TomZH TomZH
  • 13-07-2022
  • Mathematics
contestada

can i have the x,y,z for this question? THanks!

can i have the xyz for this question THanks class=

Respuesta :

wafflefly21
wafflefly21 wafflefly21
  • 13-07-2022

Answer:

x=4, y=-2, p=14

Step-by-step explanation:

First simplify the LHS:

[tex]LHS = (16p^8)^\frac{3}{2} \times (216p^{-3})^{-\frac{2}{3}}[/tex]

[tex]= 64 p^{12} \times \frac{1}{36} p^{2}[/tex]

[tex]=\frac{16}{9}p^{14}[/tex]

[tex]=\frac{2^4}{3^2}p^14\\=2^43^{-2}p^{14}[/tex]

Hence x=4, y=2, p=14

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