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What r the multiples of 3?
the multiples of 3 are........0,3,6,9,12,15,18,21,24,27.... and more
What are the numbers that are divisible by 5 in the range 1-100?
1 ÷ 5 = 0 r 1 → first multiple of 5 in the range 1-100 is 1 x 5 = 5 100 ÷ 5 = 20 → last multiples of 5 in the range 1-100 is 20 x 5 = 100 → want the first 20 multiples of 5, namely: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.
The sum of the first 5 terms of an arithmetic series is 85. The sum of the first 6 terms is 123. Determine the first four terms of the series.?
Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24
What r multiples of 4?
4, 8, 12, 16, 20, 24, 28...
R² plus r-20 equals 0?
r2 + r - 20 = 0(r + 5)(r - 4) = 0r + 5 = 0 or r - 4 = 0r = -5 or r = 4
What are the sum of the first five multiples of 9 in summation notation?
.....5 .....Σ 9r ...r=1
What r multiples?
multiples are like timesing like 7 x 5=40
What r the first nonzero multiples of 4?
4, 8, 12, 16
What are the first 3 common multiples of 6 9 10?
they r 90, 180 and 270
What r the multiples of 3?
the multiples of 3 are........0,3,6,9,12,15,18,21,24,27.... and more
How many set of six in forty?
5 r 4
What are the first five multiples of 9 that r greater then 0?
9, 18, 27, 36 and 45.
Six letter words first letter r?
rabbit
What are the numbers that are divisible by 5 in the range 1-100?
1 ÷ 5 = 0 r 1 → first multiple of 5 in the range 1-100 is 1 x 5 = 5 100 ÷ 5 = 20 → last multiples of 5 in the range 1-100 is 20 x 5 = 100 → want the first 20 multiples of 5, namely: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.
What r the first ten multiples of 90?
90, 180, 270, 360, 450, 540, 630, 720, 810, 900
What r the multiples of 3 and 4?
Any multiple of 12.
The sum of the first 5 terms of an arithmetic series is 85. The sum of the first 6 terms is 123. Determine the first four terms of the series.?
Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24
