5015 divided by 59 = 85
12 equals 3x(4) equals 6x(2) equals (12)x1
49
One answer is: t{n} = (59n⁵ - 885n⁴ + 5015n³ - 13275n² + 15446n - 4800)/120 Which gives: t1 = (59 - 885 + 5015 - 13275 + 15446 - 4800)/120 = 1560/120 = 13 t2 = (59×32 - 885×16 + 5015×8 - 13275×4 + 15446×2 - 4800)/120 = 840/120 = 7 t3 = (59×243 - 885×81 + 5015×27 - 13275×9 + 15446×3 - 4800)/120 = 120/120 = 1 t4 = (59×1024 - 885×256 + 5015×64 - 13275×16 + 15446×3 - 4800)/120 = -600/120 = -5 t5 = (59×3125 - 885×625 + 5015×125 - 13275×25 + 15446×4 - 4800)/120 = -1320/120 = -11 t6 = (59×7776 - 885×1296 + 5015×216 - 13275×36 + 15446×5 - 4800)/120 = 5040/120 = 42 However, I expect your teacher is wanting the much simpler: t{n} = 19 - 6n which also gives t{1..5} = {13, 7, 1, -5, -11} but gives a different t6 = -17 The above formulae are only valid for n = 1, 2, ..., 5 as t6 is different.
3 times blank equals 18,000 = 54000
0.0118
5015 divided by 59 = 85
It is: 5015/59 = a base length of 85 units
85 times.
334.3 repeating
59 divided by 5015 = 0.01176470588
12 equals 3x(4) equals 6x(2) equals (12)x1
What equals 59 from multiplying in the 3
49
One possible answer: Blank 1 = 3150 Blank 2 = 10 Blank 3 = (2/315) = approx 0.006349
One answer is: t{n} = (59n⁵ - 885n⁴ + 5015n³ - 13275n² + 15446n - 4800)/120 Which gives: t1 = (59 - 885 + 5015 - 13275 + 15446 - 4800)/120 = 1560/120 = 13 t2 = (59×32 - 885×16 + 5015×8 - 13275×4 + 15446×2 - 4800)/120 = 840/120 = 7 t3 = (59×243 - 885×81 + 5015×27 - 13275×9 + 15446×3 - 4800)/120 = 120/120 = 1 t4 = (59×1024 - 885×256 + 5015×64 - 13275×16 + 15446×3 - 4800)/120 = -600/120 = -5 t5 = (59×3125 - 885×625 + 5015×125 - 13275×25 + 15446×4 - 4800)/120 = -1320/120 = -11 t6 = (59×7776 - 885×1296 + 5015×216 - 13275×36 + 15446×5 - 4800)/120 = 5040/120 = 42 However, I expect your teacher is wanting the much simpler: t{n} = 19 - 6n which also gives t{1..5} = {13, 7, 1, -5, -11} but gives a different t6 = -17 The above formulae are only valid for n = 1, 2, ..., 5 as t6 is different.
3 times blank equals 18,000 = 54000