WEEK_1,_Section_C,_Discussion[1]

WEEK_1,_Section_C,_Discussion[1] - c Evaluate-1 3 = 1 d...

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C. THREADED DISCUSSION PROBLEMS – Algebra – WEEK # 1 1) Algebra a) 4(x + 3) – 2(2) = 1.5x + 25.5 4x + 12 – 4 = 1.5x + 25.5 4x + 8 = 1.5x + 25.5 4x – 1.5x = 25.5 – 8 2.5x = 17.5 Divide both sides by 2.5 x = 7 2) Common Denominator b) x/4 + 2x/3 + x/6 = ½ ____________= ____ 12 3x + 8x + 2x = 6 Since all terms on both sides have the same denominator you can eliminate 12 it thus 3x + 8x + 2x = 6 Combine like terms 13x = 6 Divide both sides by 13 x = 6/13 HINT: You need a common denominator for each term in the equation. If there was a whole number in this problem like "8", "8" must be made into a fraction by placing it over "1" such as 8/1. Then the common denominator must be obtained for 8/1 just like all other terms in the equation. 3) Evaluate Exponents

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Unformatted text preview: c) Evaluate (-1) 3 = - 1 d) Evaluate –1 3 = - 1 e) Evaluate (-1) 2 = 1 f) Evaluate –1 2 =- 1 4) Math Problem g) 10 3/5 x 20/25 = Convert 1 st term to an improper fraction and reduce 2 nd fraction to lowest terms thus 53/5 x 4/5 = Multiply across 212/25 or 8 12/25 Hint: The mixed number 10 3/5 must be converted into an improper fraction before any calculations can be done. This is accomplished by multiplying the denominator "5" times the whole number "10" and adding that result to the numerator "3". Then place that result over the original denominator "5". Therefore the improper fraction is 53/5 . The new problem is 53/5 x 20/25 . Cancel between the numerators and denominators if possible. Then multiply the fractions....
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WEEK_1,_Section_C,_Discussion[1] - c Evaluate-1 3 = 1 d...

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