Let N be any non-zero integer. Then 58*N divided by N will always give 58 as their quotient.
t(n) = (5n3 - 30n2 + 85n - 48)/6 , n = 1, 2, 3, ...
To find the sum of the first 48 terms of an arithmetic sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, a1 = 2, n = 48, and an = 2 + (48-1)*2 = 96. Plugging these values into the formula, we get: S48 = 48/2 * (2 + 96) = 24 * 98 = 2352. Therefore, the sum of the first 48 terms of the given arithmetic sequence is 2352.
t(n) = 3*2n-1 Or just 3*2n
You can either do this by trial-and error, or solve the equation n + (n + 2) + (n + 4) = 48.
georgetown
38° 58′ 48″ N, 1° 25′ 48″ E38.98, 1.43
n = 6
Latitude: N 45° 48' 53.6832" Longitude: E 15° 58' 42.6522"
32° 46′ 58″ N, 96° 48′ 14″ W32.782778, -96.803889
6n + 48
Latitude: 32°46′58″N Longitude: 96°48′14″W
32°46′58″N 96°48′14″W
The equation is abs(58 - n) = 34 or |58 - n| = 34 |58 - n| = 34 implies 58 - n = ± 34 so that 58 - n = -34 or 58 - n = +34 that is, n = 58 + 34 or 58 - 34 so n = 92 or 24
58% : 766 = 80% : n 58%n/58% = (766 * 80%)/58% n = 1056.55
6° 18′ 48″ N, 10° 48′ 5″ W6.313333, -10.801389
We're going to use 'n' to represent the number. 8n = 48 (same thing as 8 x n = 48) 8n/8 = 48/8 (divide both sides by 8) n = 6 (times 8 and divided by 8 cancel each other 8, leaving n) 8 x 6 = 48