The _____ were victims of attempted genocide ?

the lawn order lawnmower factory can produce 12 lawnmowers in 8 hours . How many hours will take the factory to produce 30 lawnm

Radda [10]

1 lawnmower= 8hours/12 =2/3 hour
30 lawnmowers= 30*2/3hour= 20 hours

7 0

3 years ago

Read 2 more answers

QUESTION IN THE PICTURE!

Lelechka [254]

Answer:

32

Step-by-step explanation:

i used my calculator

3 0

3 years ago

Find the slope of the line 7x+4y=10

Fittoniya [83]

Answer:

The slope is -7/4

Step-by-step explanation:

∵ 7x + 4y = 10

∵ y = mx + c ⇒ where m is the slope of the line

∴ Re-arrange the equation

∴ 4y = 10 - 7x ⇒ ÷4

∴ y = 10/4 - (7/4) x

∴ y = 5/2 - (7/4) x

∴ The slope is -7/4

3 0

4 years ago

The system of equation is shown on the graph.

Fudgin [204]

E. (4,4) is the answer

8 0

3 years ago

How to prove tan z is analytic using cauchy-riemann conditions

Basile [38]

A function $f(z)=f(x+iy)=u(x,y)+i v(x,y)$ is analytic if the C-R conditions are satisfied:

$\begin{cases}u_x=v_y\\u_y=-v_x\end{cases}$

We have

$f(z)=\tan z=\tan(x+iy)$

Recall the angle sum formula for tangent:

$\tan(x+iy)=\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}$

Now recall that

$\tan iy=\dfrac{\sin iy}{\cos iy}=\dfrac{-\frac{e^y-e^{-y}}{2i}}{\frac{e^y+e^{-y}}2}=\dfrac{i\sinh y}{\cosh y}=i\tanh y$

So we have

$\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}=\dfrac{\tan x+i\tanh y}{1-i\tan x\tanh y}$
$=\dfrac{(\tan x+i\tanh y)(1+i\tan x\tanh y)}{(1-i\tan x\tanh y)(1+i\tan x\tanh y)}$
$=\dfrac{\tan x+i\tanh y+i\tan^2x-\tan x\tanh^2y}{1+\tan^2x\tanh^2y}$
$=\dfrac{\tan x(1-\tanh^2y)}{1+\tan^2x\tanh^2y}+i\dfrac{\tanh y(1+\tan^2x)}{1+\tan^2x\tanh^2y}$
$=\underbrace{\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}}_{u(x,y)}+i\underbrace{\dfrac{\tanh y\sec^2x}{1+\tan^2x\tanh^2y}}_{v(x,y)}$

We could stop here, but taking derivatives may be messy and would be easier to do if we can write this in terms of sines and cosines.

$u(x,y)=\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}=\dfrac{\frac{\sin x}{\cos x\cosh^2x}}{1+\frac{\sin^2x\sinh^2y}{\cos^2x\cosh^2y}}=\dfrac{\sin x\cos x}{\cos^2x\cosh^2y+\sin^2x\sinh^2y}$
$u(x,y)=\dfrac{\sin2x}{2\cos^2x\cosh^2y+2(1-\cos^2x)(\cosh^2y-1)}=\dfrac{\sin2x}{2\cosh^2y-1+2\cos^2x-1}$
$u(x,y)=\dfrac{\sin2x}{\cos2x+\cosh2y}$

With similar usage of identities, we can find that

$v(x,y)=\dfrac{\sinh2y}{\cos2x+\cosh2y}$

Now we check the C-R conditions.

$u_x=\dfrac{2\cos2x(\cos2x+\cosh2y)-(-2\sin2x)\sin2x}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cos2x\cosh2y}{(\cos2x+\cosh2y)^2}$
$v_y=\dfrac{2\cosh2y(\cos2x+\cosh2y)-(2\sinh2y)\sinh2y}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cosh2y\cos2x}{(\cos2x+\cosh2y)^2}$
$\implies u_x=v_y$

Similarly, you can check that $u_y=-v_x$ , hence the C-R conditions are satisfied, and so $\tan z$ is analytic.

6 0

3 years ago