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Troyanec [42]
3 years ago
9

find two positive even consecutive integers such that a square of the smaller integer is 10 more than the larger integer

Mathematics
1 answer:
nalin [4]3 years ago
6 0

Answer:

the two positive consecutive integers are 4 and 6.

Step-by-step explanation:

Let the smaller integer be s; then s^2 = (s + 2) + 10.

Simplifying, s^2 - s - 2 - 10 = 0, or

s^2 - s - 12 = 0.

Solve this by factoring:  (s - 4)(s + 3) = 0.

Then s = 4 and s = -3.

If the first even integer is 4, the next is 6.  We omit s = -3 because it's not even.

The smaller integer is 4.  Does this satisfy the equation s^2 = (s + 2) + 10?

4^2 = (4 + 2) + 10  True or False?

16 = 6 + 10 = 16.

True.

So the two positive consecutive integers are 4 and 6.

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          \rule{50}{1}\large\textsf{\textbf{\underline{Question:-}}}\rule{50}{1}

        <em>What is the number sentence for "1 less than n is the sum of z and 2?"</em>

<em> </em>

<em>          </em>\rule{50}{1}\large\blue\textsf{\textbf{\underline{Answer and how to solve:-}}}\rule{50}{1}

  ❖ First, Notice it says "1 less than n". This expression indicates that

        we subtract 1 from n, which  looks as follows:-

                                                     \Large\textit{n-1}

❖ Now, It also says "the sum of z and 2". This indicates that we add z and

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❖ Now, Combine these two little expressions into one equation:-

\implies\sf{n-1=z+2}

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Answer:

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Step-by-step explanation:

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