COURSE SYLLABUS SICS 1533: FOUNDATIONS OF COMPUTER SCIENCE "Whatever you vividly imagine‚ ardently desire‚ sincerely believe and enthusiastically act upon must inevitably come to pass!" Paul J. Meyer a "To be successful‚ you must decide exactly what you want to accomplish‚ then resolve to pay the price to get it." - Bunker Hunt b [Academic Year / Semester] 2013 / 2014‚ First Semester [Class Location] City Campus‚ Computer Lab [Class Meeting Time(s)] (Depending
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Explanation : Using the Prime factorization‚ we have 556920 = 2 x 2 x 2 x 3 x 3 x 5 x 7 x 13 x 17 = 23 x 32 x 5 x 7 x 13 x 17 Q.2 Use Euclid’s division algorithm to find the HCF of 210 and 55. (1 Mark) (Ans) 5 Explanation: 5 ‚ Given integers are 210 and 55 such that 210 > 55. Applying Euclid’s division leema to 210 and 55‚ we get 210 = 55 x 3 + 45 ……….(1) 55 = 45 x 1 +10 ………(2) 45 = 10 x 4 + 5 ………..(3) 10 = 5 x 2 + 0 ………..(4) we consider the new divisor 10 and the new remainder
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3 1) Number Properties i) Integers Numbers‚ such as -1‚ 0‚ 1‚ 2‚ and 3‚ that have no fractional part. Integers include the counting numbers (1‚ 2‚ 3‚ …)‚ their negative counterparts (-1‚ -2‚ -3‚ …)‚ and 0. ii) Whole & Natural Numbers The terms from 0‚1‚2‚3‚….. are known as Whole numbers. Natural numbers do not include 0. iii) Factors Positive integers that divide evenly into an integer. Factors are equal to or smaller than the integer in question. 12 is a factor of 12‚ as are 1‚ 2
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1. The set of all integers x such that |x – 3| < 2 is equal to (a) {1‚ 2‚ 3‚ 4‚ 5} (b) {1‚ 2‚ 3‚ 4} (c) {2‚ 3‚ 4} (d) {-4‚ -3‚ -2} 2. The Range of the function f(x) = x 2 2 x − − is (a) R (b) R – {1} (c) (-1) (d) R – {-1} 3. The value of (i)i is (a) ω (b) ω2 (c) e-π/2 (d) 2√2 4. ( ) ( ) 4 5 cos isin icos sin θ + θ θ + θ is equal to (a) cos− isin θ (b) cos9θ − isin9θ (c) sin θ − icosθ (d) sin 9θ − icos9θ 5. The roots of the quadratic equation ax2 + bx + c = 0 will be reciprocal
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The date of birth I used was mine 7/19/86 I will now do all three question that was asked a = 7 b = -19 c = 86 The INTEGERS above are needed to solve each given expressions. A) A^3 – B^3 (7^3) – (-2^3) 343-(-6859) =7‚202 This is the given expression with VARIABLES A and B and raised to the EXPONENTS of 3 on each of them. By substituting the integers in the variables and raising them to the 3rd power gives the answer of B) (a – b)(a2 + ab + b2) (7-(-19) (7^2+(7)(-19)+(-19^2)
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Centre Number Candidate Number General Certificate of Secondary Education 2013 Mathematics Unit T4 (With calculator) Higher Tier *GMT31* [GMT41] *GMT41* TUESDAY 11 JUNE‚ 9.15 am – 11.15 am TIME 2 hours. INSTRUCTIONS TO CANDIDATES Write your Centre Number and Candidate Number in the spaces provided at the top of this page. You must answer the questions in the spaces provided. Do not write outside the box‚ around each page‚ on blank pages or tracing paper. Complete
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exponents by converting many of the large numbers and small numbers into a shorter form. For example: 10‚000‚000‚000‚000 can be written as (10)13. Here‚ 10 is called the base and 13 is called the exponent. For any non-zero integer a‚ a m 1 ‚ where m is a positive integer. am a–m is called the multiplicative inverse of am and vice-versa. Decimal numbers can be written in expanded form using exponents. For example: The number‚ 32845.912 can be written in an expanded form as follows.
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Principles of Algebra 1A 1-1 1-2 1-3 1-4 1-5 1-6 Expressions and Integers Evaluating Algebraic Expressions Writing Algebraic Expressions Integers and Absolute Value Adding Integers Subtracting Integers Multiplying and Dividing Integers 1B 1-7 1-8 LAB 1-9 Solving Equations Solving Equations by Adding or Subtracting Solving Equations by Multiplying or Dividing Model Two-Step Equations Solving Two-Step Equations KEYWORD: MT8CA Ch1 Firefighters can use algebra to find out how fast
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Cambridge Secondary 1 Mathematics Curriculum Framework Contents Introduction Stage 7 .....................................................................................................1 Welcome to the Cambridge Secondary 1 Mathematics curriculum framework. Stage 8 .....................................................................................................7 Stage 9 ................................................................................................... 14
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135 and 225‚ 225 is larger integer. Using Euclid’s division algorithm‚ [Where 135 is divisor‚ 90 is remainder] 225 = 135 × 1 + 90 Since‚ remainder 90 ≠ 0 ‚ by applying Eudid’s division algorithm to 135 and 90 ∴ 135 = 90 × 1 + 45 Again since‚ remainder 45 ≠ 0 ‚ by applying Eudid’s division algorithm to 90 and 45 ∴ 90 = 45 × 2 + 0 Now‚ the remainder is zero so‚ our procedure stops. Hence‚ HCF of 135 and 225 is 45. (ii) In 196 and 38220‚ 38220 is larger integer. Using Euclid’s division algorithm
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