Matemáticas, pregunta formulada por elizabethorozco1710, hace 5 meses

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Adjuntos:

Respuestas a la pregunta

Contestado por ANACF30
1

Respuesta:

SIGNO

 - 5.9 {a}^{2}  {b}^{3} c = menos \:

  - \frac{ \sqrt{3} }{3}  {h}^{4}  {k}^{2}  = menos

abc = mas

 \frac{x {y}^{2} }{4}  = mas

 - 8 {a}^{4}  {c}^{2}  {d}^{3}  = menos

COEFICIENTES

 - 5.9 {a}^{2}  {b}^{ 3} c = 5.9

- \frac{ \sqrt{3} }{3}  {h}^{4}  {k}^{2}  = \frac{  \sqrt{3}  }{3}

abc = 1

 \frac{x {y}^{2} }{4} =  \frac{1}{4}

 - 8 {a}^{4}  {c}^{2}  {d}^{3}  =  8

PARTE LITERAL

 - 5.9 {a}^{2}  {b}^{ 3} c = {a}^{2}  {b}^{3} c

- \frac{ \sqrt{3} }{3}  {h}^{4}  {k}^{2}   =  {h}^{4}  {k}^{2}

abc = abc

 \frac{x {y}^{2} }{4} = x {y}^{2}

 - 8 {a}^{4}  {c}^{2}  {d}^{3} =  {a}^{4}  {c}^{2}  {d}^{3}

EXPONENTES

 - 5.9 {a}^{2}  {b}^{ 3} c = 2 + 3 + 1 = 6

- \frac{ \sqrt{3} }{3}  {h}^{4}  {k}^{2} = 4 + 2 = 6

abc = 1 + 1 + 1 = 3

\frac{x {y}^{2} }{4} = 1 + 2 = 3

 - 8 {a}^{4}  {c}^{2}  {d}^{3} = 4 + 2 + 3 = 7

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