2 + (9*1) = 2 + 9 = 11
3
The sum of the whole numbers from 1 to 630 can be calculated using the formula for the sum of an arithmetic series: ((n/2) \times (first\ term + last\ term)). Here, (n = 630), the first term is 1, and the last term is 630. Thus, the sum is ((630/2) \times (1 + 630) = 315 \times 631 = 198165). Therefore, the sum is 198,165.
It means you multiply some quantity times 2.It means you multiply some quantity times 2.It means you multiply some quantity times 2.It means you multiply some quantity times 2.
The sum of the counting numbers from 1 to 60 can be calculated using the formula for the sum of an arithmetic series: ( S_n = \frac{n}{2} \times (a + l) ), where ( n ) is the number of terms, ( a ) is the first term, and ( l ) is the last term. Here, ( n = 60 ), ( a = 1 ), and ( l = 60 ). Plugging in these values, the sum ( S_{60} = \frac{60}{2} \times (1 + 60) = 30 \times 61 = 1860 ). Therefore, the sum of the counting numbers from 1 to 60 is 1860.
area triangle = 1/2 base times height area trapezoid = 1/2 (sum of bases) times height
3
The sum of the whole numbers from 1 to 630 can be calculated using the formula for the sum of an arithmetic series: ((n/2) \times (first\ term + last\ term)). Here, (n = 630), the first term is 1, and the last term is 630. Thus, the sum is ((630/2) \times (1 + 630) = 315 \times 631 = 198165). Therefore, the sum is 198,165.
1, 2 and 3 -1, -2, and -3 6 or -6
What is 1/2 of four times Y plus the quantity of Y and 3
It means you multiply some quantity times 2.It means you multiply some quantity times 2.It means you multiply some quantity times 2.It means you multiply some quantity times 2.
2 times the quantity 4 greater than a number
"3 times" 3 * "the quantity" ( something ) "2 plus a" 2 + a. Now combine. "Three times the quantity 2 plus a" 3*(2+a) multiply 2 and a by 3. (distribute) 6 + 3a
3
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n/2 times (n + 1)
Sum of first n integers is n/2 times n + 1 ie 27.5 x 56 which is 1540
The sum of the counting numbers from 1 to 60 can be calculated using the formula for the sum of an arithmetic series: ( S_n = \frac{n}{2} \times (a + l) ), where ( n ) is the number of terms, ( a ) is the first term, and ( l ) is the last term. Here, ( n = 60 ), ( a = 1 ), and ( l = 60 ). Plugging in these values, the sum ( S_{60} = \frac{60}{2} \times (1 + 60) = 30 \times 61 = 1860 ). Therefore, the sum of the counting numbers from 1 to 60 is 1860.