Prove that for integers m, n, p if m n and n p are both odd then m p must be an even integer.(Direct)

Answer and Step-by-step explanation:

Solution:

Given:

m, n, and p are integers

m + n and n + p are odd integers.

Then prove that m + p is even integer.

Proof:

Properties of odd and even integer:

If x is an odd integer, then exist an integer y such that x = 2y + 1.

If x is an even integer, then there exist an integer y such that x = 2y.

Let m + n and n + p be odd integers, then there exist integer y and z such that :

       m + n = 2y + 1

and   n + p = 2z + 1

let add the two odd integers:

m + n+ n + p = 2y + 1 + 2z + 1

m + 2n + p = 2y + 2z + 2

we are interested in m + p , so subtract 2n from both sides:

m + 2n – 2n + p = 2y + 2z+ 2 – 2n

m + p = 2y + 2z – 2n -2

by taking common 2 from right side:

m + p = 2 ( y + z – n + 1)

since y, z and n are integers, y + z – n is also an integer and thus m + p is an even integer.

Hence proved if m, n, and p are integers and m + n and n + p are odd integers. Then, m + p is even integer..

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