Let the two consecutive integers be x and x+1. The product of these two integers is x(x+1). Setting this equal to 132 gives us the equation x(x+1) = 132. By expanding the left side of the equation, we get x^2 + x = 132. Rearranging terms, we have x^2 + x - 132 = 0. This is a quadratic equation that can be factored as (x+12)(x-11) = 0. Therefore, the two consecutive integers are 11 and 12, and their product is 132.
-130
22 x 3 x 11 = 132
Pythagoras! Diagonal = sqrt (900 + 900) = 42.43'
By Lomg Multiplication 12 x 11 120 ( 10 x 12 = 120) 12 ( 1 x 12 = 12) 132 ====
Any rectangle that measures 12-in x 13-in has a diagonal ofsqrt(122 + 132) = 17.692 inches (rounded)
If 22x = 132,x = 132/22 = 6
11 x 12 = 132
12 x 11 = 132.
270 x 132 = 35640
132 66,2 33,2,2 11,3,2,2 2 x 2 x 3 x 11 = 132
2 x 2 x 3 x 11 = 132 The prime factors of 132 are 2, 3 and 11.
The prime factorization of 132 is 2 x 2 x 3 x 112 x 2 x 3 x 11 = 132
132 - 3% = 132 x (1 - (3/100)) = 132 x 0.97 = 128.04
A 9 x 12 rectangle has a diagonal of 15.
The factors of 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, and 132 The prime factors of 132 are 2,2,3,11.
2 x 2 x 3 x 11 = 132