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Greeley [361]
3 years ago
12

(01.05 MC)The total charge on 7 particles is -42 units. All the particles have the same charge.

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
4 0

Answer:

Unit\ Charge = -6\ units

Step-by-step explanation:

<em>Question is incomplete; However, the missing part of the question will most likely be to calculate the charges on each particle.</em>

<em></em>

If that is the case;

Tota\l Charges = -42\ units

We'll get the unit charge when we divide by 7.

This is so because, they all have the same charge.

Unit\ Charge = -42\ units/7

Unit\ Charge = -6\ units

<em>Hence;</em>

<em>the unit charge is -6 units</em>

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X/z+y^2-x evaluate where x=-2 y=4 and z=16
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Answer:

Solution

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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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Good!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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