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wlad13 [49]
3 years ago
9

Solve this for fun! I already solved it, so you can give a simple answer.

Mathematics
2 answers:
Vladimir [108]3 years ago
8 0
The answer is abcdefghijklmnopqrstuvwxyz
Ivanshal [37]3 years ago
4 0

is this 8th grade is it is it

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Al factorizar el trinomio cuadrado perfecto, obtenemos el siguiente resultado: (que no se como resolver) xd algun pro que sepa r
PolarNik [594]

Answer:

\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8=\left(\frac{10}{9}m^4p^{6}q^{8}z-\frac{1}{7}mpz^4\right)^2

Step-by-step explanation:

<u>Trinomio Cuadrado Perfecto</u>

El producto notable llamado cuadrado de un binomio se expresa como:

(a-b)^2=a^2-2ab+b^2

Si se tiene un trinomio, es posible convertirlo en un cuadrado perfecto si cumple con las condiciones impuestas en la fórmula:

* El primer término es un cuadrado perfecto

* El último término es un cuadrado perfecto

* El segundo término es el doble del proudcto de los dos términos del binomio.

Tenemos la expresión:

\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8

Calculamos el valor de a como la raiz cuadrada del primer término del trinomio:

\displaystyle a=\sqrt{\frac{100}{81}m^8p^{12}q^{16}z^2}

\displaystyle a=\frac{10}{9}m^4p^{6}q^{8}z

Calculamos el valor de a como la raiz cuadrada del primer término del trinomio:

\displaystyle b=\sqrt{\frac{1}{49}m^2p^2z^8

\displaystyle b=\frac{1}{7}mpz^4

Nos cercioramos de que el término central es 2ab:

\displaystyle 2ab=2\frac{10}{9}m^4p^{6}q^{8}z\frac{1}{7}mpz^4

Operando:

\displaystyle 2ab=\frac{20}{63}m^5p^7q^8z^5

Una vez verificado, ahora podemos decir que:

\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8=\left(\frac{10}{9}m^4p^{6}q^{8}z-\frac{1}{7}mpz^4\right)^2

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A function can be reflected about an axis by multiplying by negative one. To reflect about the y-axis, multiply every x by -1 to get -x. To reflect about the x-axis, multiply f(x) by -1 to get -f(x).

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