a)A fruitseller bought 50kg of the fruits. He sold 30kg of fruits for the cost price of 35kg of fruits and he sold the remaining quantity for the cost Price of 18kg of fruits. calculate his profit or loss percent in the total transaction.​

A fruitseller bought 50kg of the fruits. He sold 30kg of fruits for the cost price of 35kg of fruits and he sold the remaining quantity for the cost Price of 18kg of fruits. calculate his profit or loss percent in the total transaction.

let the cost price be 50x

→he sells 30kg of fruits on it's CP of 35 kg

→CP of 30kg fruits = 30x

→SP of 35kg fruits = 35x

→remaing fruits are 20kg

→he sells 20kg of fruits on CP of 16kg

→CP of 20kg fruits = 20x

→SP of 20kg fruits = 16x

→total CP is = 50x

→total SP is = (35 + 16) = 51x

→SP > CP (it means profit)

→profit = SP-CP

→ 51-50

→ 1

→ Profit% = gain/CP × 100

→ Profit% = 1/50 × 100

→ 2%

Hence the fruit seller had a profit% of 2%.

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

the correct answer is 1: 2

step-by-step explanation:

hope this your welcome : )