Respuesta :
1.
By given conditions, we have:
[tex] 2y\leq 12\\
\implies y \leq \frac{12}{2} \\
\implies y\leq 6[/tex]
Range of $y$ is $y \in (-\infty , 6]$
2.
$32x^2y-8yz^2$
$= 8y(4x^2-z^2)$
$=\boxed{8y(2x-z)(2x+z)}$
Answer:
Step-by-step explanation:
1. The certain number = y
2y<_ 12( sorry, I can't write the symbol on my phone but the less than or equal to symbol is < then an horizontal stroke underneath)
2y/2<_ 12/ 2
Y _< 6
There fore the range is ( -1, 0,1,2,3,4,5)
2. 8y(4x^2 - z^2) apply difference if two squares
8y[(2x) ^2 - (z) ^2 ]
8y[ ( 2x - z) ( 2x + z) ]