(9 x 1) + (9 x 4) = 9 x 5 = 45
To use the distributive property, multiply the term outside the parentheses by each term inside the parentheses. For example, in the expression ( a(b + c) ), you would calculate it as ( ab + ac ). This property helps simplify expressions and solve equations by distributing a common factor across terms. It's particularly useful when dealing with addition or subtraction within parentheses.
The distributive property states that for any real numbers a, b, and c, a(b + c) = ab + ac. In this case, applying the distributive property to 20 + 32, we get 1(20) + 1(32) = 20 + 32. The greatest common factor (GCF) for 20 and 32 is 4, as 4 is the largest number that divides both 20 and 32 evenly without leaving a remainder.
Suppose x and y are two terms with GCF k where the assumption (in this context) is that k is greater than 1. That implies that x = pk and y = qk where p and q are coprime terms. Then x + y = pk + qk and, using the distributive property, this is k*(p + q).
Greatest Common Factor .... largest expression which (evenly) divides into the given expressions.
2xy
An unnecessary one. 42 + 96 = 138
(6 x 7) + (6 x 16) = 6 x 23 = 138
The GCF is 16.
(21 x 2) + (25 x 2) = 46 x 2 = 92
(4 x 4) + (4 x 9) = 4 x 13 = 52
The GCF is 6.
The GCF of 24 and 30 is 6. The distributive property states that 24 x 30 = (20 x 30) + (4 x 30)
You need at least two numbers to find a GCF.
The greatest common factor (GCF) refers to a factor that is COMMON to two or more numbers. YOu have only one term "8 + W" in the question so there cannot be a GCF. It is not possible to find any factor (other than 1 and 8 + W, itself, without further information about W.The distributive property has no relevance to this question.
18 + 24 = (3 x 6) + (4 x 6) = 7 x 6 = 42
The GCF is 18. The Distributive property states that 36 x (50 + 4) = (36 x 50) + (36 x 4)
Using my knowledge of greatest common factors and the distributive property, I can tell that 45 + 63 is equal to 9(5 + 7) or 9 x 12, which is 108. Of course, I could also tell that from my knowledge of addition.