8. (8 marks) prove that for all integers m and n, m + n and m-n are either both even or both odd

Answer with explanation:

Let m and n are integers

To prove that m+n and m-n are either both even or both odd.

1. Let m and n are both even

We know that sum of even number is even and difference of even number is even.

Suppose m=4 and n=2

m+n=4+2=6 =Even number

m-n=4-2=2=Even number

Hence, we can say m+n and m-n are both even .

2. Let m and n are odd numbers .

We know that sum of odd numbers is even and difference of odd numbers is even.

Suppose m=7 and n=5

m+n=7+5=12=Even number

m-n=7-5=2=Even number

Hence, m+n and m-n are both even .

3. Let m is odd and n is even.

We know that sum of an odd number and an even number is odd and difference of an odd and an even number is an odd number.

Suppose m=7 , n=4

m+n=7+4=11=Odd number

m-n=7-4=3=Odd number

Hence, m+n and m-n are both odd numbers.

4.Let m is even number and n is odd number .

Suppose m=6, n=3

m+n=6+3=9=Odd number

m-n=6-3=3=Odd number

Hence, m+n and m-n are both odd numbers.

Therefore, we can say for all inetegers m and n , m+n and m-n are either both even or both odd.Hence proved.

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet

50-60=10

step-by-step explanation: