Answer:
[tex]-\frac{1}{32}[/tex]
Step-by-step explanation:
Given Equation:
[tex]\frac{1}{2^-^2*x^-^3*y^5}[/tex]
Multiplying numerator and denominator by [tex]2^2*x^3[/tex] gives:
[tex]\frac{2^2*x^3}{y^5}[/tex]
We have to evaluate the equation at [tex]x=2 and y=-4[/tex]
Putting the values of x and y in the equation:
= [tex]\frac{2^2*2^3}{(-4)^5}[/tex]
= [tex]\frac{4*8}{(-1024)}[/tex]
= [tex]-\frac{32}{1024} \\[/tex]
= [tex]-\frac{1}{32}[/tex]
On putting the value of 'x' and 'y' in the equation the solution of the equation is [tex]-\frac{1}{32}[/tex]
