Publicidad
Publicidad

MΓ‘s contenido relacionado

Publicidad

Math 8 Enhancement Activity.docx

  1. Directions: Write the Rational Algebraic Expressions in simplest form. Given Solution Scratch 1.) 24π‘š2𝑝2 36π‘šπ‘2 = πŸπŸπ’Žπ’‘πŸ(2π‘š) πŸπŸπ’Žπ’‘πŸ(3) Numerator 24π‘š2 𝑝3 = (2) (2) (2) (3) (m) (m) (p) (p) Denominator 36π‘š 𝑝3 = (2) (2) (3) (3) (m) (p) (P) GCMF = (2) (2) (3) (m) (p) (p) = πŸπŸπ’Žπ’‘πŸ = πŸπŸπ’Žπ’‘πŸ(2π‘š) πŸπŸπ’Žπ’‘πŸ(3) Simplified Form = 2π‘š 3 2.) 4𝑏𝑝2+6𝑏2𝑝 6𝑝2+2𝑝 = πŸπ’ƒπ’‘(πŸπ’‘ +πŸ‘π’ƒ) πŸπ’‘(πŸ‘π’‘ +𝟏) Numerator 4𝑏𝑝2 = (2) (2) b (p) (p) 6𝑏2 𝑝 = (2) (3) (b) (b) (p) GCMF = (2) (b) (p) = πŸπ’ƒπ’‘ Factored Form of the Numerator = πŸπ’ƒπ’‘(2p +3b) = πŸπ’ƒπ’‘(πŸπ’‘ +πŸ‘π’ƒ) πŸπ’‘(πŸ‘π’‘ +𝟏) Simplified Form = 𝒃(πŸπ’‘ +πŸ‘π’ƒ) (πŸ‘π’‘ +𝟏) Denominator 6𝑝2 = (2) (3) (p) (p) 2𝑝 = (2) (p) GCMF = (2) (p) = πŸπ’‘ Factored Form of the Numerator = πŸπ’‘(3p +1)
  2. Given Solution Scratch 3.) 4π‘Ÿ2βˆ’ 9𝑐2 8π‘Ÿ3+ 27𝑐3 = (πŸπ’“+πŸ‘π’„)(2π‘Ÿβˆ’3𝑐) (2π‘Ÿ+3𝑐 )( 4π‘Ÿ2 – 6π‘Ÿπ‘+9𝑐2) Numerator Difference of Two Squares Formula π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏) 4π‘Ÿ2 βˆ’ 9𝑐2 = (2π‘Ÿ + 3𝑐)(2π‘Ÿ βˆ’ 3𝑐) (2π‘Ÿ)(2π‘Ÿ) (3𝑐)(3𝑐) 𝒂 𝒃 Denominator Sum of Two Cubes 8π‘Ÿ3 + 27𝑐3 = (2π‘Ÿ)3 + (3𝑐)3 (2r)(2π‘Ÿ)(2π‘Ÿ) (3𝑐)(3𝑐)(3𝑐) π’‡π’Šπ’“π’”π’• 𝒃𝒂𝒔𝒆 𝒔𝒆𝒄𝒐𝒏𝒅 𝒃𝒂𝒔𝒆 = (2π‘Ÿ 3𝑐 )( ) = (2π‘Ÿ 3𝑐 )( 2π‘Ÿ.2π‘Ÿ ) = (2π‘Ÿ 3𝑐 )( 4π‘Ÿ2 ) = (2π‘Ÿ 3𝑐 )( 4π‘Ÿ2 2π‘Ÿ. 3𝑐 ) = (2π‘Ÿ 3𝑐 )( 4π‘Ÿ2 6π‘Ÿπ‘ ) = (2π‘Ÿ 3𝑐 )( 4π‘Ÿ2 6π‘Ÿπ‘ 3𝑐. 3𝑐) ) = (2π‘Ÿ 3𝑐 )( 4π‘Ÿ2 6π‘Ÿπ‘ 9𝑐2 ) = (2π‘Ÿ + 3𝑐 )( 4π‘Ÿ2 – 6π‘Ÿπ‘ + 9𝑐2 ) S O AP = (πŸπ’“+πŸ‘π’„)(2π‘Ÿβˆ’3𝑐) (2π‘Ÿ+3𝑐 )( 4π‘Ÿ2 – 6π‘Ÿπ‘+9𝑐2) Simplified Form = 2π‘Ÿβˆ’3𝑐 4π‘Ÿ2 – 6π‘Ÿπ‘+9𝑐2
  3. Given Solution Scratch 4.) 4𝑑2+2π‘‘βˆ’ 6 2𝑑2+5𝑑+ 3 = (2𝑑+3)(2π‘‘βˆ’2) (2𝑑+3)(𝑑+1) Numerator Trinomial (2𝑑 + 3) (2𝑑 βˆ’ 2) Factors of the Trinomial (2𝑑 + 3)(2𝑑 βˆ’ 2) Denominator Trinomial (2𝑑 + 3) (𝑑 + 1) Factors of the Trinomial (2𝑑 + 3)(𝑑 + 1) = (2𝑑+3)(2π‘‘βˆ’2) (2𝑑+3)(𝑑+1) Simplified Form = 2π‘‘βˆ’2 𝑑+1
  4. Given Solution Scratch 5.) 2π‘‘βˆ’ 6 6 βˆ’ 2𝑑 = (2π‘‘βˆ’6) βˆ’1(2π‘‘βˆ’6) Retain Numerator 2𝑑 βˆ’ 6 Change the Denominator 6 βˆ’ 2𝑑 a. 2𝑑 6 b. 2𝑑 βˆ’ 6 c. βˆ’1(2𝑑 βˆ’ 6) = (2π‘‘βˆ’6) βˆ’1(2π‘‘βˆ’6) Simplified Form = 1 βˆ’1 = βˆ’1
Publicidad