1. If a certain number y is doubled , the result is less than or equal to 12 .Find the range of value of the number. 2. Factorise 32x²y - 8yz²
Please help me.

1.

By given conditions, we have:

Range of $y$ is $y \in (-\infty , 6]$

2.

$32x^2y-8yz^2$

$= 8y(4x^2-z^2)$

$=\boxed{8y(2x-z)(2x+z)}$

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) ]

1. 22

2. 1

3. 7

4. 18

step-by-step explanation

e2020 got it right