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What is 1 1 2 x 2 3?

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Ammarali9212fb0248

Lvl 1
∙ 10y ago
Updated: 10/17/2024

If you mean y+3 = 2(x-1) then y = 2x-5

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Madie Volkman

Lvl 9
∙ 2y ago

What else can I help you with?

Related Questions

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

Solve simultaneously y equals 2-x and x2 plus 2xy equals 3?

if y = 2- x then x^2 + 2x(2 - x) = 3 ie x^2 + 4 x -2x^2 = 3 ie 4x - 3 = x^2 x^2 - 4x +3 = 0 = (x - 1)(x - 3) so x = 1 or 3 Check: y= 2 - 1 =1; x^2 = 1, 2xy = 2 x 1 x 1 = 2, 1 +2 = 3. y = 2 - 3 = -1; x^2 = 9, 2xy = 2 x 3 x -1 = -6, 9 + -6 = 3. QED.

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

Differentiate 3 Sq root x?

y = 3√x y = 3x^(1/2) y' = 3(1/2)x^(1/2 -1) y'= (3/2)x^(-1/2) y' = 3/[2x^(1/2)] y' = 3/(2√x)

What is the solution of x2 plus x plus 1 equals 0?

x2 + x + 1 = 0 ∴ x2 + x + 1/4 = -3/4 ∴ (x + 1/2)2 = -3/4 ∴ x + 1/2 = ± √(-3/4) ∴ x = - 1/2 ± (i√3) / 2 ∴ x = (-1 ± i√3) / 2

Does fermat's last Theorem transformer a monster?

Attention: Only (N+). Fermat's last Theorem z ^ 3 = x ^ 3 + y ^ 3 is capable exists a solution if fully meet the following conditions: First step: (1+2+3+4+........+a)^2+(1+2+3+4+........+b)^2=v^2. In fact, using the computer, this equation has the ability to survive. Second step: (1+2+3+4+........+a+1)^2+(1+2+3+4+........+b+1)^2=s^2. Third step: v=1+2+3+4+........+c. In fact, using the computer, this equation has the ability to survive. Fourth step: s=1+2+3+4+........d. Fifth step: d=c+1 If all five steps are satisfied.This equation is capable of existence. [z(z+1)/2]^2 - [z(z-1)/2]^2=[x(x+1)/2]^2- [x(x-1)/2]^2+[y(y+1)/2]^2 - [y(y-1)/2]^2. Because: z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2. Mean this equation is capable of existence. z^3=x^3+y^3. However, too hard to satify all five equations in same time.. And more: Attention about series of number: 1,3,6,10,15,21,28,36,45........ Recognize: 10 and 15 are two number consecutive which belong this string. Having: 15^2 - 10^2=5^3. Or: z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2. Impossible in same time exist both: [z(z+1)/2]^2=[x(x+1)/2]^2+[y(y+1)/2]^2 And [z(z-1)/2]^2=[x(x-1)/2]^2+[y(y-1)/2]^2 Attention: All numbers as z(z+1)/2 and x(x+1)/2 and y(y+1)/2 and z(z-1)/2 and x(x-1)/2 and y(y-1)/2 are belong this string and they are Pythagorean This is main proof: z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2 Define: x<x+a<y. x^3+y^3=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+a-1)^3+(x+a)^3+(x+a+1)^3+........+(y-1)^3] Suppose: z^3=x^3+y^3. Because also: (x+a)^3= [(x+a)(x+a+1)/2]^2 - [(x+a)(x+a-1)/2]^2. Therefore a system of equations is generated. [z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+a-1)^3+(x+a+1)^3+........+(y-1)^3] [z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+b)(x+b+1)/2]^2 + [(x+b)(x+b-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+b-1)^3+(x+b+1)^3+........+(y-1)^3] [z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+c)(x+c+1)/2]^2 + [(x+c)(x+c-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+c-1)^3+(x+c+1)^3+........+(y-1)^3] [z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+d)(x+d+1)/2]^2 + [(x+d)(x+d-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+d-1)^3+(x+d+1)^3+........+(y-1)^3]. ........ Can not count the number of equations And more; Using the two formulas by rotation affect each other: z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2 And [z(z+1)/2]^2=1^3+2^3+........+z^3. This method makes root equation z^3=x^3+y^3 is structured as a Transfiguration equation which having unlimited formats as Robot Bumblebee Transformer. Similar,this method is used for general case: z ^ n = x ^ n + y ^ n. Mean: z ^ (n-3) ​​* z ^ 3 = x ^ (n-3) ​​* x ^ 3 + y ^ (n-3) ​​* y^ 3.  You can structure this equation according to your own discretion when use z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2 then use [z(z+1)/2]^2=1^3+2^3+........+z^3 then continue by same to become your own format. ADIEU.

Why -1-1 equals 1?

-1*-1 (-2+1)(-2+1) -2*-2+1*-2+1*-2+1*1 since multiplication is repeated addition then -1*-1*4+(-2)+(-2)+(1) let -1*-1=x -1*-1=-1*-1*4-2-2+1 x=x*4-4+1 x=4x-3 3=4x-x 3=3x 3/3=x 1=x thus -1*-1=x=1

What is 1 0ver 2 x 2 over 3?

1/2 x 2/3 = 1/3

What is 2 times negative 2 minus 3 times 3 plus negative 1?

2 x -2 - 3 x 3 + -1 = (2 x -2) - (3 x 3) + (-1) = -4 - 9 - 1 = -14.

Factor x3 1?

x^3 + 1 = (x + 1)(x^2 - x + 1) x^3 - 1 = (x - 1)(x^2 + x + 1)

Is Fermat's last Theorem a equation metamorphosis?

Attention: Only (N+). Fermat's last Theorem z ^ 3 = x ^ 3 + y ^ 3 is capable exists a solution if fully meet the following conditions: First step: (1+2+3+4+........+a)^2+(1+2+3+4+........+b)^2=v^2. In fact, using the computer, this equation has the ability to survive. Second step: (1+2+3+4+........+a+1)^2+(1+2+3+4+........+b+1)^2=s^2. Third step: v=1+2+3+4+........+c. In fact, using the computer, third step and first step have the ability to survive in same time. Fourth step: s=1+2+3+4+........d. Fifth step: d=c+1 If all five steps are satisfied.This equation is capable of existence. [z(z+1)/2]^2 - [z(z-1)/2]^2=[x(x+1)/2]^2- [x(x-1)/2]^2+[y(y+1)/2]^2 - [y(y-1)/2]^2. Because: z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2. Mean this equation is capable of existence. z^3=x^3+y^3. However, too hard to satisfy all five conditions in same time.. And an other solution: Attention about series of number: 1,3,6,10,15,21,28,36,45........ Recognize: 10 and 15 are two number consecutive which belong this string. Having: 15^2 - 10^2=5^3. Or: z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2. Impossible in same time exist both: [z(z+1)/2]^2=[x(x+1)/2]^2+[y(y+1)/2]^2 And [z(z-1)/2]^2=[x(x-1)/2]^2+[y(y-1)/2]^2 Attention: All numbers as z(z+1)/2 and x(x+1)/2 and y(y+1)/2 and z(z-1)/2 and x(x-1)/2 and y(y-1)/2 are belong this string and they are Pythagorean This is main proof: z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2 Define: x<x+a<y. x^3+y^3=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+a-1)^3+(x+a)^3+(x+a+1)^3+........+(y-1)^3] Suppose: z^3=x^3+y^3. Because also: (x+a)^3= [(x+a)(x+a+1)/2]^2 - [(x+a)(x+a-1)/2]^2. Therefore a system of equations is generated. [z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+a-1)^3+(x+a+1)^3+........+(y-1)^3] [z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+b)(x+b+1)/2]^2 + [(x+b)(x+b-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+b-1)^3+(x+b+1)^3+........+(y-1)^3] [z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+c)(x+c+1)/2]^2 + [(x+c)(x+c-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+c-1)^3+(x+c+1)^3+........+(y-1)^3] [z(z+1)/2]^2 - [z(z-1)/2]^2=[y(y+1)/2]^2 - [x(x-1)/2]^2 - [(x+d)(x+d+1)/2]^2 + [(x+d)(x+d-1)/2]^2 - [(x+1)^3+(x+2)^3+........+(x+d-1)^3+(x+d+1)^3+........+(y-1)^3]. ........ Can not count the number of equations because number (a) can change to infinity. And finally a great equation metamorphosis: Using the two formulas by rotation affect each other: z^3=[z(z+1)/2]^2 - [z(z-1)/2]^2 And [z(z+1)/2]^2=1^3+2^3+........+z^3. This method makes the original equation z ^ 3 = x ^ 3 + y ^ 3 is structured like a Robot Bumblebee Transformer. Similar,this method is used for general case: z ^ n = x ^ n + y ^ n. Mean: z ^ (n-3) ​​* z ^ 3 = x ^ (n-3) ​​* x ^ 3 + y ^ (n-3) ​​* y^ 3.  You can structure this equation according to your own discretion. ADIEU.

How do you factor x square plus one?

(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

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