Question

(3xyz)(4/9x²z)(-27/2z) and verify the result for x=2 y=3and z=-1​

Answer 1

The product of (3xyz)($\frac{4}{9}$·$x^{2}$·z) × (9·$x^{2}$·z) × ($\frac{- 27}{2z}$) is "- 18 × $x^{3}$ × yz".

Step-by-step explanation:

We have,

(3xyz)($\frac{4}{9}$·$x^{2}$·z) × (9·$x^{2}$·z) × ($\frac{- 27}{2z}$)

= (3 × $\frac{4}{9}$ × $\frac{- 27}{2}$) × (xyz ×[/tex]·$x^{2}$·z × $\frac{1}{z}$)

= - 18 × $x^{3}$ × yz

Hence, the product of (3xyz)($\frac{4}{9}$·$x^{2}$·z) × (9·$x^{2}$·z) × ($\frac{- 27}{2z}$) is "- 18 × $x^{3}$ × yz".

Put x = 2, y = 3 and z = -1, we get

- 18 × $x^{3}$ × yz

=  - 18 × $2^{3}$ × 3 × (- 1)

=  432

The value is "432".

Answer 2

Answer:

18xy²z

Step-by-step explanation:

-3xyz × 4/9x²z × -27/2xy²z

= (-3×4/9×-27/2) (x×x²×x) (y×y²) (z×z×z)

= 18×x⁴×y³×z³

= 18x⁴y³z³

This is the correct answer.

Hope it will help...

&

I don't know how to verify. Actually, I am also finding for verification in the form of L.H.S. & R.H.S.

If anybody know please answer me also...