All categories

3 years ago

Envision geometry 2018 question 1.3.9 answer

Mathematics

1 answer:

kozerog [31]3 years ago

8 0

Answer:

Jajajja

Step-by-step explanation:

Nejsjs

You might be interested in

Find the measure of angle JCA if J is the incenter of the triangle ABC

melamori03 [73]

It would be two times the angle ABC

7 0

3 years ago

A. 20<br> b. 5<br> c. 40<br> d. 45

Rina8888 [55]

Answer:

C, 40

Step-by-step explanation:

You can set up the equation as 20+4x=180

180 is the total angle of the line

4x=160 and divide each side by 4

x=40

6 0

2 years ago

Read 2 more answers

AYO CAN YOU BE MY BAKA (HELP ME WITH MATH)

nikitadnepr [17]

Answer:

Addition

Step-by-step explanation:

1) Add 275 to both sides.

$800 + 275 = p$

2) Simplify 800 + 275 to 1075.

$1075 = p$

3) Switch sides.

$p = 1075$

We need to add in order to solve the equation.

so therefor, the answer is option G.

5 0

2 years ago

Read 2 more answers

The Flushing River has a current of 3mph. A motorboat takes as long to go 12 miles downstream as to go 8 miles upstream. What is

Nata [24]

Answer: 1.3

Step-by-step explanation:

12-8 = 4

4/3 = 1.3

3 0

2 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