Asked by Anonymous
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)
(x+y)(x+y+z)
x(y+1)-x
Answers
Answered by
victoria
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 ..
(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 ..
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