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IceJOKER [234]
3 years ago
6

Find three consecutive odd integers whose sum is 117

Mathematics
2 answers:
irina [24]3 years ago
8 0
37, 39, 41 is the answer
Delvig [45]3 years ago
6 0

First odd integer: 2k+1 (note: using 2k+1 insures it is an odd integer)

Second odd integer: 2k+3

Third odd integer: 2k+5

First + Second + Third = 117

2k+1 + 2k+3 + 2k+5 = 117 <em>substituted for First, Second, and Third</em>

6k + 9 = 117 <em>added like terms</em>

6k = 108 <em> subtracted 7 from both sides</em>

k = 18

First: 2k + 1 = 2(18) + 1 = 36 + 1 = 37

Second: 2k + 3 = 2(18) + 3 = 36 + 3 = 39

Third: 2k + 5 = 2(18) + 5 = 36 + 5 = 41

Answer: 37, 39, 41

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11. Find the solutions of x^2-x-30 = 0.
asambeis [7]

To find the solutions of quadratic equation, there are two ways to do which are:

  • Factorize
  • Quadratic Formula

<u>S</u><u>t</u><u>e</u><u>p</u><u> </u><u>1</u>

- Factor the expression.

To factor the expression, refer below:

\displaystyle \large{ (x - a)(x - b) =  {x}^{2}  - bx - ax + ab}

For bx and ax, both can be common-factored. Therefore

\displaystyle \large{ (x - a)(x - b) =  {x}^{2}  - (b  +  a)x + ab}

From the above, we conclude that:

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  • The last term is a×b
  • Thus, we have to find two numbers that satisfy a+b and a×b

From the expression, 30 comes from 5×6 and when 5-6 = -1. Therefore, a can be 5 and b can be 6.

\displaystyle \large{{x}^{2}  - x - 30 = (x  +  5)(x - 6)}

Because in the middle term, it is -x which is negative, we have to let the highest number become negative.

From the factored expression:

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  • The last term = 5 × (-6) = -30

Then we replace the standard equation with factored form.

\displaystyle \large{ (x  +  5)(x - 6) = 0}

For this part, we solve like a linear equation where we isolate x. Just think you are solving two linear equations!

Hence

\displaystyle \large{ x =  - 5, 6}

Therefore, the solutions are x = -5, 6.

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3 years ago
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