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What is the greater of two consecutive negative integers whose product is 132?
Let x = the lesser integer and let x + 1 = the greater integer. Because product is a key word for multiplication, the equation is x(x + 1) = 132. Multiply by using the distributive property on the left side of the equation: x2 + x = 132. Put the equation in standard form and set it equal to zero: x2 + x - 132 = 0. Factor the trinomial: (x - 11)(x + 12) = 0. Set each factor equal to zero and solve: x - 11 = 0 or x + 12 = 0; x = 11 or x = -12. Since you are seems for a negative integer and reject the x-value of 11. Thus, x = -12 and x + 1 = -11. The greater negative integer is -11.
Determine if the following sequences are monotonic and/or bounded. (a) {-n 2 } ∞ n=0 (b) {( -1) n+1 } ∞ n=1 (c) {2/n 2 } ∞ n=5 Solution {-n 2 } ∞ n=0
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Tammi's latest printer can print 13.5 pages a minute. How many pages can it print in 4 minutes? Multiply 13.5 by 4 to ?nd out the number of copies made; 13.5 × 4 = 54 copies.
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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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