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MrRissso [65]
3 years ago
5

Determina el termino que hace falta en cada trinomio, de modo que sea trinomio cuadrado perfecto. luego, factorizalo

Mathematics
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

1) El término faltante es b^{2} = 9. La forma factorizada del polinomio es (\pm 3 \cdot n^{3} \mp 3)^{2}, 2) El término faltante es \pm 10^{\frac{3}{2} }\cdot x^{\frac{5}{2}\cdot n }. La forma factorizada puede ser (\pm \sqrt{10} \cdot x^{\frac{5}{2}\cdot n} \pm 5)^{2} o (\pm \sqrt{10} \cdot x^{\frac{5}{2}\cdot n} \mp 5)^{2}, 3) El término faltante es \frac{1}{4} \cdot y ^{2\cdot n}. La forma factorizada es (\pm 3 \cdot x^{n} \pm \frac{1}{2}\cdot y^{n})^{2}.

Step-by-step explanation:

(This exercise is presented in Spanish and therefore explanation will be held in such language)

Un trinomio cuadrado perfecto es una expresión algebraica que responde a la siguiente identidad:

(a + b)^{2} = a^{2} + 2 \cdot a \cdot b + b^{2}

Este ejercicio pide completar cada trinomio cuadrado perfecto con el componente faltante:

1) 9\cdot n^{6} - 18 \cdot n^{3}...

El término central implica el producto

a = \pm 3\cdot n^{3}

2\cdot a \cdot b = - 18 \cdot n^{3}

2\cdot (\pm 3\cdot n^{3})\cdot b = -18\cdot n^{3}

\pm 6\cdot n^{3}\cdot b = -18 \cdot n^{3}

b = \mp \frac{18\cdot n^{3}}{6\cdot n^{3}}

b = \mp 3

El término faltante es b^{2} = 9. La forma factorizada del polinomio es (\pm 3 \cdot n^{3} \mp 3)^{2}.

2) 10\cdot x^{5\cdot n} + ... + 25

a^{2} = 10\cdot x^{5\cdot n}

a_{1} = +\sqrt{10}\cdot x^{\frac{5}{2}\cdot n } o a_{2} = -\sqrt{10}\cdot x^{\frac{5}{2}\cdot n }

b^{2} = 25

b_{1} = 5 o b_{2} = -5

Existen dos posibles soluciones (y cuatro combinaciones posibles):

Posibilidad 1:

2\cdot a_{1}\cdot b_{1} =  2\cdot (\sqrt{10}\cdot x^{\frac{5}{2}\cdot n }) \cdot  (5)

2\cdot a_{1}\cdot b_{1} = 10^{\frac{3}{2} }\cdot x^{\frac{5}{2}\cdot n }

Posibilidad 2:

2\cdot a_{2}\cdot b_{1} =  2\cdot (-\sqrt{10}\cdot x^{\frac{5}{2}\cdot n }) \cdot  (5)

2\cdot a_{1}\cdot b_{1} = -10^{\frac{3}{2} }\cdot x^{\frac{5}{2}\cdot n }

El término faltante es \pm 10^{\frac{3}{2} }\cdot x^{\frac{5}{2}\cdot n }. La forma factorizada puede ser (\pm \sqrt{10} \cdot x^{\frac{5}{2}\cdot n} \pm 5)^{2} o (\pm \sqrt{10} \cdot x^{\frac{5}{2}\cdot n} \mp 5)^{2}.

3) 9\cdot x^{2\cdot n} + 3\cdot x^{n}\cdot y^{n} +...

a = \pm 3\cdot x^{n}

2\cdot a \cdot b = 3\cdot x^{n}\cdot y^{n}

2\cdot (\pm 3 \cdot x^{n})\cdot b = 3 \cdot x^{n}\cdot y^{n}

\pm 6 \cdot x^{n}\cdot b = 3 \cdot x^{n}\cdot y^{n}

b = \pm \frac{3\cdot x^{n}\cdot y^{n}}{6\cdot x^{n}}

b = \pm \frac{1}{2}\cdot y^{n}

b^{2} = \frac{1}{4}\cdot y^{2\cdot n}

El término faltante es \frac{1}{4} \cdot y ^{2\cdot n}. La forma factorizada es (\pm 3 \cdot x^{n} \pm \frac{1}{2}\cdot y^{n})^{2}.

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Answer:

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Step-by-step explanation:

As the table of the relation is given as follows:

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<em>10               5</em>

<em>15               9</em>

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Domain of a relation consists of all the x-coordinates (first elements) of order pairs.

Range of a relation consists of all the y-coordinates (second elements) of ordered pairs.

So, domain  and range of relation will be as follows:

                                  Domain : {10, 15, 19, 32}

                                  Range: {-1, 5, 9}

<em>Note: If there is any </em><em>duplicate</em><em> element in any x or y-coordinate of any ordered pair, it will be written only </em><em>once </em><em>when we determine domain and range. Here, in this example, 5 is duplicate, so, it will be mentioned only one time when we determine the range of this relation.</em>

Keywords:  domain, relation, range

Learn more about domain and range of a relation from brainly.com/question/11422136

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