rename each of the following using the distributive property of multiplication over addition so that there are no parentheses in the final answer simplify when possible.
3(x+y+5)
(x+y)(x+y+z)
x(y+1)-x
3(x+y+5)= 3x+3Y+15
(x+y)(x+y+z)= x2+y +x+y2 + xz+yz (x2 means x squared, y2 means y2)
x(y+1)-x = xy+x-x
im pretty sure these are correct but you might want to double check
all you do is take the number that's out of the ()from the left and distribute/multiply it to the numbers or variables inside the () .. like this : 3(x+4) would be 3x+12 ..
To rename each of the given expressions using the distributive property of multiplication over addition and simplify, follow these steps:
1. 3(x + y + 5):
Distribute the 3 to each term inside the parentheses:
3x + 3y + 15
2. (x + y)(x + y + z):
Distribute the (x + y) to each term inside the second set of parentheses:
x(x + y + z) + y(x + y + z)
Simplify each term:
x^2 + xy + xz + xy + y^2 + yz
Combine like terms:
x^2 + 2xy + y^2 + xz + yz
3. x(y + 1) - x:
Distribute the x to each term inside the parentheses:
xy + x - x
Combine like terms (x - x = 0):
xy
So, the renamed expressions without parentheses are:
1. 3x + 3y + 15
2. x^2 + 2xy + y^2 + xz + yz
3. xy
To rename each of the following expressions using the distributive property of multiplication over addition, we need to apply the multiplication to each term inside the parentheses:
1. 3(x+y+5):
Distributing the 3 to each term inside the parentheses:
3x + 3y + 15
2. (x+y)(x+y+z):
Using the distributive property twice:
(x+y) * (x+y+z) = x(x+y+z) + y(x+y+z)
Expanding further:
x² + xy + xz + yx + y² + yz
3. x(y+1)-x:
Distributing the x to each term inside the parentheses:
xy + x - x
Simplifying the expression:
xy