{"id":2102,"date":"2023-10-02T09:11:10","date_gmt":"2023-10-02T09:11:10","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2102"},"modified":"2023-10-02T09:14:57","modified_gmt":"2023-10-02T09:14:57","slug":"week-5-ss1-first-term-general-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-5-ss1-first-term-general-mathematics-notes\/","title":{"rendered":"Week 5 – SS1 First Term General Mathematics Notes"},"content":{"rendered":"

\u00a0WEEK 5<\/p>\n

STANDARD FORM AND APPROXIMATION
\n<\/h2>\n

STANDARD FORM: Is a convenient way of writing very large or small numbers. It is the product of the numbers in powers of 10 to determine the position of the decimal pointand it\u00b4s writing between 1 and 9 i.e. a X . Where ais the number between 1and 9 and n is the position of the decimal point.
\nEXAMPLE:
\nExpress the following numbers in standard form;<\/p>\n

    \n
  1. \n
    9 (b) 54.6 (c) 570200\n<\/div>\n

    SOLUTION.<\/p>\n

      \n
    1. \n
      9 = 9 x (b) 54.6 = 5.46 x (c) 570200 = 5.702 x\n<\/div>\n

      NEGATIVE POWERS (NUMBERS LESS THAN 1 )
      \nExpress the following numbers in standard form;<\/p>\n

        \n
      1. \n
        0.02 (b) 0.000175\n<\/div>\n

        SOLUTION<\/p>\n

          \n
        1. 0.02 = 2 x (b) o.000175 =1.75 x .\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

          ADDITION AND SUBTRACTION OF NUMBERS IN STANDARD FORM\u00a0\u00a0\u00a0\u00a0
          \n<\/h3>\n

          Simplify the following;<\/p>\n

            \n
          1. 5.14 x + 2.842 x (b) 5.24 x 4.33 x\n<\/li>\n<\/ol>\n

            SOLUTION<\/p>\n

              \n
            1. \n
              5.14 x = 5.14 x 10x10x10x10x10x10x10 = 51400000\n<\/div>\n

              \"\"\/2.842 x = 2.842 x 10x10x10x10x10 = 284200 +
              \n\"\"\/ 51684200
              \n51684200 =5.16842 x\n<\/li>\n

            2. \n
              5.24 x = 5.24 x= = 0.0000524\n<\/div>\n

              \u00a04.33 x = 4.33 x = = 0.00000433
              \n 0.0000524 0.00000433 =0.00004803
              \n0.00004807 =4.807 x
              \nMULTIPLICATION AND DIVISION OF STANDARD FORM
              \nSimplify the following:<\/p>\n

                \n
              1. \n
                8.222 x x 6.32 x (b) 2.8 x 1.5 x\n<\/div>\n

                SOLUTION<\/p>\n

                  \n
                1. \n
                  (8.222 x 6.32) x () = 51.96304 x\n<\/div>\n

                  5.196304 x =5.196304 x\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                  (b) (2.8 1.5) x ( ) = 1.86667 x
                  \n 1.86667 x <\/p>\n

                  APPROXIMATION (SIGNIFICANT FIGURE, DECIMAL PLACES AND ROUNDING OFF)
                  \n<\/h3>\n

                  PPROXIMATION: Is calculating values or numbers only to a certain degree of accuracy.
                  \nSIGNIFICANT FIGURE: Is a process in which each of the digits in a number that are needed to a given accuracy is presented.
                  \nDECIMAL: is the writing of numbers to a required fractional part.
                  \nEXAMPLE:
                  \nApproximate the following to the nearest (i) ten (ii) hundred (iii) thousand <\/p>\n

                    \n
                  1. 7562 (b)907235 (c)8991\n<\/li>\n<\/ol>\n

                    SOLUTION<\/p>\n

                      \n
                    1. \n
                      7562 = 7560 (nearest ten)\n<\/div>\n

                      = 7600 (nearest hundred)
                      \n =8000 (nearest thousands)\n<\/li>\n

                    2. \n
                      907235 = 907240 (nearest ten)\n<\/div>\n

                      = 907200 (nearest hundred)
                      \n =907000 (nearest thousand)\n<\/li>\n<\/ol>\n

                      \u00a9 8991 = 8990 (nearest ten)
                      \n = 9000 (nearest hundred)
                      \n = 9000(nearest thousands)
                      \nSIGNIFICANT FIGURE: Express the following numbers to (i) 1s.f (ii) 2s.f (iii) 3s.f<\/p>\n

                        \n
                      1. 78602 (b) 702.976 (c) 0.000057849\n<\/li>\n<\/ol>\n

                        SOLUTION<\/p>\n

                          \n
                        1. 78602 = 80000 (1s.f), 79000 (2s.f), 78600(3s.f)\n<\/li>\n
                        2. 702.976 = 700 (1s.f), 700(2s.f) 703 (3s.f)\n<\/li>\n
                        3. 0.000057849 = 0.00006(1s.f), 0.000058(2s.f), 0.0000578(3s.f)\n<\/li>\n<\/ol>\n

                          DECIMAL PLACES: Express the following numbers to ; (i) 2d.p (ii)3d.p<\/p>\n

                            \n
                          1. 67.3994 (b) 0.00749\n<\/li>\n<\/ol>\n

                            SOLUTION<\/p>\n

                              \n
                            1. 67.3994 = (i) 67.40(2d.p) , (ii) 67.399(3d.p)\n<\/li>\n
                            2. 0.00749 = (i) 0.01(2d.p), (ii) 0.007(3d.p)\n<\/li>\n<\/ol>\n

                              ASSESSMENT: Solve the following questions;<\/p>\n

                                \n
                              1. \n
                                Write the following in standard form:\n<\/div>\n
                                  \n
                                1. 7560 (ii)7560000000 (iii) 0,00756 (iv) 0.000000756\n<\/li>\n<\/ol>\n<\/li>\n
                                2. \n
                                  Find each of the following ,leaving your answers in standard form:\n<\/div>\n
                                    \n
                                  1. (4 x ) x (2.6 x ) (ii) (6.4 x (8 x\n<\/li>\n<\/ol>\n<\/li>\n
                                  2. Evaluate (9.05 x ) + (6.5 x ) \u2013 (6.5 x ) giving your answer in standard form.\n<\/li>\n
                                  3. Express each of the following numbers to : (i) 1s.f, (ii) 3s.f, (iii) 2d.p (iv) 3d.p\n<\/li>\n
                                  4. 799.8094 (b) 6.3006 (c) 0.0014698\n<\/li>\n<\/ol>\n

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

                                    \u00a0WEEK 5 STANDARD FORM AND APPROXIMATION STANDARD FORM: Is a convenient way of writing very…<\/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-2102","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\/2102","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=2102"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2102\/revisions"}],"predecessor-version":[{"id":2103,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2102\/revisions\/2103"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2102"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2102"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2102"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}