct answer.
MAVERICK [17]

Answer:crane and engine I guess

Explanation:

8 0
3 years ago
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,
babunello [35]

Answer:

Qx = 9.109.10^5 \times 10^{-6} m³/s  

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × 10^6 Pa

viscosity n = 100 Pas

angle = 18°

so  Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA   ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 \times 10^{-3} )²× 9  \times 10^{-3}  × sin18 × cos18

Qd = 94.305 × 10^{-6} m³/s

and

Qb = p × π × D × dc³ × sin²A ÷  12  × n × L    ............2

Qb = 11 × 10^{6} × π × 85 \times 10^{-3}  × ( 9  \times 10^{-3} )³ × sin²18 ÷  12  × 100 × 2

Qb = 85.2 × 10^{-6} m³/s

so here

volume flow rate Qx = Qd - Qb   ..............3

Qx =  94.305 × 10^{-6}  - 85.2 × 10^{-6}  

Qx = 9.109.10^5 \times 10^{-6} m³/s  

8 0
3 years ago
What is the resultant force on one side of a 25cm diameter circular plate standing at the bottom of 3m of pool water?
Tom [10]

Answer:

F=1.47 KN

Explanation:

Given that

Diameter of plate = 25 cm

Height of pool h = 3 m

We know that force can be given as

F= P x A

P=ρ x g x h

Now by putting the values

P=1000 x 10 x 3

P= 30 KPa

A=\dfrac{\pi}{4}\times 0.25^2\ m^2

A=0.049\ m^2

F= 30 x 0.049 KN

F=1.47 KN

So the force on the plate will be 1.47 KN.

4 0
3 years ago
This is it dont anwser this is for my other account
Nezavi [6.7K]

Answer:

thanks for the poiunts

Explanation:

6 0
3 years ago
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