Contest for White Rabbit

[ivo] ivo
Joined 18/03/2017
Posts 2,566
03:41 PM 23/03/2019
The first person to solve the equation, showing the steps they used (which cannot including help from a non-human), wins:

Find ∂²z/∂x∂y of z = 4x² −xy +y² −x³

Good luck.


alfons
Joined 20/01/2019
Posts 5,065
03:42 PM 23/03/2019
unanswerable


gooqbye
Joined 25/02/2019
Posts 23,305
03:43 PM 23/03/2019
I didn’t learn thzt yet


Kaiokeno
Joined 02/11/2017
Posts 5,819
03:43 PM 23/03/2019
1.


Winar
Joined 12/08/2018
Posts 7,230
03:43 PM 23/03/2019
Quote from alfons , 03:42 PM 23/03/2019
unanswerable
i think alfons won
also, 0


Voodoo
Joined 14/02/2019
Posts 1,845
03:43 PM 23/03/2019
Given equations are,

2x² - xy - y² + 3x + 2y - 3 = 0. ...(1)

x² - 3x + 2 = 0. ...(2)

From equation (2),

x² - 3x + 2 = 0
or, x² - 2x - x + 2 = 0
or,. ( x - 2) ( x - 1 ) = 0

So, values of x are

x = 1, 2.

On putting the values of x in equation(1), we get

For x = 1, the value of y is

2(1)² - (1)y - y² + 3(1) + 2y - 3 = 0

- y² + y + 2 = 0
or,
y² - y - 2 = 0

y = (1±3)/2

On taking +ve sign,

y = 2

On -ve sign,

y = -1.

And for x = 2, the value of y is

2(4) - 2y - y² + 3(2) + 2y - 3 = 0

8 - y² + 6 - 3 = 0
or, y² = 11
or, y = ±√11

So, the values of x&y are

x = 1. y = -1, 2.
x = 2. y = ±√11

How i did it: searched it on quora


[heck] Afonso64
Joined 06/07/2018
Posts 3,930
03:44 PM 23/03/2019
Yo @Executive i have the Answer:



A-Bruh are we in school


[ivo] ivo
Joined 18/03/2017
Posts 2,566
03:44 PM 23/03/2019
Quote from Voodoo , 03:43 PM 23/03/2019
Given equations are,

2x² - xy - y² + 3x + 2y - 3 = 0. ...(1)

x² - 3x + 2 = 0. ...(2)

From equation (2),

x² - 3x + 2 = 0
or, x² - 2x - x + 2 = 0
or,. ( x - 2) ( x - 1 ) = 0

So, values of x are

x = 1, 2.

On putting the values of x in equation(1), we get

For x = 1, the value of y is

2(1)² - (1)y - y² + 3(1) + 2y - 3 = 0

- y² + y + 2 = 0
or,
y² - y - 2 = 0

y = (1±3)/2

On taking +ve sign,

y = 2

On -ve sign,

y = -1.

And for x = 2, the value of y is

2(4) - 2y - y² + 3(2) + 2y - 3 = 0

8 - y² + 6 - 3 = 0
or, y² = 11
or, y = ±√11

So, the values of x&y are

x = 1. y = -1, 2.
x = 2. y = ±√11

How i did it: searched it on quora
Hint: you're not mean to find the values of x, y or z.


alfons
Joined 20/01/2019
Posts 5,065
03:44 PM 23/03/2019
i did with a human

Given equations are,

2x² - xy - y² + 3x + 2y - 3 = 0. ...(1)

x² - 3x + 2 = 0. ...(2)

From equation (2),

x² - 3x + 2 = 0
or, x² - 2x - x + 2 = 0
or,. ( x - 2) ( x - 1 ) = 0

So, values of x are

x = 1, 2.

On putting the values of x in equation(1), we get

For x = 1, the value of y is

2(1)² - (1)y - y² + 3(1) + 2y - 3 = 0

- y² + y + 2 = 0
or,
y² - y - 2 = 0

y = (1±3)/2

On taking +ve sign,

y = 2

On -ve sign,

y = -1.

And for x = 2, the value of y is

2(4) - 2y - y² + 3(2) + 2y - 3 = 0

8 - y² + 6 - 3 = 0
or, y² = 11
or, y = ±√11

So, the values of x&y are

x = 1. y = -1, 2.
x = 2. y = ±√11

How i did it: copied


Pres. Jordan
Joined 02/04/2018
Posts 3,042
03:44 PM 23/03/2019
2, 3, 1


Winar
Joined 12/08/2018
Posts 7,230
03:45 PM 23/03/2019
Quote from ivo , 03:44 PM 23/03/2019
Quote from Voodoo , 03:43 PM 23/03/2019
Given equations are,

2x² - xy - y² + 3x + 2y - 3 = 0. ...(1)

x² - 3x + 2 = 0. ...(2)

From equation (2),

x² - 3x + 2 = 0
or, x² - 2x - x + 2 = 0
or,. ( x - 2) ( x - 1 ) = 0

So, values of x are

x = 1, 2.

On putting the values of x in equation(1), we get

For x = 1, the value of y is

2(1)² - (1)y - y² + 3(1) + 2y - 3 = 0

- y² + y + 2 = 0
or,
y² - y - 2 = 0

y = (1±3)/2

On taking +ve sign,

y = 2

On -ve sign,

y = -1.

And for x = 2, the value of y is

2(4) - 2y - y² + 3(2) + 2y - 3 = 0

8 - y² + 6 - 3 = 0
or, y² = 11
or, y = ±√11

So, the values of x&y are

x = 1. y = -1, 2.
x = 2. y = ±√11

How i did it: searched it on quora

Hint: you're not mean to find the values of x, y or z.
answer is nothing

1 2 3 4 5 6