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kobusy [5.1K]
3 years ago
9

Lucio quiere repartir su colección de estampillas entre la mayor cantidad de amigos pero de manera que todos reciban la misma ca

ntidad tiene parar repartir 30 estampillas de animales 75 estampillas de flores y 160 de ciudades del mundo cuantos grupos iguales puede armar
Mathematics
1 answer:
Daniel [21]3 years ago
4 0

Answer:

Lucio podrá armar 5 grupos iguales de estampillas.

Step-by-step explanation:

El problema plantea un caso de distribución mediante la aplicación del divisor común mayor a todos los números involucrados. Para ello, primero debemos analizar los divisores de cada uno de estos números:

-30: divisible por 2, 3, 5, 6, 10, 15 y 30.

-75: divisible por 3, 5, 15, 25 y 75.

-160: divisible por 2, 4, 5, 8, 10, 16, 20, 32, 40, 80 y 160.

Así, podemos ver que el divisor común mayor a los tres números es 5, ya que 30, 75 y 160 dan como resultado un número entero tras ser divididos por 5.

Entonces, para poder repartir equitativamente sus estampillas entre la mayor cantidad de amigos posibles, deberá repartir 6 estampillas de animales, 15 estampillas de flores y 32 estampillas de ciudades a 5 amigos. De esta manera, Lucio podrá armar 5 grupos iguales de estampillas.

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nikdorinn [45]

Answer:

<em>In 5 years the product of their ages will be 208</em>

Step-by-step explanation:

The age of two children is 11 and 8 years.

Let's call x the number of years ahead.

We need to find when the product of their future ages is 208. The 11 years old child will be 11+x years old and the other child will be 8+x years, thus:

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Answer:

a

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2

+

b

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+

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b

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−

4

a

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a

(

1

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4

x

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3

=

0

cmp

a

x

2

+

b

x

+

c

=

0

a

=

1

b

=

4

c

=

3

(

2

)

Substitute these numbers into eh formula

x

=

−

4

±

√

4

2

−

(

4

×

1

×

3

)

2

×

1

(

3

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x

=

−

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±

√

16

−

12

2

x

=

−

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±

√

4

2

x

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−

4

±

2

2

now calculate the two separate solutions

x

1

=

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+

2

2

=

−

2

2

=

−

1

x

2

=

−

4

−

2

2

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−

6

2

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