{"id":2203,"date":"2023-10-02T10:53:31","date_gmt":"2023-10-02T10:53:31","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2203"},"modified":"2023-10-02T11:02:26","modified_gmt":"2023-10-02T11:02:26","slug":"week-1-ss1-second-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-1-ss1-second-term-mathematics-notes\/","title":{"rendered":"Week 1 – SS1 Second Term Mathematics Notes"},"content":{"rendered":"

SECOND TERM E-LEARNING NOTE
\n<\/strong>
\n\u00a0SUBJECT: MATHEMATICS\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0CLASS: SS 1\u00a0\u00a0\u00a0\u00a0
\n<\/strong>
\n\u00a0SCHEME OF WORK
\n<\/strong>
\n\u00a0WEEK\u00a0\u00a0\u00a0\u00a0TOPIC
\n<\/strong><\/p>\n

    \n
  1. \n
    Quadratic Equation by (a) Factorization (b) Completing the square method\n<\/div>\n<\/li>\n
  2. \n
    General Form of Quadratic Equation leading to Formular Method\n<\/div>\n

    from ax2<\/sup> + bx + c = 0\n<\/li>\n

  3. \n
    Solutions of Quadratic Equation by Graphical Methods:\n<\/div>\n
      \n
    1. \n
      Reading the Roots from the Graph\n<\/div>\n<\/li>\n
    2. \n
      Determination of the Minimum and Maximum Values\n<\/div>\n<\/li>\n
    3. \n
      Line of Symmetry.\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n
    4. \n
      Idea of Sets:\n<\/div>\n
        \n
      1. \n
        Universal Sets, Finite and Infinite Sets, Empty Set, Subset\n<\/div>\n<\/li>\n
      2. \n
        Idea of Notation for Union and Intersection of Sets\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n
      3. \n
        Complements of Sets:\n<\/div>\n
          \n
        1. \n
          Disjoints of Null.\n<\/div>\n<\/li>\n
        2. \n
          Venn Diagramand its Use in Solving Problems Involving two and three Sets Relation to Real Life Situations.\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n
        3. \n
          Review of the First Half Term’s Work and Periodic Test\n<\/div>\n<\/li>\n
        4. \n
          Trigonometric Ratios\n<\/div>\n
            \n
          1. \n
            Sine, Cosine, Targentof Acute Angles\n<\/div>\n<\/li>\n
          2. \n
            Use of Tables of Trigonometric Ratios\n<\/div>\n<\/li>\n
          3. \n
            Determination of Length of Chord\n<\/div>\n<\/li>\n
          4. \n
            Using Trigonometric Ratios\n<\/div>\n<\/li>\n
          5. \n
            Graph of Sine and Cosine for Angles 0o<\/sup> = x\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n
          6. \n
            (a) Application of Sine, Cosine and Tangent, Simple Problems with Respect to Right Angle Triangles.\n<\/div>\n

            (b) Angles of Elevation and Depression
            \n(c) Bearing and Distances of Places Strictly Application of Trigonometric Ratio.\n<\/li>\n

          7. \n
            (a) Introduction of Circle and its Properties\n<\/div>\n

            (b) Calculation of Length of Arc and Perimeter of a Sector
            \n(c) Area of Sectors and Segments. Area of triangles\n<\/li>\n

          8. \n
            Logic\n<\/div>\n
              \n
            1. \n
              Simple True and False Statements\n<\/div>\n<\/li>\n
            2. \n
              Negative and Contra Positive of Simple Statement.\n<\/div>\n<\/li>\n
            3. \n
              Antecedents, Consequence and Conditional Statement (implication)\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

              \u00a0REFERENCE BOOK
              \n<\/strong><\/p>\n

                \n
              • New General Mathematics SSS 1 M.F. Macrae et al\n<\/li>\n
              • WABP Essential Mathematics For Senior Secondary Schools 1 A.J.S Oluwasanmi\n<\/li>\n<\/ul>\n

                WEEK ONE<\/strong>
                \n\tTopic<\/strong>: Quadratic equation by (a) Factorization (b) Completing the square method
                \n<\/strong>Quadratic Equations
                \n<\/strong>A quadratic equation contains an equal sign and an unknown raised to the power 2. For example:
                \n2x2<\/sup> \u2013 5x \u2013 3 = 0
                \nn2<\/sup> + 50 = 27n
                \n0 = (4a – 9)(2a + 1)
                \n49 = k2<\/sup><\/p>\n

                \u00a0Are all quadratic equations.
                \nDiscussion: can you see why
                \n0 = (4a \u2013 9)(2a + 1) is a quadratic equation?
                \nOne of the main objectives of the chapter is to find ways of solving quadratic equations,
                \ni.e. finding the value(s) of the unknown that make the equation true. <\/p>\n

                \u00a0Solving Quadratic Equations
                \n<\/strong>One way of solving quadratic equation is to apply the following argument to a quadratic expression that has been factorized.
                \nIf the product of two numbers is 0, then one of the numbers (or possibly both of them) must be 0. For example,
                \n3 0 = 0, 0 5 = 0 and 0 0 = 0
                \nIn general, if a b = 0
                \n\u00a0\u00a0\u00a0\u00a0Then either a = 0
                \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Or b = 0
                \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Or both a and b are 0 <\/p>\n

                \u00a0Example 1
                \n<\/strong>Solve the equation (x \u2013 2)(x + 7) = 0.
                \nIf (x \u2013 2)(x + 7) = 0
                \nThen either x \u2013 2 = 0 or x + 7 = 0
                \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x = 2 or -7 <\/p>\n

                \u00a0Example 2
                \n<\/strong>Solve the equation d(d \u2013 4)(d + 62<\/sup>) = 0.
                \n(3a + 2)(2a \u2013 7) = 0, then any one of the four factors of the LHS may be 0,
                \ni.e d = 0 or d \u2013 4 = 0 or d + 6 = 0 twice.
                \n d = 0, 4 or -6 twice. <\/p>\n

                \u00a0EVALUATION
                \n<\/strong>Solve the following equations.<\/p>\n

                  \n
                1. 3d2<\/sup>(d \u2013 7) = 0\n<\/li>\n
                2. (6 \u2013 n)(4 + n) = 0\n<\/li>\n
                3. A(2 \u2013 a)2<\/sup>(1 + a) = 0\n<\/li>\n<\/ol>\n

                  \u00a0Solving quadratic equations using factorization method
                  \n<\/strong>The LHS of the quadratic equation m2<\/sup> \u2013 5m \u2013 14 = 0 factorises to give (m + 2)(m \u2013 7) = 0. <\/p>\n

                  \u00a0Example 1
                  \n<\/strong>Solve the equation 4y2<\/sup> + 5y \u2013 21 = 0
                  \n4y2<\/sup> + 5y \u2013 21 = 0
                  \n (y + 3)(4y \u2013 7) = 0
                  \neither y + 3 = 0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0or \u00a0\u00a0\u00a0\u00a04y \u2013 7 = 0
                  \n\u00a0\u00a0\u00a0\u00a0y = – 3 \u00a0\u00a0\u00a0\u00a0or \u00a0\u00a0\u00a0\u00a04y = 7
                  \n\u00a0\u00a0\u00a0\u00a0y = – 3\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a0y = 7\/4
                  \n\u00a0\u00a0\u00a0\u00a0y = -3\u00a0\u00a0\u00a0\u00a0or\u00a0\u00a0\u00a0\u00a01
                  \ncheck: by substitution:
                  \nif y = -3
                  \n4y2<\/sup> + 5y \u2013 21 = 36 \u2013 15 \u2013 21 = 0
                  \nIf y = 1,
                  \n\u00a0\u00a0\u00a0\u00a04y2<\/sup> + 5y \u2013 21 = 4 x 7\/4 x 7\/4 + 5 x 7\/4 \u2013 21
                  \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= \u2013 21 = 0 <\/p>\n

                  \u00a0Example 2
                  \n<\/strong>Solve the equation m2<\/sup> = 16
                  \nRearrange the equation.
                  \nIf m2<\/sup> = 16
                  \nThen m2<\/sup> \u2013 16 = 0
                  \nFactorise (difference of two squares)
                  \n(m – 4)(m + 4) = 0
                  \nEither \u00a0\u00a0\u00a0\u00a0m \u2013 4 = 0\u00a0\u00a0\u00a0\u00a0or \u00a0\u00a0\u00a0\u00a0m + 4 = 0
                  \n\u00a0\u00a0\u00a0\u00a0m = +4\u00a0\u00a0\u00a0\u00a0or \u00a0\u00a0\u00a0\u00a0m = -4
                  \n\u00a0\u00a0\u00a0\u00a0m = 4 <\/p>\n

                  \u00a0EVALUATION
                  \n<\/strong>Solve the following quadratic equations:<\/p>\n

                    \n
                  1. h2<\/sup> \u2013 15h + 54 = 0\n<\/li>\n
                  2. 12y2<\/sup> + y \u2013 35 = 0\n<\/li>\n
                  3. 4a2<\/sup> \u2013 15a = 4\n<\/li>\n
                  4. v2<\/sup> + 2v \u2013 35 = 0\n<\/li>\n<\/ol>\n

                    \u00a0GENERAL EVALUATION
                    \n<\/strong>Solve the following equations:<\/p>\n

                      \n
                    1. y2<\/sup>(3 + y) = 0\n<\/li>\n
                    2. x2<\/sup>(x + 5)(x – 5) = 0\n<\/li>\n
                    3. (v – 7)(v – 5)(v – 3) = 0\n<\/li>\n
                    4. 9f2<\/sup> + 12f + 4 = 0\n<\/li>\n<\/ol>\n

                      \u00a0WEEKEND ASSIGNMENT
                      \n<\/strong>Solve the following equations. Check the results by substitution. <\/p>\n

                        \n
                      1. (4b – 12)(b – 5) = 0 A. \u00bd, 4 B. 3, 5 C. 4, 6 D.5, 3\n<\/li>\n
                      2. (11 \u2013 4x)2<\/sup> = 0 \u00a0\u00a0\u00a0\u00a0A., 3 B.2, 3 C. 2 twice D. 2 twice\n<\/li>\n
                      3. (d \u2013 5)(3d \u2013 2) = 0 A. 5, B. 4, 5 C. 5, 9 D. , 5\n<\/li>\n<\/ol>\n

                        Solve the following quadratic equations
                        \n<\/strong><\/p>\n

                          \n
                        1. u2<\/sup> \u2013 8u \u2013 9 = 0A. \u2013 9, 1 B. -1, 9 C. 1, 8 D. 9 , -1\n<\/li>\n
                        2. c2<\/sup> = 25 A. 5 B. -5 C.+5 D.5\n<\/li>\n<\/ol>\n

                          \u00a0THEORY
                          \n<\/strong>Solve the equation <\/p>\n

                            \n
                          1. 2x2<\/sup> = 3x + 5\n<\/li>\n
                          2. a2<\/sup> \u2013 3a = 0\n<\/li>\n
                          3. p2<\/sup> + 7p + 12 = 0\n<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                            SECOND TERM E-LEARNING NOTE \u00a0SUBJECT: MATHEMATICS\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0CLASS: SS 1\u00a0\u00a0\u00a0\u00a0 \u00a0SCHEME OF WORK \u00a0WEEK\u00a0\u00a0\u00a0\u00a0TOPIC Quadratic Equation by…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,190],"tags":[],"class_list":["post-2203","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss1-general-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2203","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2203"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2203\/revisions"}],"predecessor-version":[{"id":2204,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2203\/revisions\/2204"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2203"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2203"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2203"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}