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Helga [31]
2 years ago
6

Solve the inequality for u.​

Mathematics
1 answer:
daser333 [38]2 years ago
6 0

Answer:

u ≤ 8

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtract Property of Equality

Step-by-step explanation:

<u>Step 1: Define</u>

32 ≥ 4u

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

  1. Divide both sides by 4:                     8 ≥ u
  2. Rewrite:                                              u ≤ 8

Here we see that any value <em>u</em> smaller than or equal to 8 would work as a solution to the inequality.

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Interval Notation: ( -2, ∞)

Step-by-step explanation:

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Step-by-step explanation:

\sqrt{ x^{2}-10x+25}+25+12\sqrt{x} =15\sqrt{x} \\  \\  \therefore \: \sqrt{ x^{2}-10x+ {5}^{2} }+25 =15\sqrt{x} - 12\sqrt{x}\\  \\  \therefore \: \sqrt{( x - 5)^{2}}+25 =3\sqrt{x}  \\  \\ \therefore \:  x - 5+ 25 = 3 \sqrt{x}  \\  \\ \therefore \: x + 20 = 3 \sqrt{x}  \\  \\ squaring \: both \: sides \\ (x + 20)^{2}  = ( {3 \sqrt{x} })^{2}  \\ \therefore \:  {x}^{2}  + 40x + 400 = 9x \\ \therefore \:  {x}^{2}  + 40x + 400  - 9x = 0 \\  \therefore \:  {x}^{2}  + 31x + 400  = 0 \\

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Answer:

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Step-by-step explanation:

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