Answer:
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!
U= 9.4m/s
v= -7.4m/s (Negative sign because it is in the opposite direction as he is rolling back)
t= ?
s= ?
a= ?
Now, a= v-u/t
= -7.4-9.4÷3
=-5.6
By the second equation of motion.
s= ut+1÷2at*2 ( *2 is the power)
s= 9.4×3+1÷2×-5.6×3*2
= 28.2 +(-25.2)
=3
Therefore s or the distance travelled is 3m.
Answer:
This one is also B my friend
Answer:
1800 or D.
Step-by-step explanation:
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