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.
11, 12 -11, -12
12 x 11 = 132.
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
The product of two consecutive integers is 132. Find the two integers. They are: 11*12 = 132
The product is 132
17,556.
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
169 = 132
The least common multiple of 3, 11 and 12 is 132.
11 and 12