{"id":2104,"date":"2023-10-02T09:12:20","date_gmt":"2023-10-02T09:12:20","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2104"},"modified":"2023-10-02T09:14:57","modified_gmt":"2023-10-02T09:14:57","slug":"week-6-7-and-8-ss1-first-term-general-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-6-7-and-8-ss1-first-term-general-mathematics-notes\/","title":{"rendered":"Week 6, 7 and 8 – SS1 First Term General Mathematics Notes"},"content":{"rendered":"

WEEK 6
\n<\/em> INDICES
\nINDICES: are numbers expressed in powers on ten i.e. . The analysis and simplification of indices depends on the basic interpretation and rules of indices as enumerated below.
\nLAWS OF INDICES<\/p>\n

    \n
  1. \n\t\t<\/li>\n
  2. \n\t\t<\/li>\n
  3. = 1\n<\/li>\n
  4. = (\n<\/li>\n
  5. ( =\n<\/li>\n
  6. =\n<\/li>\n<\/ol>\n

    EXAMPLES:
    \nWrite down the values of the following in index form:<\/p>\n

      \n
    1. x x 4x 2 (iii) 16 (iv) ()\n<\/li>\n<\/ol>\n

      SOLUTION<\/p>\n

        \n
      1. \n\t\t<\/li>\n
      2. x 4x 2 = (5 x 4 x 2) = 40 =\n<\/li>\n
      3. = (16 = 4\n<\/li>\n
      4. =\n<\/li>\n
      5. = = = 1\u00f7 = 1 x = or 2.25 or 2\n<\/li>\n<\/ol>\n

        Simplify the following:<\/p>\n

          \n
        1. x ( (b) 3\u00f7 6\n<\/li>\n<\/ol>\n

          SOLUTION<\/p>\n

            \n
          1. x = x =\n<\/li>\n
          2. (3\u00f76) = () = .\n<\/li>\n<\/ol>\n

            ASSESSMENT
            \nSimplify the following questions:
            \n(1(2)(3). x \u00f7 (4) -10\u00f7 (-5) (5) x x (6)
            \nASSIGNMENT: MAN Mathematics for senior secondary school 1<\/p>\n

              \n
            1. Page 11 Exercise B1 numbers 8, 10, 17,20 and 30.\n<\/li>\n
            2. Page 12, exercise B3 e,f,I,k,r,t ,v and z\n<\/li>\n
            3. Page 13 exercise B4 a, b, c, d, e, g, h and i.\n<\/li>\n<\/ol>\n

              \u00a0
              \n\u00a0WEEK 7 Review of first half and periodic test
              \nWeek 8
              \nLOGARITHMS OF WHOLE NUMBERS
              \nThe logarithms of any number N to any base M is the index or power to which the base must be raised, to equal the number N.
              \n The logarithms of any given number consist of two parts called the characteristics and the mantissa.The characteristics is a whole number which can either be positive, zero or negative integers, While the Mantissa is the decimal (fractional) part of the integers always from the table values.
              \nEXAMOLE
              \n399 = 2.6010. 2 Is the characteristics of the number and 6010 from table is the Mantissa or 3.99 x
              \nFind the Logarithms of the following numbers:<\/p>\n

                \n
              1. 8615 (b) 690460 (c) 1.607\n<\/li>\n<\/ol>\n

                SOLUTION<\/p>\n

                  \n
                1. 8615 = 8.615 x : in mathematics table, check logarithm of 86 under 1 difference 5 = 9350+3\n<\/li>\n
                2. 690460 = 6.90460 x\n<\/li>\n
                3. 1.607 = 1.607 x\n<\/li>\n<\/ol>\n

                  ANTILOGARITHM: Is the opposite of logarithm.
                  \nFind the original number of the following logarithms numbers:<\/p>\n

                    \n
                  1. (b) (c) 6.3892\n<\/li>\n<\/ol>\n

                    SOLUTION<\/p>\n

                      \n
                    1. = 1.862, from antilogarithm table check 27 under zero since there is no third value and the zero before the point (characteristics) determines where the point occupies in the number. Add onto every positive characteristics to determine your value\n<\/li>\n
                    2. = 3698.0 or 3698\n<\/li>\n
                    3. 2450000.0\n<\/li>\n<\/ol>\n

                      MULTIPLICATION OF NUMBERS
                      \nWhen multiplying numbers in logarithms, their table values are been added before checking antilogarithms for its solutions.
                      \nEXAMPLE
                      \nEvaluate the following using table:<\/p>\n

                        \n
                      1. 143.8 x 23.46 (b) 8234 x 70000\n<\/li>\n<\/ol>\n

                        \"\"\/SOLUTION
                        \n\"\"\/(a)143.8 x 23.46 = NUMBER LOGARITHM
                        \n 143.8
                        \n\"\"\/ 23.46 +<\/p>\n

                        \tAntilogarithm of 5280 = 3374
                        \n143.8 x 23.46 = 3374.0<\/p>\n

                        \u00a0
                        \n\u00a0\"\"\/\"\"\/(b) 8234 x70000 = NO LOG
                        \n 8234
                        \n\"\"\/ 70000 +
                        \n =
                        \nAntilog of 7609 = 5766 characteristics is 8+1 =9 numbers before point
                        \n8234 x70000 = 576600000
                        \nDIVISION OF NUMBERS IN LOGARITHMS: When dividing numbers in logarithms we subtract their values
                        \nEXAMPLE
                        \nEvaluate the following numbers using table:<\/p>\n

                          \n
                        1. 912.4 \u00f7 30.42 (b) 36.75 x 284.7 \u00f7 26.45\n<\/li>\n<\/ol>\n

                          \"\"\/SOLUTION<\/p>\n

                            \n
                          1. \n
                            \"\"\/912.4 \u00f7 30.42 = NO LOG\n<\/div>\n

                            912.4
                            \n\"\"\/ 30.42 –
                            \n =\n<\/li>\n<\/ol>\n

                            Antilog of 4770 = 2999
                            \n\"\"\/ 912.4 \u00f7 30.42 = 29.99.
                            \n\"\"\/ (b) 36.75 x284.7 \u00f726.45 = NO LOG
                            \n 36.75
                            \n\"\"\/ 284.7 +<\/p>\n

                            \t\"\"\/ 26.45 –<\/p>\n

                            \t Antilog of 5973 =3957
                            \n 36.75 x 284.7 \u00f7 26.45 = 395.7
                            \nASSESSMENT: Using table evaluate the following numbers:<\/p>\n

                              \n
                            1. (a)497.2 x 8.789 (b) 89 x34.56 x2.094 (c) 8050 \u00f7 20.15 (d) 45.08 \u00f7 5.462\n<\/li>\n
                            2. \n
                              (a) 98.45 x 56 \u00f7 30.8 (b) (c)\n<\/div>\n

                              \u00a0<\/li>\n<\/ol>\n

                              3. Find the antilogarithms of the following numbers: <\/p>\n

                                \n
                              1. (b) (c) 0.5971 (d) 7.8903 (e) 2.0079\n<\/li>\n<\/ol>\n

                                \u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

                                WEEK 6 INDICES INDICES: are numbers expressed in powers on ten i.e. . The analysis…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,183],"tags":[],"class_list":["post-2104","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss1-general-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2104","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=2104"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2104\/revisions"}],"predecessor-version":[{"id":2105,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2104\/revisions\/2105"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2104"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2104"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2104"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}