An employment agency specializing in temporary construction help pays heavy equipment operation $138 per day and general laborers $93 dollars per day. If thirty-one people were hired and the payroll was $3648, how many heavy equipment operators were employed? How many laborers?

Solution:

Let x = number of heavy equipment operators employed

Then,

31 - x = number of laborers employed

From given,

Pay of heavy equipment operators = $ 138 per day

Pay of general labours = $ 93 per day

Payroll was $3648

Therefore,

number of heavy equipment operators x Pay of heavy equipment operators + number of laborers employed x Pay of general labours = 3648

138x + 93(31 - x) = 3648

138x + 2883 - 93x = 3648

45x = 3648 - 2883

45x = 765

x = 17

Thus,

31 - x = 31 - 17 = 14

Therefore, 17 heavy equipment operators and 14 general labourers were employed

you can not drow on this

step-by-step explanation:

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