WebAug 16, 2016 · Explanation: Using synthetic division and the fact that x = − 1 is obviously a solution we find that we can expand this to: (x +1)(x2 −x +1) = 0. In order to have LHS=RHS need one of the brackets to be equal to zero, ie. (x +1) = 0 1. (x2 − x + 1) = 0 2. From 1 we note that x = −1 is a solution. WebAlgebra. Solve for x x^3-1=0. x3 − 1 = 0 x 3 - 1 = 0. Add 1 1 to both sides of the equation. x3 = 1 x 3 = 1. Subtract 1 1 from both sides of the equation. x3 − 1 = 0 x 3 - 1 = 0. Factor the …

Solved Solve the boundary-value problem \[ y^{\prime - Chegg

Webx 2( 3+1)x+ 3=0x 2− 3x−x+ 3=0x(x− 3)−1(x− 3)=0(x− 3)(x−1)=0⇒x=1 or x= 3∴ roots are (1, 3) Webadditional constraints: does adding 1 or 2 below complicate the problem? 1. no more than half of total power is in any 10 lamps 2. no more than half of the lamps are on (pj > 0) • answer: with (1), still easy to solve; with (2), extremely difficult • moral: (untrained) intuition doesn’t always work; without the proper how to donate a timeshare unit to charity

How do you solve #(x-1)(x-2)(x-3)>=0#? - Socratic.org

WebAug 22, 2016 · Explanation: f (x) = (x −1)(x − 2)(x − 3) = x3 − 6x2 + 11x − 6. is a cubic with positive leading coefficient and zeros of multiplicity 1 at x = 1, x = 2 and x = 3. As a result: f (x) is continuous. f (x) is positive for large positive values of x. f (x) is negative for large negative values of x. f (x) changes sign at each of its zeros. Web6sin (π/2⋅x) = 3. sin (π/2⋅x) = 1/2. From the unit circle, the sine values that equal 1/2 have the solution set x = π/6+2πn, 5π/6+2πn. π/2⋅x = π/6+2πn, 5π/6+2πn. x = 1/3+4n, 5/3+4n. … WebSolve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. learn to read and write spanish free