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3) 2 x ( b 3) 2 x ( c 3) 8 x ( d 3 9 x 11 The simplified form of 12 7 6 2 2 x x x x is a 2 b 2 c 4 d 4 4 x x 4 x x 2 x x 2 x x 12 A linear equation in the variable x is an equation that can be written in the form 0 b ax, where b is constant and a should not be equal to a 0 b 1 c 1 d ∞Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!
1/x 1 2/x 2 4/x 4 solve for x-1
1/x 1 2/x 2 4/x 4 solve for x-1-Notice how5 solves the associated equality For 12 (be aware here that 1 and 2 are not solutions, since they yield a value of 1 on the LHS), we reduce to 2x3>=4, which becomes x>= 35 So Ex 42, 12 By using properties of determinants, show that 8(1&x&x2@x2&1&x@x&x2&1) = (1 – x3)2 Solving LHS 8(1&x&x2@x2&1&x@x&x2&1) Applying R1 → R1 R2 R3 = 8(𝟏𝐱𝟐𝐱&𝐱𝟏𝐱𝟐&𝐱𝟐𝐱𝟏@x2&1&x@x&x2&1) Taking (1 x x2) Common from 1st row = (𝟏𝐱𝐱𝟐) 8(1&1&1@x2&1&x@x&x2
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Transcript Misc 8 Solve the inequalities and represent the solution graphically on number line 2(x – 1) < x 5, 3(x 2) > 2 – x Solving 2(x – 1) < x 5 2x – 2 < x 5 2x – x < 5 2 x < 7 Solving 3(x 2) > 2 – x 3x 6 > 2 – x 3x x > 2 – 6 4x > – 4 x > (−4)/4 x > −1 Thus, x < 7 & x > – 1 The graphical representation is Here both 7 & 1 are not included in the4 x ( x 2), 1 4 x ( x 2), 1 Since 4 x ( x 2), 1 4 x ( x 2), 1 contain both numbers and variables, there are four steps to find the LCM Find LCM for the numeric, variable, and compound variable parts Then, multiply them all together Steps to find the LCM for 4 x ( x 2), 1 4 x ( x 2), 1 are 1X^2 WolframAlpha Volume of a cylinder?
Ln (x2)ln (x1)=1 \square! Ex 75, 11 Integrate the function 5𝑥/((𝑥 1) (𝑥2− 4) ) We can write the integrand as 5𝑥/((𝑥 1) (𝑥2− 4) ) = 5𝑥/((𝑥 1) (𝑥 − 24(x1)=2x4 We simplify the equation to the form, which is simple to understand 4(x1)=2x4 Reorder the terms in parentheses (4x4)=2x4 Remove unnecessary parentheses 4x4=2x4 We move all terms containing x to the left and all other terms to the right 4x2x=44 We simplify left and right side of the equation 2x=0 We divide both sides
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Solve for x cot^1(1/(x^2 1)) tan^1(x^2 1) = π/2 asked in Sets, relations and functions by Raghab ( 505k points) inverse trigonometric functionsX ≠ 0, 1,2 asked Apr 30 in Quadratic Equations by Eeshta ( 323k points) quadratic equations
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