X= fancy, y= plain
x+y= 7
28x+ 15y=131
solve by substitution
6(10+z+3)
= (6)(10 + z +3)
= (6)(10) + (6)(x) + (6)(3)
= 60 + 6x + 18
= 6x + 78
Hi! I don’t know what any of this means but I hope you have an amazing day and I hope god/allah/etc. blesses you :)
sorry I couldn’t answer
Answer:
y^ ⁻⁹/⁴ * z^⁽⁻³⁾
Step-by-step explanation:
(Y^3 z^4)^(-3/4) = (y^3)^(-3/4) * (z^4)^(-3/4)
= y^3* ⁽⁻³/⁴⁾ * z^4 *⁽⁻³/⁴⁾
=y^ ⁻⁹/⁴ * z^⁽⁻³⁾
