The answer: The 3 (three) consecutive odd integers are: -3, -1, 1.
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Explanation:
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To represent 3 (three consecutive odd integers):
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Let "x" be the first odd integer.
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Let "(x+2)" be next consecutive odd integer.
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Let "(x+4") be the third odd integer.
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The sum of these three consecutive odd integers:
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x + (x + 2) + (x + 4) = x + x + 2 + x + 4 = 3x + 6 ;
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Six ("6") times the sum of these 3 (three) consecutive odd integers =
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6*(3x+6) = 6(3x + 6) = -18 (as given in the problem).
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Given: 6(3x + 6) = -18 ; We can divide EACH SIDE of the equation by "6", to cancel the "6" on the left-hand side into a "1";
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{6(3x + 6) } / 6 = -18 / 6 ; to get:
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3x + 6 = -3 ; Now, we can subtract "6" from EACH SIDE of the equation:
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3x + 6 - 6 = -3 - 6 ; to get:
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3x = -9 ; Now, we can divide EACH SIDE of the equation by "3"; to isolate "x" on one side of the question; and solve for "x" ;
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3x / 3 = -9 / 3 ; x = - 3 ;
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Remember, from above:
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Let "x" be the first odd integer. We know that "x = -3".
Is this an odd integer? Yes!
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Let "(x+2)" be next consecutive odd integer. So (x+2) = (-3+2) = -1.
Is this an odd integer? Yes! Is this "{-1}" the next consecutive odd integer with respect to "{-3}"? Yes!
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Let "(x+4") be the third odd integer. So (x+4) = (-3+4) = 1.
Is this an odd integer? Yes! Is this "{1"} the next consecutive odd integer with respect to "{-1}"? Yes!
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So, our 3 (three) consecutive odd integers are: -3, -1, 1.
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To check our work: Is 6 times the sum of our 3 consecutive odd integers, equal to "(-18)" ?
The sum of our 3 consecutive odd integers = -3 + (-1) + 1 = -3 - 1 + 1 = -3.
6 * -3 = ? -18? Yes!
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I got 18 and 30 for my answer
Answer: 90 textbooks
Step-by-step explanation:
everytime you multiply the number of students you multiply the books by the same number. to find how many times 72 was multiplied you divide it by 12.
72 ÷ 12 = 6
then multiply the original number of books (15) by 6.
15 * 6 = 90
so for every 72 students the school orders 90 textbooks
Answer:
mmm
Step-by-step explanation:
Answer:
Step-by-step explanation:
To prove divisibility, we need to factor the divident such that one of its factors matches the divisor.
(I use the notation x|y to denote that x divides y)
(A)
(B)
In this case, it is easier to also factor the divisor to primes:
Both of these factor must be matched in the dividend in order to prove divisibility, and that indeed turns out to be true:
