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If x exceeds y by 1 and y exceeds z by 3 how are x and z related?
Danger108
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.
Wiki User
∙ 14y ago
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Three times the square of a certain positive number exceeds six times the number by nine Find the number?
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
Solve x squared plus 3x minus 1 by completing the square?
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.
What number 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.
What number exceeds its square by the maximum amount?
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.
How do you factor x squared plus x minus one half?
(x^2+x-1/2)= x(x+1)-1/2 [x + (1 - square root of 3)/2][x + (1 + square root of 3)/2] = 0 Check it: x^2 + x/2 + (square root of 3)x)/2 + x/2 + 1/4 + (square root of 3)/4 - (square root of 3)x/2 - (square root of 3)/4 - 3/4 = 0 x^2 + x/2 + x/2 + [(square root of 3)x]/2 - [(square root of 3)x]/2 + (square root of 3)/4 - (square root of 3)/4 + 1/4 - 3/4 = 0 x^2 + x - 2/4 = 0 x^2 + x - 1/2 = 0 How to find this roots: Using the completing the square method: x^2 + x - 1/2 = 0 x^2 + x = 1/2 x^2 + x + 1/4 = 1/2 + 1/4 (x + 1/2)^2 = 3/4 x + 1/2 = (plus & minus)(square root of 3/4) x = -1/2 + (square root of 3)/2 x = - 1/2 - (square root of 3)/2
If x exceeds y by 1 and y exceeds z by 3 how are x and y related?
x exceeds y by one
What kind of questions are related to multiplication?
1 x 1 2 x 2 3 x 3 4 x 4 ............. 1(2)=2
Three times the square of a certain positive number exceeds six times the number by nine. Find the number?
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.
The unit digit of a two digit number exceeds twice the tens by one find the number is the sum of its digits is ten?
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
X squared-4x equals -3 using completing the square?
-1
Three times the square of a certain positive number exceeds six times the number by nine Find the number?
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
Solve x squared plus 3x minus 1 by completing the square?
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.
Simplify by dividing x3 plus 2x2 plus 3x-6 over x-1?
(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
What number 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.
What number exceeds its square by the maximum amount?
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.
What is 54.99 plus 3 x 1 plus 4 x 1 plus 3 x 3 plus 4 x 1 plus 3 x 1 plus 3 x 2 plus 3 x 3 plus 4 x 1 plus 3 x 1 x 3 x 3?
It is 123.99
Who do solve FLT some lines?
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.
