The expression "16y 0 16y" appears to be a typographical error or misrepresentation of a mathematical operation, as it lacks clear context. If you meant to refer to a property involving the number 16y, such as factoring or distribution, please clarify the intended operation. In algebra, properties like the distributive property or factoring can be applied to expressions involving variables and coefficients. If you have a specific operation in mind, please provide more details for a precise answer.
y2 - 16y + 64 = 0 (y - 8)2 = 0 y - 8 = 0 y = 8
4y2+16y+16=0 Divide by 4 on both sides: y2+4y+4=0 Factorise: (y+2)(y+2)=0 Therefore y= -2
16y-9y3 = 0 y*(16-9y2) = 0 y*(4-3y)*(4+3y) = 0 So y = 0 or (4-3y) = 0 or (4+3y) = 0 ie y = 0 or y = 4/3 or y = -4/3
28y - 16y = 12y.
80
y2 - 16y + 64 = 0 (y - 8)2 = 0 y - 8 = 0 y = 8
6xy/16y = 3x/8 (as long as y≠0) = 3/8x
4y2+16y+16=0 Divide by 4 on both sides: y2+4y+4=0 Factorise: (y+2)(y+2)=0 Therefore y= -2
16y-9y3 = 0 y*(16-9y2) = 0 y*(4-3y)*(4+3y) = 0 So y = 0 or (4-3y) = 0 or (4+3y) = 0 ie y = 0 or y = 4/3 or y = -4/3
It is: 28y -16y = 12y
It is: 28y -16y = 12y
16y -64 as an inequality = -48
28y - 16y = 12y.
80
4xy/0.25x = 16y
x^2 plus 16y?
16y^2(8y)=0? Factor out an 8y to give you 8y(2y(2))=0 Use order of operations to give you 8y(4y)=0 Distribute to give you 32y=0 and ultimately the answer will be 0 If there is a missing plus or minus operator between 16y2 and the 8y, then it can be factorised by taking out common factors: 16y2 ± 8y = 0 ⇒ 8y(2y ± 1) = 0