Let the consecutive integers be 'n' & 'n+1'. Their product is n(n+1) Their sum is n + n+ 1 = 2n + 1 Eight timres the sum is 8(2n+1) Their product being 8 less that their sum is n(n+1) = 8(2n+1) - 8 Hence n^(2) + n = 16n + 8 - 8 n^(2) + n = 16n n^(2) = 15n ( Cancel down by 'n'). n = 15 n+1 = 16 Hence '15' & '16' are the two numbers.
Product of 320 and 16 is 5120.
The product of 416 and 16 is 6656.
let n = {...-2,-1,0,1,2,3,4,...} then n+1 is the consecutive number. Their product is n(n+1) = n2 + n
(N/8)=(12/16) N= (12*8/16) N=6
Let the number be 'm' & 'n' Hence Product ; mn = 80 Sum ; m + n = 21 m = 21 - n Substitute (21 - n) n = 80 21n - n^2 = 80 n^2 - 21 n + 80 = 0 It is now in quadratic form to factor. ( n - 16)(n - 5) = 0 Hence the numbers are 5 & 16
Let the consecutive integers be 'n' & 'n+1'. Their product is n(n+1) Their sum is n + n+ 1 = 2n + 1 Eight timres the sum is 8(2n+1) Their product being 8 less that their sum is n(n+1) = 8(2n+1) - 8 Hence n^(2) + n = 16n + 8 - 8 n^(2) + n = 16n n^(2) = 15n ( Cancel down by 'n'). n = 15 n+1 = 16 Hence '15' & '16' are the two numbers.
n + 16 = 25 Subtract '16' from both sides. n + 16 - 16 = 25 - 16 Collect terms n = 9
A product of 3 and N would be 3 and N multiplied together, so the product would be 3N. To get a numeric answer, you would first need to find what the value of N is.
(n+16+16+18)/4 = 17 n+16+16+18 = 17*4 n+50 = 68 n = 68-50 n = 18
The product refers to the answer. 2 X 8 = 16 16 is the product
Product of 320 and 16 is 5120.
The product of 416 and 16 is 6656.
16-n
let n = {...-2,-1,0,1,2,3,4,...} then n+1 is the consecutive number. Their product is n(n+1) = n2 + n
(N/8)=(12/16) N= (12*8/16) N=6
It is expressed as n+16. For example, for n = 2 the result is 18. For n = 3 the result is 19.