How do you do number 1?
whenever i tried to answer it, i always get fraction. me.

The pairs are (13,15) and (-15,-13).

Step-by-step explanation:

If n is an odd integer, the very next odd integer will be n+2.

n+1 is even (so we aren't using this number)

The sum of the squares of (n) and (n+2) is 394.

This means

(n)^2+(n+2)^2=394

n^2+(n+2)(n+2)=394

n^2+n^2+4n+4=394               since (a+b)(a+b)=a^2+2ab+b^2

Combine like terms:

2n^2+4n+4=394

Subtract 394 on both sides:

2n^2+4n-390=0

Divide both sides by 2:

n^2+2n-195=0

Now we need to find two numbers that multiply to be -195 and add up to be 2.

15 and -13 since 15(-13)=-195 and 15+(-13)=2

So the factored form is

(n+15)(n-13)=0

This means we have n+15=0 and n-13=0 to solve.

n+15=0

Subtract 15 on both sides:

n=-15

n-13=0

Add 13 on both sides:

n=13

So if n=13 , then n+2=15.

If n=-15, then n+2=-13.

Let's check both results

(n,n+2)=(13,15)

13^2+15^2=169+225=394.  So (13,15) looks good!

(n,n+2)=(-15,-13)

(-15)^2+(-13)^2=225+169=394.  So (-15,-13) looks good!

the answer is a

step-by-step explanation: