{"id":3880,"date":"2023-10-06T06:16:49","date_gmt":"2023-10-06T06:16:49","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3880"},"modified":"2023-10-06T06:20:05","modified_gmt":"2023-10-06T06:20:05","slug":"week-1-ss3-first-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-1-ss3-first-term-mathematics-notes\/","title":{"rendered":"Week 1 – SS3 First Term Mathematics Notes"},"content":{"rendered":"

SCHEME OF WORK FOR SS3 MATHEMATICS
\n<\/strong>FIRST TERM
\n<\/strong><\/p>\n

    \n
  1. Revision: Indices and Logarithm\n<\/li>\n
  2. Surds\n<\/li>\n
  3. Surds in relation to Trigonometry\n<\/li>\n
  4. Matrices and Determinants\n<\/li>\n
  5. Linear and Quadratic Equations\n<\/li>\n
  6. Surface Area and Volume of Sphere and Hemispherical shapes\n<\/li>\n
  7. Mid-term Test\n<\/li>\n
  8. Longitude and Latitude\n<\/li>\n
  9. Longitude and Latitude\n<\/li>\n
  10. Arithmetic of Finance\n<\/li>\n
  11. Revision\n<\/li>\n
  12. Examination\n<\/li>\n
  13. Vacation\n<\/li>\n<\/ol>\n

    \u00a0
    \n\u00a0
    \n\u00a0WEEK ONE\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
    \n<\/strong>TOPIC: BASIC CONCEPT & APPLICATION OF LAWS OF INDICES
    \n<\/strong>CONTENT:
    \n<\/strong><\/p>\n

      \n
    • Basic Concept of Laws of Indices\n<\/li>\n
    • Application of Laws of Indices\n<\/li>\n<\/ul>\n

      Basic Concept of Laws of Indices
      \n<\/strong>A number of the form am<\/sup> where a is a real number, a is multiplied by itself m times
      \nThe number a is called the base and the super script m is called the index (plural indices) or exponent.
      \n1.\u00a0\u00a0\u00a0\u00a0A m<\/sup> x A n<\/sup> = Am + n<\/sup> ——————–Multiplication law
      \nExample: X3 <\/sup>xX2 <\/sup> =( X x Xx X) x (X x X) = X 5<\/sup>
      \n\tOr X3<\/sup> x X2<\/sup> = X3 + 2<\/sup> = X5
      \n<\/sup>2.\u00a0\u00a0\u00a0\u00a0Am<\/sup> \u00f7 An<\/sup>=Am-n <\/sup> ———————Division law
      \nExample: X6<\/sup> \u00f7 X4 <\/sup>= X6-4<\/sup> = X2<\/sup>
      \n\t3.\u00a0\u00a0\u00a0\u00a0(a m <\/sup>)n <\/sup>= amn<\/sup> —————-Power law
      \nExample: (x3<\/sup>)2<\/sup> = X3<\/sup> x X3<\/sup> = X3+3<\/sup> =X6
      \n<\/sup>Or X3X2<\/sup> = X6<\/sup>
      \n\t4.\u00a0\u00a0\u00a0\u00a0am<\/sup> \u00f7 am<\/sup> = am-m<\/sup> = a0 <\/sup>=1
      \nam <\/sup>\u00f7am<\/sup> = am<\/sup>\/am<\/sup> = ao<\/sup> = 1
      \na0<\/sup>\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.Zero Index
      \n:. Any number raised to power zero is 1
      \nExample: 3o<\/sup> = 1, co<\/sup> = 1, yo<\/sup> = 1
      \n5.\u00a0\u00a0\u00a0\u00a0(ab)m<\/sup> =am<\/sup>bm <\/sup> ————-Product power law
      \ne.g. (2xy)2<\/sup>= 4x2<\/sup>y2<\/sup>
      \n\t6.\u00a0\u00a0\u00a0\u00a0Negative index\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
      \n a \u2013m<\/sup> = 1\/am<\/sup> ————- Negative Index
      \nExample: 2 -1<\/sup> = \u00bd, and 3 -2<\/sup> = 1\/3 2<\/sup> = 1\/9
      \n7.\u00a0\u00a0\u00a0\u00a0a1\/n <\/sup> =n<\/sup>\u221aa ————–Root power law
      \nExample :<\/strong> 9 \u00bd<\/sup> =\u221a9=3
      \n\t\t<\/sup>
      \n\u00a0\"\"\/27 1\/3<\/sup> =3<\/sup>\u221a27 = 3 ie (3)3 <\/sup>= 3
      \n8.\u00a0\u00a0\u00a0\u00a0a m\/n<\/sup> = (a 1\/n<\/sup>) m<\/sup> = (n<\/sup>\u221aa)m<\/sup> ———–Fraction Index
      \nor a m\/n<\/sup> = (am<\/sup>) 1\/n<\/sup> = (n<\/sup>\u221aa)m<\/sup>
      \n\tExample : 272\/3<\/sup> = 3\u221a27=32<\/sup>=9.
      \nEVALUATION
      \n<\/strong>1. 275\/3<\/sup> 2. 10000000000<\/sup>
      \n\tApplication of Laws of Indices
      \n<\/strong>Solve the following
      \n(i)\u00a0\u00a0\u00a0\u00a032 3\/5 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sup>(ii)\u00a0\u00a0\u00a0\u00a0343 2\/3<\/sup>\u00a0\u00a0\u00a0\u00a0(iii) 64 2\/3<\/sup>(iv) 0.001\u00a0\u00a0\u00a0\u00a0(v) 14 0<\/sup>
      \n\tSolution:
      \ni)\u00a0\u00a0\u00a0\u00a032 3\/5<\/sup> = (32 1\/5<\/sup>) 3<\/sup> = (5<\/sup>\u221a32) 3<\/sup>
      \n\t= 2 3<\/sup> = 8
      \nii)\u00a0\u00a0\u00a0\u00a0343 2\/3<\/sup> = (343 1\/3<\/sup> )2<\/sup> = (3<\/sup>\u221a343)2<\/sup>
      \n\t = (7 3<\/sup>)1\/3<\/sup>)2<\/sup>
      \n\t\u00a0\u00a0\u00a0\u00a0= 72<\/sup> = 49
      \niii)\u00a0\u00a0\u00a0\u00a064 2\/3<\/sup> = (64 1\/3<\/sup>)2<\/sup> = (4 3<\/sup>)1\/3<\/sup>)2<\/sup> = 4 2<\/sup>
      \n\tiv)\u00a0\u00a0\u00a0\u00a0(0.001)3<\/sup> = (1\/100)3<\/sup> = (1\/10)3<\/sup>)3<\/sup> = (10 -3<\/sup>)3<\/sup>
      \n\t= 10 -9<\/sup> = 1\/10 9<\/sup>
      \n\tv)\u00a0\u00a0\u00a0\u00a014 0<\/sup> = 1
      \nEVALUATION
      \n<\/strong>1)\u00a0\u00a0\u00a0\u00a0Simplify the following
      \na)\u00a0\u00a0\u00a0\u00a0216 4\/3<\/sup>\u00a0\u00a0\u00a0\u00a0b) 25 1.5<\/sup>\u00a0\u00a0\u00a0\u00a0c) (0.00001)2<\/sup>
      \n\td)\u00a0\u00a0\u00a0\u00a0d 32 2\/5 <\/sup> e) 81 \u00be<\/sup>
      \n\tReading Assignment<\/strong>: F\/Maths project book 1(New third edition).Chapter 2 pg.4 – 6
      \n\t\t\t<\/strong>WEEKEND ASSIGNMENT
      \n<\/strong>1)\u00a0\u00a0\u00a0\u00a0Evaluate 3 x <\/sup> = 1\/8 1 (a) 4 (b) -4 (c) -2 (d) 2
      \n2)\u00a0\u00a0\u00a0\u00a0Simplify\u00a0\u00a0\u00a0\u00a02r5<\/sup> X 9r3<\/sup>\u00a0\u00a0\u00a0\u00a0(a) P2<\/sup>\u00a0\u00a0\u00a0\u00a0(b) 2p4<\/sup> (c) P3<\/sup> d)18r8<\/sup>
      \n\t3)\u00a0\u00a0\u00a0\u00a0Solve 3-y<\/sup> = 243\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(a) -5 \u00a0\u00a0\u00a0\u00a0(b) 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) 3 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) -3
      \n4)\u00a0\u00a0\u00a0\u00a0Solve 25-5n<\/sup> = 625\u00a0\u00a0\u00a0\u00a0(a) 1\/5 \u00a0\u00a0\u00a0\u00a0(b) 2\/5\u00a0\u00a0\u00a0\u00a0(c) 1 1\/5 \u00a0\u00a0\u00a0\u00a0(d) \u2013 2\/5
      \n5)\u00a0\u00a0\u00a0\u00a0Simplify (0.0001)2<\/sup>= \u00a0\u00a0\u00a0\u00a0(a) 10-5<\/sup>\u00a0\u00a0\u00a0\u00a0(b) 10 -3<\/sup>\u00a0\u00a0\u00a0\u00a0(c) 10-8<\/sup>\u00a0\u00a0\u00a0\u00a0(d) 10-10
      \n<\/sup>Theory
      \n1. 163\/2 <\/sup>x 82\/<\/sup>3<\/sup>2. 3X2<\/sup> x 4X3<\/sup>
      \n\t\t 321\/5<\/sup>6X7<\/sup>
      \n\t
      \n\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

      SCHEME OF WORK FOR SS3 MATHEMATICS FIRST TERM Revision: Indices and Logarithm Surds Surds in…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,302],"tags":[],"class_list":["post-3880","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss3-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3880","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=3880"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3880\/revisions"}],"predecessor-version":[{"id":3881,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3880\/revisions\/3881"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3880"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3880"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3880"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}