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3 years ago

Need help to bring up my grade.

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

7 0

Answer:

I'll help you

Step-by-step explanation:

just send me the questions clearly please.

You might be interested in

A farm has chickens and cows. All the cows have 4 legs and all the chickens have 2 legs.

Sladkaya [172]

Answer:

oh dude this is such a cool riddle i think the answer is ..... 62

Step-by-step explanation:

if there are 82 legs .... and you said there are 5 cows and cows have four legs so that is gonna be 5x4 which equals to 20 and then we can subtract 20 with 82 and that will give the answer as ..... 62 !!!!!!! yeaaaaaa

make me brainliest or whatever .... i.d.k .... just kidding make me brainliest NOW !!!!!! please

5 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

What's the equation for slope (0,5) and (4,1)

Zielflug [23.3K]

Answer: y = -x+5

Step-by-step explanation:

7 0

3 years ago

Francis surveyed a random sample of 70 students at Franklin High School about their favorite season. Of the students surveyed, 1

Ymorist [56]

Answer:

467 is the correct answer.

(This is not rounded, and should be one of your options as the answer.)

7 0

3 years ago

Read 2 more answers

Kinda need some help:)

Natali [406]

1.
f(x) = x
g(x) = x + 3

For each x value you use in f(x) and g(x), the corresponding y value in g(x) is always 3 more than in f(x). If each y-coordinate is 3 more, that means the graph is shifted 3 units up. The graph of f(x) is shifted 3 units up to create the graph of g(x).

2.
The first term is -0.5.
Then each term goes up by 0.25.

1st term: -0.5
2nd term: -0.5 + 0.25 = -0.5 + 0.25(1)
3rd term: -0.5 + 0.25 + 0.25 = -0.5 + 0.25(2)
4th term: -0.5 + 0.25 + 0.25 + 0.25 = -0.5 + 0.25(3)
nth term: -0.5 + 0.25(n - 1)

-0.5 + 0.25(n - 1) = -0.5 + 0.25n - 0.25 = -0.75 + 0.25n = 0.25n - 0.75

Answer is C.

5 0

3 years ago