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anygoal [31]
3 years ago
11

The sum of two consecutive whole numbers is 57. What is the value of the greater of the two numbers

Mathematics
1 answer:
KIM [24]3 years ago
5 0

n,\ n+1-\text{ two consecutive whole numbers}\\57-\text{the sum}\\\\n+n+1=57\qquad\text{subtract 1 from both sides}\\2n=56\qquad\text{divide both sides by 2}\\n=28\\\\n+1=28+1=29\\\\Answer:\ \boxed{\text{The greater of the two numbers is 29}}

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3 years ago
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xz_007 [3.2K]

Answer:

L.S = R.S ⇒ Proved down

Step-by-step explanation:

Let us revise some rules in trigonometry

  1. sin²α + cos²α = 1
  2. sin2α = 2 sin α cosα
  3. cscα = 1/sinα

To solve the question let us find the simplest form of the right side and the left side, then show that they are equal

∵ L.S = csc2α + 1

→ By using the 3rd rule above

∴ L.S = \frac{1}{sin2\alpha} + 1

→ Change 1 to \frac{sin2\alpha}{sin2\alpha}

∴ L.S = \frac{1}{sin2\alpha} + \frac{sin2\alpha}{sin2\alpha}

→ The denominators are equal, then add the numerators

∴ L.S = \frac{1+sin2\alpha}{sin2\alpha}

∵ R. S = \frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}

∵ (sinα + cosα)² = sin²α + 2 sinα cosα + cos²α

∴ (sinα + cosα)² = sin²α + cos²α + 2 sinα cosα

→ By using the 1st rule above, equate sin²α + cos²α by 1

∴ (sinα + cosα)² = 1 + 2 sinα cosα

→ By using the 2nd rule above, equate 2 sinα cosα by sin2α

∴ (sinα + cosα)² = 1 + sin2α

→ Substitute it in the R.S above

∴ R. S = \frac{1+sin2\alpha}{sin2\alpha}

∵ L.S = R.S

∴ csc 2α + 1 = \frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}

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Step-by-step explanation:

2x + 2x + 2 = 4x + 2           <em>combine like terms</em>

4x + 2 = 4x + 2          <em>subtract 4x from both sides</em>

4x - 4x + 2 = 4x - 4x + 2

2 = 2   TRUE

<h2>CONCLUSION:</h2><h3>This equation has an infinitely many solutions. The solution to this equation is any real number.</h3>
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Answer:

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