📅 Free live lesson: Tuesday 17th NovemberSave your free place →

Mixed Topic Worksheet Grades 8/9: Day 058

Grades 8/9Calculator

  1. Question 1. x is an acute angle and sin x = sin 100^°. Work out x

    Show worked answerHide worked answer
    1. B1: sin 100^° = sin(180^° - 100^°), so x = 80^°
    Answer: x = 80^°
    Calculating Values of Trig Functions Using Graphs
    [1 mark]

  2. Question 2. Show that x³ - 5x + 1 = 0 has a solution between x = 2.1 and x = 2.2

    Show worked answerHide worked answer
    1. M1: work out f(2.1) and f(2.2) for f(x) = x³ - 5x + 1
    2. A1: -0.239 and 0.648, a sign change, so a solution lies between them
    Answer: Change of sign shows a solution between 2.1 and 2.2
    Change of Signs
    [2 marks]

  3. Question 3. The graph of y = x² - 6x is translated by the vector beginpmatrix -3 9 endpmatrix Find the equation of the new graph in the form y = ax² + bx + c

    Show worked answerHide worked answer
    1. M1: new equation is y = f(x + 3) + 9
    2. M1: y = (x + 3)² - 6(x + 3) + 9
    3. A1: y = x²
    Answer: y = x²
    Translations Along the x- & y-Axes
    [3 marks]

  4. Question 4. ABCD is a cyclic quadrilateral.
    Angle BAD = (3x + 10)^° and angle BCD = (2x + 20)^°.
    Work out angle BAD. Give reasons. [3]

    Show worked answerHide worked answer
    1. B1: opposite angles of a cyclic quadrilateral sum to 180^°: 5x + 30 = 180
    2. M1: x = 30
    3. A1: angle BAD = 100^°
    Answer: 100^°
    Circle Theorems
    [3 marks]

  5. Question 5. Rationalise the denominator of dfrac64 + √2
    Give your answer in its simplest form.

    Show worked answerHide worked answer
    1. M1: multiply the numerator and the denominator by 4 - √2
    2. M1: denominator 16 - 2 = 14
    3. A1: dfrac12 - 3√27
    Answer: dfrac12 - 3√27
    Rationalising Denominators of the Form a ± √b
    [3 marks]
📅 Free lesson: Tue 17th NovSave your place →

Mixed Topic Worksheet Grades 8/9: Day 058

Grades 8/9Calculator

  1. Question 1. x is an acute angle and sin x = sin 100^°. Work out x

    [1 mark]
    Show worked answerHide worked answer
    1. B1: sin 100^° = sin(180^° - 100^°), so x = 80^°
    Answer: x = 80^°
    Calculating Values of Trig Functions Using Graphs

  2. Question 2. Show that x³ - 5x + 1 = 0 has a solution between x = 2.1 and x = 2.2

    [2 marks]
    Show worked answerHide worked answer
    1. M1: work out f(2.1) and f(2.2) for f(x) = x³ - 5x + 1
    2. A1: -0.239 and 0.648, a sign change, so a solution lies between them
    Answer: Change of sign shows a solution between 2.1 and 2.2
    Change of Signs

  3. Question 3. The graph of y = x² - 6x is translated by the vector beginpmatrix -3 9 endpmatrix Find the equation of the new graph in the form y = ax² + bx + c

    [3 marks]
    Show worked answerHide worked answer
    1. M1: new equation is y = f(x + 3) + 9
    2. M1: y = (x + 3)² - 6(x + 3) + 9
    3. A1: y = x²
    Answer: y = x²
    Translations Along the x- & y-Axes

  4. Question 4. ABCD is a cyclic quadrilateral.
    Angle BAD = (3x + 10)^° and angle BCD = (2x + 20)^°.
    Work out angle BAD. Give reasons. [3]

    [3 marks]
    Show worked answerHide worked answer
    1. B1: opposite angles of a cyclic quadrilateral sum to 180^°: 5x + 30 = 180
    2. M1: x = 30
    3. A1: angle BAD = 100^°
    Answer: 100^°
    Circle Theorems

  5. Question 5. Rationalise the denominator of dfrac64 + √2
    Give your answer in its simplest form.

    [3 marks]
    Show worked answerHide worked answer
    1. M1: multiply the numerator and the denominator by 4 - √2
    2. M1: denominator 16 - 2 = 14
    3. A1: dfrac12 - 3√27
    Answer: dfrac12 - 3√27
    Rationalising Denominators of the Form a ± √b
📅 Free live lesson: Tuesday 17th NovemberSave your free place →

Mixed Topic Worksheet Grades 8/9: Day 058

Grades 8/9Calculator

  1. Question 1. x is an acute angle and sin x = sin 100^°. Work out x

    Show worked answerHide worked answer
    1. B1: sin 100^° = sin(180^° - 100^°), so x = 80^°
    Answer: x = 80^°
    Calculating Values of Trig Functions Using Graphs
    [1 mark]

  2. Question 2. Show that x³ - 5x + 1 = 0 has a solution between x = 2.1 and x = 2.2

    Show worked answerHide worked answer
    1. M1: work out f(2.1) and f(2.2) for f(x) = x³ - 5x + 1
    2. A1: -0.239 and 0.648, a sign change, so a solution lies between them
    Answer: Change of sign shows a solution between 2.1 and 2.2
    Change of Signs
    [2 marks]

  3. Question 3. The graph of y = x² - 6x is translated by the vector beginpmatrix -3 9 endpmatrix Find the equation of the new graph in the form y = ax² + bx + c

    Show worked answerHide worked answer
    1. M1: new equation is y = f(x + 3) + 9
    2. M1: y = (x + 3)² - 6(x + 3) + 9
    3. A1: y = x²
    Answer: y = x²
    Translations Along the x- & y-Axes
    [3 marks]

  4. Question 4. ABCD is a cyclic quadrilateral.
    Angle BAD = (3x + 10)^° and angle BCD = (2x + 20)^°.
    Work out angle BAD. Give reasons. [3]

    Show worked answerHide worked answer
    1. B1: opposite angles of a cyclic quadrilateral sum to 180^°: 5x + 30 = 180
    2. M1: x = 30
    3. A1: angle BAD = 100^°
    Answer: 100^°
    Circle Theorems
    [3 marks]

  5. Question 5. Rationalise the denominator of dfrac64 + √2
    Give your answer in its simplest form.

    Show worked answerHide worked answer
    1. M1: multiply the numerator and the denominator by 4 - √2
    2. M1: denominator 16 - 2 = 14
    3. A1: dfrac12 - 3√27
    Answer: dfrac12 - 3√27
    Rationalising Denominators of the Form a ± √b
    [3 marks]