Pregunta
upstudy study bank question image url

20. \( -5 x(-2 x y z)^{0} \) 21. \( \frac{7 a^{2} b^{-3} \times 3 a^{-3} b^{0}}{14 a^{-1} b} \) 22. \( \frac{16 a b^{-2} \times 125 a^{-1} b^{3}}{80 a^{2} b^{5}} \) 23. \( 5\left(2 p q^{3}\right)\left(3 p q^{-2}\right) \) 24. \( \sqrt{5 a p^{2}} \times \sqrt{20 a^{3}} \) 25. \( \frac{8 x^{2} y^{-1} \times 3 x^{3} y^{-5}}{} \)

Ask by Schwartz Macdonald. in South Africa
Feb 26,2025

Solución de inteligencia artificial de Upstudy

Respuesta verificada por el tutor

Responder

Here are the simplified results for each expression: 20. \( -5x \) 21. \( \frac{3}{2b^{4}} \) 22. \( \frac{25}{a^{2}b^{4}} \) 23. \( 30p^{2}q \) 24. \( 10pa^{2} \) 25. \( \frac{24x^{5}}{y^{6}} \)

Solución

Simplify the expression by following steps: - step0: Solution: \(\left(8x^{2}y^{-1}\times 3x^{3}y^{-5}\right)\) - step1: Evaluate: \(8x^{2}y^{-1}\times 3x^{3}y^{-5}\) - step2: Multiply the terms: \(24x^{2}y^{-1}x^{3}y^{-5}\) - step3: Multiply the terms: \(24x^{2+3}y^{-1}\times y^{-5}\) - step4: Add the numbers: \(24x^{5}y^{-1}\times y^{-5}\) - step5: Multiply the terms: \(24x^{5}y^{-1-5}\) - step6: Subtract the numbers: \(24x^{5}y^{-6}\) - step7: Simplify: \(\frac{24x^{5}}{y^{6}}\) Calculate or simplify the expression \( 5 * (2 * p * q^{3}) * (3 * p * q^{-2}) \). Simplify the expression by following steps: - step0: Solution: \(5\left(2pq^{3}\right)\left(3pq^{-2}\right)\) - step1: Evaluate: \(5\times 2pq^{3}\left(3pq^{-2}\right)\) - step2: Remove the parentheses: \(5\times 2pq^{3}\times 3pq^{-2}\) - step3: Multiply the terms: \(30pq^{3}pq^{-2}\) - step4: Multiply the terms: \(30p^{2}q^{3}\times q^{-2}\) - step5: Multiply the terms: \(30p^{2}q^{3-2}\) - step6: Subtract the numbers: \(30p^{2}q\) Calculate or simplify the expression \( -5 * x * (-2 * x * y * z)^{0} \). Simplify the expression by following steps: - step0: Solution: \(-5x\left(-2xyz\right)^{0}\) - step1: Multiply the terms: \(-5x\) Calculate or simplify the expression \( (16 * a * b^{-2} * 125 * a^{-1} * b^{3}) / (80 * a^{2} * b^{5}) \). Simplify the expression by following steps: - step0: Solution: \(\frac{\left(16ab^{-2}\times 125a^{-1}b^{3}\right)}{\left(80a^{2}b^{5}\right)}\) - step1: Evaluate: \(\frac{\left(16ab^{-2}\times 125a^{-1}b^{3}\right)}{80a^{2}b^{5}}\) - step2: Remove the parentheses: \(\frac{16ab^{-2}\times 125a^{-1}b^{3}}{80a^{2}b^{5}}\) - step3: Multiply: \(\frac{2000b}{80a^{2}b^{5}}\) - step4: Reduce the fraction: \(\frac{25}{a^{2}b^{4}}\) Calculate or simplify the expression \( \sqrt(5 * a * p^{2}) * \sqrt(20 * a^{3}) \). Simplify the expression by following steps: - step0: Solution: \(\sqrt{5ap^{2}}\times \sqrt{20a^{3}}\) - step1: Simplify the root: \(p\sqrt{5a}\times \sqrt{20a^{3}}\) - step2: Simplify the root: \(p\sqrt{5a}\times 2a\sqrt{5a}\) - step3: Calculate: \(p\times 2a\times 5a\) - step4: Multiply the terms: \(10pa\times a\) - step5: Multiply the terms: \(10pa^{2}\) Calculate or simplify the expression \( (7 * a^{2} * b^{-3} * 3 * a^{-3} * b^{0}) / (14 * a^{-1} * b) \). Simplify the expression by following steps: - step0: Solution: \(\frac{\left(7a^{2}b^{-3}\times 3a^{-3}b^{0}\right)}{\left(14a^{-1}b\right)}\) - step1: Remove the parentheses: \(\frac{7a^{2}b^{-3}\times 3a^{-3}b^{0}}{14a^{-1}b}\) - step2: Multiply by \(a^{-n}:\) \(\frac{7a^{2}b^{-3}\times 3a^{-3}b^{0}ab^{-1}}{14}\) - step3: Evaluate the power: \(\frac{7a^{2}b^{-3}\times 3a^{-3}\times 1\times ab^{-1}}{14}\) - step4: Calculate: \(\frac{7a^{2}b^{-3}\times 3a^{-3}\times ab^{-1}}{14}\) - step5: Multiply: \(\frac{21b^{-4}}{14}\) - step6: Reduce the fraction: \(\frac{3b^{-4}}{2}\) - step7: Rewrite the expression: \(\frac{\frac{3}{b^{4}}}{2}\) - step8: Simplify: \(\frac{3}{2b^{4}}\) Here are the simplified results for each expression: 20. For the expression \( -5 x(-2 x y z)^{0} \): - Since any non-zero number raised to the power of 0 is 1, we have: \[ -5 x(-2 x y z)^{0} = -5 x \cdot 1 = -5x \] 21. For the expression \( \frac{7 a^{2} b^{-3} \times 3 a^{-3} b^{0}}{14 a^{-1} b} \): - First, simplify the numerator: \[ 7 a^{2} b^{-3} \times 3 a^{-3} b^{0} = 21 a^{2 - 3} b^{-3 + 0} = 21 a^{-1} b^{-3} \] - Now, simplify the entire expression: \[ \frac{21 a^{-1} b^{-3}}{14 a^{-1} b} = \frac{21}{14} \cdot \frac{a^{-1}}{a^{-1}} \cdot \frac{b^{-3}}{b} = \frac{3}{2b^{4}} \] 22. For the expression \( \frac{16 a b^{-2} \times 125 a^{-1} b^{3}}{80 a^{2} b^{5}} \): - Simplifying the numerator: \[ 16 a b^{-2} \times 125 a^{-1} b^{3} = 2000 a^{1 - 1} b^{-2 + 3} = 2000 b^{1} \] - Now, simplify the entire expression: \[ \frac{2000 b}{80 a^{2} b^{5}} = \frac{2000}{80} \cdot \frac{b^{1}}{b^{5}} = \frac{25}{a^{2}b^{4}} \] 23. For the expression \( 5\left(2 p q^{3}\right)\left(3 p q^{-2}\right) \): - Simplifying: \[ 5 \cdot 2 p q^{3} \cdot 3 p q^{-2} = 30 p^{1 + 1} q^{3 - 2} = 30 p^{2} q \] 24. For the expression \( \sqrt{5 a p^{2}} \times \sqrt{20 a^{3}} \): - Using the property of square roots: \[ \sqrt{5 a p^{2} \times 20 a^{3}} = \sqrt{100 a^{4} p^{2}} = 10 a^{2} p \] 25. For the expression \( \frac{8 x^{2} y^{-1} \times 3 x^{3} y^{-5}}{} \): - Simplifying the numerator: \[ 8 x^{2} y^{-1} \times 3 x^{3} y^{-5} = 24 x^{2 + 3} y^{-1 - 5} = 24 x^{5} y^{-6} \] - Since there is no denominator, the result is: \[ \frac{24 x^{5}}{y^{6}} \] In summary, the simplified results are: - 20: \( -5x \) - 21: \( \frac{3}{2b^{4}} \) - 22: \( \frac{25}{a^{2}b^{4}} \) - 23: \( 30p^{2}q \) - 24: \( 10pa^{2} \) - 25: \( \frac{24x^{5}}{y^{6}} \)

Respondido por UpStudy AI y revisado por un tutor profesional

error msg
Explicar
Simplifique esta solución

Bonus Knowledge

Let's start with an engaging tidbit: Did you know that the concept of exponents dates back to ancient civilizations? The Babylonians were already utilizing early forms of exponential notation as far back as 2000 BC! They weren't using letters like we do now, but their methods for calculations laid the groundwork for what would eventually evolve into modern algebra. And here's a fun puzzle for you: When simplifying fractions like the ones in your equations, remember that adding and subtracting exponents isn't as tricky as it seems! A common mistake is misapplying the laws, such as mixing up multiplication with addition. Always keep in mind that in multiplication you add exponents, while in division you subtract them—this little tip can save you a lot of time and frustration!

¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad