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tekilochka [14]
2 years ago
11

Someone help What is 3.5=12s-55

Mathematics
2 answers:
Vaselesa [24]2 years ago
6 0

Answer:

4.875 = s

Step-by-step explanation:

3.5=12s-55

Add 55 to each side

3.5+55=12s-55+55

58.5 = 12s

Divide each side by 12

58.5/12 = 12s/12

4.875 = s

Brrunno [24]2 years ago
5 0

Answer:

s = 4.875

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS
  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define equation</u>

3.5 = 12s - 55

<u>Step 2: Solve for </u><em><u>s</u></em>

  1. Add 55 to both sides:                    58.5 = 12s
  2. Divide 12 on both sides:                4.875 = s
  3. Rewrite:                                           s = 4.875

<u>Step 3: Check</u>

<em>Plug in s into the original equation to verify it's a solution.</em>

  1. Substitute in <em>s</em>:                    3.5 = 12(4.875) - 55
  2. Multiply:                               3.5 = 58.5 - 55
  3. Subtract:                              3.5 = 3.5

Here we see that 3.5 does indeed equal 3.5.

∴ s = 4.875 is a solution of the equation.

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From which we have;

x^3 + \dfrac{1}{x^3} is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

x + \dfrac{1}{x}

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}

By simplification of the cube of the given integer expressions, we have;

\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )

Therefore, we have;

\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}

By rearranging, we get;

x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )

Given that  x + \dfrac{1}{x} is an integer, from the closure property, the product of two integers is always an integer, we have;

\left ( x + \dfrac{1}{x} \right) ^3 is an integer and 3\cdot \left (x + \dfrac{1}{x} \right ) is also an integer

Similarly the sum of two integers is always an integer, we have;

\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right  ) is an integer

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From which we have;

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Step-by-step explanation:

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