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Grade 11Ac Mathematics Assignment 7 Due Date A. In the diagram below, 3 straight line graphs and 1 parabola are drawn. Lines A and B are parallel. 𝒚 C. Each row, column and 3x3 diagonal in the square below adds up to 3. D 𝑥−4 𝑥+1 𝑥 B A −𝟐 𝒙 𝟑 𝟐 𝟏 −𝟑 𝑥+𝑦 C 𝟓 3𝑥 − 2𝑦 Which pair of simultaneous equations has: Equation pair a solution with a positive 𝒙-value and negative 𝒚-value ? A C no solution ? A B a solution with 𝒚-value 𝟎 ? B C two distinct solutions ? C D one repeated root solution ? B D B. Use the Completing the Square method to solve the following quadratic equations, leaving your answers in surd form: (a) 𝑥 2 + 6𝑥 − 23 = 0 (b) 𝑥 2 + 8𝑥 + 4 = 0 (c) 𝑥 2 − 3𝑥 − 2 = 0 2𝑥 𝟎 −𝟏 𝟒 (a) Find the value of 𝑥 and 𝑦 From top row deduce that 𝒙 = 𝟐. Using 𝑥 = 2 and left side column, deduce that 𝒚 = 𝟑. (b) Complete the table with possible expressions using 𝑥 and 𝑦 for each empty square. Any expressions that correctly equate to the missing values, shown in blue in the square above. D. A paint mixture is made up of blue (b) and yellow (y) and paint and must satisfy: 2𝑏 + 𝑦 < 14 , 𝑏 ≥ 3, 𝑦>4 (a) Sketch the inequalities on the grid below. 𝒚 9 (See next page) 𝟏𝟒 A ball is shot vertically up into the air from ground level. The ball’s distance (in meters) above ground level, 𝒔, is given by the formula: 𝒔 = 𝟐𝟒. 𝟓𝒕 − 𝟒. 𝟗𝒕𝟐 where 𝑡 is time, in seconds. (d) At what times will the ball be exactly 19.6 m above ground level? 19.6 = 24.5𝑡 − 4.9𝑡 2 4.9𝑡 2 − 24.5𝑡 + 19.6 = 0 ÷ 4.9 to give … 𝑡 2 − 5𝑡 + 4 = 0 (𝑡 − 4)(𝑡 − 1) = 0 So, 𝒕 = 𝟒 or 𝒕 = 𝟏 seconds Self Reflection: I got 𝟒 𝟑 𝒃 (b) If the paint comes in bottles of 1 liter only, list all the possible combinations of blue and yellow 1L paint bottles that may be used to make the paint mixture. 3 blue and 5 yellow, 3 blue and 6 yellow 3 blue and 7 yellow, 4 blue and 5 yellow I did best at correct. 𝟕 My target for improvement is Question B (a) (b) 𝑥 2 + 6𝑥 − 23 = 0 𝑥 2 + 8𝑥 + 4 = 0 𝑥 2 + 6𝑥 = 23 8 2 8 2 𝑥 2 + 8𝑥 + ( ) = −4 + ( ) 2 2 6 2 6 2 𝑥 2 + 6𝑥 + ( ) = 23 + ( ) 2 2 𝑥 2 + 8𝑥 + 16 = −4 + 16 𝑥 2 + 6𝑥 + 9 = 23 + 9 (𝑥 + 4)2 = 12 (𝑥 + 3)2 = 32 𝑥 = −4 ± 2√3 𝒙 = −𝟑 ± 𝟒√𝟐 𝒙 = 𝟐(−𝟐 ± √𝟑) (c) 𝑥 2 − 3𝑥 − 9 =0 2 3 2 9 3 2 𝑥 2 − 3𝑥 + ( ) = + ( ) 2 2 2 3 2 27 (𝑥 − ) = 2 4 𝑥= 3 27 3 3√3 ±√ = ± 2 4 2 2 𝒙= 𝟑 (𝟏 ± √𝟑) 𝟐