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If x exceeds y by 1 and y exceeds z by 3 how are x and y related?
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∙ 14y ago
x exceeds y by one
Wiki User
∙ 14y ago
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X squared-4x equals -3 using completing the square?
-1
What is 1 over x plus 3 plus 1 divided by 1 over x-3?
There is some ambiguity in the question because of the absence of brackets. For example, is it 1/x + 3 ... or 1/(x+3) ... ? So, this is my interpretation of the question. If it is not what you meant please resubmit using brackets. 1/(x+3) + 1/1/(x-3) = 1/(x+3) + (x - 3) = (1 + x2 - 9)/(x + 3) = (x2 - 8)/(x + 3)
What is the third to the fourth power?
1/3 x 1/3 x 1/3 x 1/3 =1/81
What is one third percent of 1500?
1/3% = (1/3) x 1% = 1/3 x 0.011/3 x 0.01 x 1,500 = 1/3 x 15 = 5
What is x2 plus 3x plus x plus 3?
X2+3x+x+3=x(x+1)+3(x+1)=(x+3)(x+1)
If x exceeds y by 1 and y exceeds z by 3 how are x and z related?
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.
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.
