SOLUTION: multiply 3xyz2 over 6y4 by 2y over xz4
My assumption of the equation is written as (for it can be written in other ways also depends on the intent of the writer):
multiply 3x
Question 337211: multiply 3xyz2 over 6y4 by 2y over xz4
My assumption of the equation is written as (for it can be written in other ways also depends on the intent of the writer):
multiply 3xyz2 over 6y4 by 2y over xz4
{(3xyz)*(2)/6y4} x {(2y)/(xz4)}
Or
{(3xyz^2)/6y^4} x {(2y)/(xz^4)}
I take it is the later or here it is:
multiply 3xyz2 over 6y4 by 2y over xz4
{(3xyz^2)/6y^4} * {(2y)/(xz^4)}=
[{((3/6) * xyz^2)/y^4} * {((2/1) *y)/xz^4)}]=
Multiply numerator with numerator and denominator with denominator. consolidate numbers separate from letters. Simplify all numbers to lowest simple numbers. 2 is same as 2 over 1 or 2/1 and 3/6 is same as ½ divisable by 3.
[{(3/6) *(2/1)} * {(xyz^2*y)/(y^4*xz^4)}]=
(1/2)(2/1) * (xy^2z^2) / (xy^4z^4) =number divisible itself is one
2/2 * x/x*y^2/y^4*z^2/z^4 =
1*1*(1/y^2)*(1/z^2)=
1/(y^2*z^2)
You can put this solution on YOUR website! My assumption of the equation is written as (for it can be written in other ways also depends on the intent of the writer):
multiply 3xyz2 over 6y4 by 2y over xz4
{(3xyz)*(2)/6y4} x {(2y)/(xz4)}
Or
{(3xyz^2)/6y^4} x {(2y)/(xz^4)}
I take it is the later or here it is:
multiply 3xyz2 over 6y4 by 2y over xz4
{(3xyz^2)/6y^4} * {(2y)/(xz^4)}=
[{((3/6) * xyz^2)/y^4} * {((2/1) *y)/xz^4)}]=
Multiply numerator with numerator and denominator with denominator. consolidate numbers separate from letters. Simplify all numbers to lowest simple numbers. 2 is same as 2 over 1 or 2/1 and 3/6 is same as ½ divisable by 3.
[{(3/6) *(2/1)} * {(xyz^2*y)/(y^4*xz^4)}]=
(1/2)(2/1) * (xy^2z^2) / (xy^4z^4) =number divisible itself is one
2/2 * x/x*y^2/y^4*z^2/z^4 =
1*1*(1/y^2)*(1/z^2)=
1/(y^2*z^2)
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