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The sum of the squares of two consecutive positive odd integers is 74. What is the value of the smaller integer?
Let x = the lesser odd integer and let x + 2 = the greater odd integer. The translation of the sentence, "The sum of the squares of two consecutive odd integers is 74," is the equation x2 + (x + 2)2 = 74. Multiply (x + 2)2 out as (x + 2)(x + 2) by using the distributive property: x2 + (x2 + 2x + 2x + 4) = 74. Combine such as terms on the left side of the equation: 2x2 + 4x + 4 = 74. Put the equation in standard form through subtracting 74 from both sides, and set it equal to zero: 2x2 + 4x - 70 = 0; factor the trinomial completely: 2(x2 + 2x - 35) = 0; 2(x - 5)(x + 7) = 0. Set each factor equal to zero and solve: 2 ≠ 0 or x - 5 = 0 or x + 7 = 0; x = 5 or x = -7. Because you are seems for a positive integer, reject the solution of x = -7. Thus, the smaller positive integer is 5.
Find the sum of a+b, a-b, a-3b, ...... to 22 terms. Ans: a + b, a - b, a - 3b, up to 22 terms d= a - b - a - b = 2b S22 =22/2 [2(a+b)+21(-2b)] 11[2a + 2b - 42b] =
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