Half of the product of two consecutive numbers is 105. To solve for n, the smaller of the two numbers, which equation can be used?

A.) n^2 + n - 210 = 0

B.) n^2 + n - 105 = 0

C.) 2n^2 + 2n + 210 = 0

D.) 2n^2 + 2n + 105 = 0

Question

Nayeli Bonilla

Mathematics

25 August, 07:48

Half of the product of two consecutive numbers is 105. To solve for n, the smaller of the two numbers, which equation can be used?

A.) n^2 + n - 210 = 0

B.) n^2 + n - 105 = 0

C.) 2n^2 + 2n + 210 = 0

D.) 2n^2 + 2n + 105 = 0

+1

Answers (1)

Know the Answer?

Chloe Pratt · Answer 1

There are several information's already given in the question. Based on those information's the answer can be detrmined.

Smaller of the two integers = n

Larger of the two integers = n + 1

Then

(1/2) * [ (n) * (n + 1) ] = 105

(1/2) * (n^2 + n) = 105

m^2 + n = 210

n^2 + n - 210 = 0

From the above deduction, it can be concluded that the correct option among all the options that are given in the question is the first option or option "A".

Half of the product of two consecutive numbers is 105. To solve for n, the smaller of the two numbers, which equation can be used?A.) n^2 + n - 210 = 0B.) n^2 + n - 105 = 0C.) 2n^2 + 2n + 210 = 0D.) 2n^2 + 2n + 105 = 0

Half of the product of two consecutive numbers is 105. To solve for n, the smaller of the two numbers, which equation can be used?

A.) n^2 + n - 210 = 0

B.) n^2 + n - 105 = 0

C.) 2n^2 + 2n + 210 = 0

D.) 2n^2 + 2n + 105 = 0