The question is down below. pleaseee

Proved below

Step-by-step explanation:

We want to prove that;

√(2yx^(-1)) × √(4xz^(-1)) × √(8zy^(-1)) = 8

Now, the √ symbol is also written by denoting the numbers attached to it as resisted to the power of ½. Thus, we have;

[(2yx^(-1))^(½)] × [(4xz^(-1))^(½)] × [(8zy^(-1))^(½)]

Simplifying this gives;

2^(½) × y^(½) × x^(-½) × 4^(½) × x^(½) × z^(-½) × 8^(½) × z^(½) × y^(-½)

Simplifying further, let's write 8 and 4 in terms of 2. Thus;

2^(½) × y^(½) × x^(-½) × (2²)^(½) × x^(½) × z^(-½) × (2³)^(½) × z^(½) × y^(-½)

This gives;

2^(½) × y^(½) × x^(-½) × (2¹) × x^(½) × z^(-½) × (2^(3/2)) × z^(½) × y^(-½)

Collecting like terms and applying laws of indices gives;

2^(½ + 1 + (3/2)) × y^(½ - ½) × x^(-½ + ½) × z^(-½ + ½)

This gives;

8y^(0) × x^(0) × y^(0)

Any Number raised to the power of zero equals 1. Thus;

8y^(0) × x^(0) × y^(0) = 8

hello,

as we don't know if g(x) is continue,

answer b

f(x)=2x²-8x+1=2(x²-4x+4)+1-8=2(x-2)²-7

vertex is (2,-7) ( a minimum since coefficient of x² is > 0)

the answer that is correct is false