Math Problem Statement

p = 11 + sqrt(120 / (2π)). Calculate p√p + 1/(p√p).

Solution

Let's go through this step-by-step.

Step 1: Understand the Given Expressions

The problem involves two parts:

  1. The value of pp: p=11+1202πp = 11 + \sqrt{\frac{120}{2\pi}}
  2. An expression to evaluate: pp+1ppp\sqrt{p} + \frac{1}{p\sqrt{p}}

Step 2: Simplify the Expression for pp

First, we simplify pp as follows:

p=11+1202π=11+60πp = 11 + \sqrt{\frac{120}{2\pi}} = 11 + \sqrt{\frac{60}{\pi}}

To proceed further, let's calculate the value of 60π\sqrt{\frac{60}{\pi}}.

Step 3: Calculate 60π\sqrt{\frac{60}{\pi}}

Assuming π3.14159\pi \approx 3.14159: 60π=603.1415919.09864.37\sqrt{\frac{60}{\pi}} = \sqrt{\frac{60}{3.14159}} \approx \sqrt{19.0986} \approx 4.37

So, we get: p11+4.37=15.37p \approx 11 + 4.37 = 15.37

Step 4: Substitute pp into the Expression

Now we need to evaluate: pp+1ppp\sqrt{p} + \frac{1}{p\sqrt{p}}

  1. Calculate ppp\sqrt{p}: p=15.373.92\sqrt{p} = \sqrt{15.37} \approx 3.92 pp15.37×3.9260.25p\sqrt{p} \approx 15.37 \times 3.92 \approx 60.25

  2. Calculate 1pp\frac{1}{p\sqrt{p}}: 1pp160.250.0166\frac{1}{p\sqrt{p}} \approx \frac{1}{60.25} \approx 0.0166

  3. Add the two results: pp+1pp60.25+0.016660.27p\sqrt{p} + \frac{1}{p\sqrt{p}} \approx 60.25 + 0.0166 \approx 60.27

Final Answer:

pp+1pp60.27p\sqrt{p} + \frac{1}{p\sqrt{p}} \approx 60.27

Would you like a more detailed explanation of any step, or do you have any further questions?


Here are five additional related questions you might find interesting:

  1. What is the significance of the square root operation in simplifying expressions?
  2. How can this approach be extended to other similar problems involving roots and powers?
  3. How would the expression change if the square root of pp were a different root, such as a cube root?
  4. What are some real-world applications of calculating expressions like pp+1ppp\sqrt{p} + \frac{1}{p\sqrt{p}}?
  5. What methods are available for approximating p\sqrt{p} without a calculator?

Tip: When working with square roots in fractions, simplifying the fraction as much as possible before taking the square root can make calculations easier.

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Math Problem Analysis

Mathematical Concepts

Algebra
Roots and Radicals
Square Root
Numerical Approximation

Formulas

p = 11 + sqrt(120 / (2π))
Expression: p√p + 1/(p√p)

Theorems

Radical Simplification
Square Root Approximation

Suitable Grade Level

Grade 10-12