Hello :
all n in N ; n(n+1)(n+2) = 3a a in N or : <span>≡ 0 (mod 3)
1 ) n </span><span>≡ 0 ( mod 3)...(1)
n+1 </span>≡ 1 ( mod 3)...(2)
n+2 ≡ 2 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 0×1×2 ( mod 3) : ≡ 0 (mod 3)
2) n ≡ 1 ( mod 3)...(1)
n+1 ≡ 2 ( mod 3)...(2)
n+2 ≡ 3 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 1×2 × 3 ( mod 3) : ≡ 0 (mod 3) , 6≡ 0 (mod)
3) n ≡ 2 ( mod 3)...(1)
n+1 ≡ 3 ( mod 3)...(2)
n+2 ≡ 4 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 2×3 × 4 ( mod 3) : ≡ 0 (mod 3) , 24≡ 0 (mod3)
Answer:
D. Meters per second
Step-by-step explanation:
here's why the answer is D even though they arent the best explanations but none the less the answer is still right.
1. Newton = force
2.kilo = mass
3. foot-pound = force
4. Meters Per sec = speed/acceleration
It would be a > 12, move all terms to one side and then solve for x. make sure if you divide then you flip the term
Answer:
n b knhkn hbk n
Step-by-step explanation:
jhuguoljhjjl;
Answer:
The amount of money in Enid bank account can be written as a linear equation.
Ye = Xe + $4*m
where Ye is the money that Enid has in her account, m is the number of months that have passed since she opened it, and Xe is the initial deposit.
For Jim, the equation is similar:
Yj = Xj + $3*m
where Yj and Xj are similar as above.
Between May 15 and December 31 of the same year, we have 7 months (where i am counting December because the deposit is made in the first day of the month).
Then we have that:
Ye = $72 = Xe + $4*7 = Xe + $28
Xe = $72 - $28 = $44
So in May 15, Enid deposited $44.
For Jim we have:
Yj = $72 = Xj + $3*7 = Xj + $21
Xj = $72 - $21 = $51
So in May 15, Jim deposited $51.