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2 years ago

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Volunteers for a political campaign gave out 21/38 of their fliers. They gave out the remaining 612 fliers in another neighborho

od. What is the total number of fliers they gave out

1 answer:

dmitriy555 [2]2 years ago

4 0

The volunteers gave 1368 fliers in total.

Step-by-step explanation:

Given,

Fliers given by volunteers in first neighborhood = $\frac{21}{38}$

Remaining fliers = 612

Fraction of remaining fliers = $1-\frac{21}{38}=\frac{38-21}{38}$

Fraction of remaining fliers = $\frac{17}{38}$

Let,

x be the total fliers given by volunteers.

$\frac{17}{38}\ of\ x=612$

$\frac{17}{38}x=612$

Multiplying both sides by $\frac{38}{17}$

$\frac{38}{17}*\frac{17}{38}x=612*\frac{38}{17}\\\\x=\frac{23256}{17}\\\\x=1368$

The volunteers gave 1368 fliers in total.

Keywords: fraction, division

Learn more about fractions at:

#LearnwithBrainly

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melamori03 [73]

Let's assume they meant C=40 degrees. With an angle like that they're asking for approximation; we'll oblige.

The circumradius is the product of the triangle sides divided by four times the area.

Here we have remaining side given by the Law of Cosines.

$AB^2 = AC^2 + BC^2 - 2\ AC \ BC \cos C$

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3 years ago

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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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