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Which expression is equivalent to (StartFraction (2 a Superscript negative 3 Baseline b Superscript 4 Baseline) squared Over (3 a Superscript 5 Baseline b) Superscript negative 2 Baseline EndFraction) Superscript negative 1? Assume mc026-2. Jpg.

Respuesta :

thohiele

Answer:

1

Explanation:

The expression that is equivalent to [(2a⁻³b⁴)²/(3a⁵b)⁻²]⁻¹ is;

1/(36a⁴b¹⁰)

  • The given expression is;

[(2a⁻³b⁴)²/(3a⁵b)⁻²]⁻¹

  • Rewriting this equation gives;

[(2a⁻³b⁴)² × (3a⁵b)²]⁻¹

Let us simplify (2a⁻³b⁴)² to get;

4b⁸/a⁶

Similarly for (3a⁵b)²;

⇒ 9a¹⁰b²

Thus, we now have;

[(4b⁸/a⁶) × 9a¹⁰b²]⁻¹

  • From laws of indices, we know that; x⁵/x³ = x⁽⁵ ⁻ ³⁾ = x²

Thus; a¹⁰/a⁶ = a⁴

Similarly  b⁸ × b² = b⁽⁸ ⁺ ²⁾ = b¹⁰

  • Thus, we now have;

[(4b⁸/a⁶) × 9a¹⁰b²]⁻¹ = (36a⁴b¹⁰)⁻¹

Finally, we now have;

(36a⁴b¹⁰)⁻¹ = 1/(36a⁴b¹⁰)

Read more about simplification of algebra fractions at; https://brainly.com/question/15514217

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