Need this QUICK!! Evaluate without using a calculator (x2y-3)0+x1/3-(4x)1/2 when x = 64 and y = 27.Show work

-11

Step-by-step explanation:

Given the expression

(x²y-3)⁰+x^1/3-(4x)^1/2 when x = 64 and y = 27.Show work

Note that anything raise to power of zero is 1, hence (x²y-3)⁰ =  1

x^1/3

= 64^1/3

= ∛64

= 4

(4x)^1/2

=  (4* 64)^1/2

= 4^1/2 * 64^1/2

= 2 * 8

= 16

Substitute;

(x²y-3)⁰+x^1/3-(4x)^1/2

=  1 + 4 - 16

= 5 - 16

= -11

Hence the required value is -11

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

a is the correct