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

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".

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TestTime4Me TestTime4Me · Answer 2 · 2020-01-21T18:57:28+01:00

TestTime4Me TestTime4Me

21-01-2020

Answer:

n^2+n-210=0

Step-by-step explanation:

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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

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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