X + Y = 23
XY = 132
Y = 23-x
xy = x (23-x) = 23x -x2= 132
x2-23x+132 = 0
(X-11)(x-12) = 0
X = 11
Y = 12
child 1 = a child 2 = b a + b = 23 a x b = 132 a = 23 - b (23 - b) x b = 132 -b^2 + 23b - 132 = 0 Now it has become a quadratic, you can use the quadratic rule, but i'll bore you no longer child b = 11 or 12 So The first born child would be 12 years old and the second born would be 11 years old
11 and 12
There is an infinite number of multiples of 132. Choose any number and multiply, this will give you a product. Which is the name given to the result of multiplying numbers together. If you choose a number less than 1 then the result will be less than 132 but is also a multiple of it.
The sum of 12 and a product of 11 can be calculated by first finding the product of 12 and 11, which equals 132. Then, adding 12 to this product gives a sum of 144. Mathematically, this can be represented as 12 + (12 x 11) = 12 + 132 = 144.
11, 12 -11, -12
child 1 = a child 2 = b a + b = 23 a x b = 132 a = 23 - b (23 - b) x b = 132 -b^2 + 23b - 132 = 0 Now it has become a quadratic, you can use the quadratic rule, but i'll bore you no longer child b = 11 or 12 So The first born child would be 12 years old and the second born would be 11 years old
11 & 12 are the integers whose product is 132.
1+131=132
17,556.
The product is 132
They are: 11 times 12 = 132
As a product of its prime factors: 2*2*3*11 = 132
2 x 2 x 3 x 11 = 132
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.
169 = 132
The least common multiple of 3, 11 and 12 is 132.
11 and 12