x + 1 = y
y + 3 = z
z
= y + 3
= (x + 1) + 3
= x + 4
Or:
x
= y - 1
= (z - 3) - 1
= z - 4
Which results in the same:
x exceeds z by 4.
The problem can be written as3x2 = 6x + 9x is the number we want to findSolution:x2 = 2x + 3x2 - 2x - 3 = 0(x-3)(x+1) = 0x = -1, 3
x² + 3x - 1 = [ x² + 3x + (3/2)²] - 1 - (3/2)² = (x + 3/2)² - 1 - (9/4) = (x + 3/2)² - (13/4) if you are still confused, i want you to follow the related link that explains the concept of completing the square clearly.
If we were to graph the number it would be: y = x If we were to graph the square it would be: y = x² The difference would be: f(x) = x - x² You want to maximize this difference, so take the derivative: f'(x) = 1 - 2x Then set it to zero: 0 = 1 - 2x Add 2x to both sides: 2x = 1 Divide both sides by 2: x = ½ Answer: ½ is the number that most exceeds its square.
If we were to graph the number it would be: y = x If we were to graph the square it would be: y = x² The difference would be: f(x) = x - x² You want to maximize this difference, so take the derivative: f'(x) = 1 - 2x Then set it to zero: 0 = 1 - 2x Add 2x to both sides: 2x = 1 Divide both sides by 2: x = ½ Answer: ½ is the number that most exceeds its square.
The derivative of x divided by 3 is 1/3. This can be found using the power rule of differentiation, where the derivative of x^n is nx^(n-1). In this case, x can be written as x^1, so the derivative is 1(1/3)*x^(1-1) = 1/3.
x exceeds y by one
1 x 1 2 x 2 3 x 3 4 x 4 ............. 1(2)=2
3x2 - 9 = 6x 3x2 - 6x - 9 = 0 3(x + 1)(x - 3) x = -1, 3 The answer is 3. Check it. 27 - 9 = 18 It checks.
let x- unit digit 2x + 1- twice the tens by one x + 2x +1 = 10 x + 2x = 10 - 9 3x= 9 x = 3 therefore : x = 3 2x + 1 = 7 the number is 37. you can try checking it: x + 2x + 1 = 10 3 + 2(3) = 10-1 3 + 6 = 9 9 = 9
-1
The problem can be written as3x2 = 6x + 9x is the number we want to findSolution:x2 = 2x + 3x2 - 2x - 3 = 0(x-3)(x+1) = 0x = -1, 3
x² + 3x - 1 = [ x² + 3x + (3/2)²] - 1 - (3/2)² = (x + 3/2)² - 1 - (9/4) = (x + 3/2)² - (13/4) if you are still confused, i want you to follow the related link that explains the concept of completing the square clearly.
(x^3 + 2x^2 + 3x - 6)/(x - 1) add and subtract x^2, and write -6 as (- 3) + (-3) = (x^3 - x^2 + x^2 + 2x^2 - 3 + 3x - 3)/(x - 1) = [(x^3 - x^2) + (3x^2 - 3) + (3x - 3)]/(x - 1) = [x^2(x - 1) + 3(x^2 - 1) + 3(x - 1)]/(x - 1) = [x^2(x - 1) + 3(x - 1)(x + 1) + 3(x - 1)]/(x - 1) = [(x - 1)(x^2 + 3x + 3 + 3)]/(x - 1) = x^2 + 3x + 6
It is 123.99
If we were to graph the number it would be: y = x If we were to graph the square it would be: y = x² The difference would be: f(x) = x - x² You want to maximize this difference, so take the derivative: f'(x) = 1 - 2x Then set it to zero: 0 = 1 - 2x Add 2x to both sides: 2x = 1 Divide both sides by 2: x = ½ Answer: ½ is the number that most exceeds its square.
If we were to graph the number it would be: y = x If we were to graph the square it would be: y = x² The difference would be: f(x) = x - x² You want to maximize this difference, so take the derivative: f'(x) = 1 - 2x Then set it to zero: 0 = 1 - 2x Add 2x to both sides: 2x = 1 Divide both sides by 2: x = ½ Answer: ½ is the number that most exceeds its square.
I like back into multidimensional space where were born to be sleeping forever thousands autumn. Pierre De Fermat's last theorem. The conditions. x,y,z,n are the integers >0 and n>2. z^n=/x^n+y^n. Assumptions z^3=x^3+y^3. Therefore z=(x^3+y^3)^1/3. I define . F(x,y)=(x^3+y^3}^1/3 - [ (x-x-1)^3+(y-x-1)^3]^1/3. Therefore [z-F(x,y)]^3={ (x^3+y^3)^1/3 - (x^3+y^3}^1/3 + [ (x-x-1)^3+(y-x-1)^3] ^1/3 }^3={ [(x-x-1)^3+(y-x-1)^3]^1/3 }^3 =(x-x-1)^3+(y-x-1)^3= (y-x-1)^3-1. Because [z-F(x,y)]^3=(y-x-1)^3-1 . Attention [(y-x-1)^3-1 ] is an integer , [(y-x-1)^3-1 ]^1/3 is an irrational number therefore [(y-x-1)^3-1 ]^2/3 is an irrational number too. Example (2^3-1) is an integer , (2^3-1)^1/3 is an irrational number and (2^3-1)^2/3 is an irrational number too. Because z-F(x,y)=[y-x-1)^3-1]^1/3. Therefore z=F(x,y)+[(y-x-1)^3-1] ^1/3. Therefore z^3=[F(x,y)]^3.+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+3F(x,y)*[(y-x-1)-1]^2/3+[(y-x-1)^3-1]. Therefore 3F(x,y)*[(y-x-1)^3-1]^2/3+[F(x,y)]^3+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3 =0 Because. z-F(x,y)=[ (y-x-1)^3-1]^1/3. Therefore F(x,y)=z - [(y-x-1)^3-1]^1/3. Therefore 3F(x,y)*[(y-x-1)^3-1]^2/3 =3z*[(y-x-1)^3-1]^2/3 -3[(y-x-1)^3-1]. Therefore. 3z*[(y-x-1)^3-1]^2/3 - 3[(y-x-1)^3-1]+[F(x,y)]^3+3[F(x,y)]^2*[y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3=0.. Named [F(x,y)]^3+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3= W. We have 3z*[(y-x-1)^3-1]^2/3 is an irrational number because z is an integer and had proved [(y-x-1)^3-1]^2/3 is an irrational number. And 3[(y-x-1)^3-1] is an integer because x,y are the integers. And 3z*[(y-x-1)^3-1]^2/3 - 3[(y-x-1)^3-1]+[F(x,y)]^3+W=0. Therefore an irrational number - an integer+W=0. Therefore W is an complex irrational number. Named 3z*[(y-x-1)^3-1]^2/3 is 3z*B And Named 3[(y-x-1)^3-1] is C . Therefore 3z*B - C +W=0. Therefore 3z*B=C-W. Because z is an integer. B is an irrational number=[(y-x-1)^3-1]^2/3 Attention (an integer)^2/3 and (an integer)^2/3 is an irrational number. W is an complex irrational number. C is an integer. Therefore. an integer*an irrational number=an complex irrational number + an integer. Unreasonable. Therefore. z^3=/x^3+y^3 Similar z^n=/x^n+y^n. ISHTAR.