Answer: The answer is C) lll.
A) I.
B) II.
C) III.
D) IV.
Step-by-step explanation:
If a < 0 and b < 0, then the point (a, b) is in Quadrant
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:6205
Step-by-step explanation:
Hello!
Since we are given that the speed of sound in the air travels about 343 meters per second, we can multiply 343 meters per second by 5, and then we convert the meters to kilometers.
343 · 5 = 1715 meters per 5 seconds.
1000 meters are equal to 1 kilometer. So, we can divide 1715 by 1000 to convert meters to kilometers.
1715 / 1000 = 1.715
Therefore, the speed of sound in air is about 1.715 kilometers per 5 seconds.
The sum of all the even integers between 99 and 301 is 20200
To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.
In this case, the first even integer will be 100 and the last even integer will be 300.
nth term of the AP = first term + (n-1) x common difference
⇒ 300 = 100 + (n-1) x 2
Therefore, n = (200 + 2 )/2 = 101
That is, there are 101 even integers between 99 and 301.
Sum of the 'n' terms in an AP = n/2 ( first term + last term)
= 101/2 (300+100)
= 20200
Thus sum of all the even integers between 99 and 301 = 20200
Learn more about arithmetic progressions at brainly.com/question/24592110
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