Selina Concise Mathematics Class 8 Icse Solutions Chapter 12 Algebraic Identities Cbse Tuts
Expand (i) (3x – 4y 5z)2 (ii) (2a – 5b – 4c)2 (iii) (5x 3y)3 (iv) (6a – 7b)33 Examples of expanding brackets Example1 Expand 3(x2) The 3 outside must multiply both terms inside the brackets 3(x2) = 3x6 Example2 Expand x(x− y)
Expand 3 x 2 y 7 square
Expand 3 x 2 y 7 square- Anyway, I mean the row that starts 1, 7 Each number is formed by adding the two numbers above to the left and right The last row gives us the coefficients we need (xy)^7 = x^77x^6y21x^5y^235x^4y^335x^3y^421x^2y^57xy^6y^7Precalculus The Binomial Theorem Pascal's Triangle and Binomial Expansion 1 Answer
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Algebra1helpcom contains invaluable info on bracket expansion calculator, syllabus for intermediate algebra and decimals and other math subject areas Any time you require guidance on factor or functions, Algebra1helpcom is without question the best site to go to!3 (x 2 y 2) = 3 × 113/2 = 339/2 = 169 ½ (iii) We know that 5 (x 2 y 2) 4 (x – y) = 5 × 113/2 4 × ± 7 By further calculation = 565/2 ± 28 We can write it as = 565/2 28 or 565/2 – 28 = 621/2 or 509/2 It can be written as = 310 ½ or 254 ½ 12 If x 2 y 2 = 34 and xy = 10 ½, find the value of 2 (x y) 2 (x – y) 2 Solution It is given thatSolution Apply the square of the difference formula in numerator 9 x2 6 x 1 (3 x 1) = (3 x 1) 2 (3 x 1) = 3 x 1 Note that using square of the difference formula is easily find the squares of large numbers without using a calculator or multiplication in a column Example 4 Apply the square of the difference formula to calculate
Answer (1 of 10) Generate formulas for x²y² (a) (x y)² = x² y² 2xy x² y² = (x y)² 2xy (1) (b) (x y)² = x² y² 2xy x² y²\begin{aligned} 4(xy) x(x2) &= 4x 4y x^2 2x \\ &= x^2 6x 4y\ _\square \end{aligned} 4 (x y) x (x 2) = 4 x 4 y x 2 2 x = x 2 6 x 4 y 95 97 99 101 What is the coefficient of x x x after you simplify the expressionExpanding Logarithms When you are asked to expand log expressions, your goal is to express a single logarithmic expression into many individual parts or componentsThis process is the exact opposite of condensing logarithms because you compress a bunch of log expressions into a simpler one The best way to illustrate this concept is to show a lot of examples
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