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mixer [17]
3 years ago
7

Si agrupo mis lápices de 2 en dos me sobra 1; si lo hago de 3 en 3, también me sobra 1 y si lo hago de 5 en 5 ocurre los mismo.¿

Cuántos lápices tengo si son más de 30 y menos de 40?
Mathematics
1 answer:
Vinvika [58]3 years ago
7 0

Answer:

31 pencils

Step-by-step explanation:

What we must do is reduce the range of responses, the first thing is to limit the numbers to the interval between 30 and 40.

If it is two by two and there is one left, it means that all the pairs are eliminated.

If it is five in five and there is one left over, the ones ending in 5 are also eliminated.

If it is three by three and there is one left over, the 33 and 39 are eliminated.

Taking this into account, the possibilities are reduced to two numbers only, 31 and 37.

37 is discarded because when it is grouped from five to five, there are 2 left over.

So the answer is 31, you have 31 pencils.

 

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pav-90 [236]

Answer:

g(n) = 10 * 3^{n}

Step-by-step explanation:

Given

See attachment for table

Required

Determine the explicit formula for the table

First, we need to check if the guppies increases at arithmetic progression or geometry progression.

For arithmetic progression:

We calculate the common difference (d)

d = g(n) - g(n-1)

Take n as 1

d = g(1) - g(1-1)

d = g(1) - g(0)

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g(1) = 30\ \&\ g(0) = 10

d = 30 - 10

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Take n as 2

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From the table:

g(1) = 30\ \&\ g(2) = 90

d = 90 - 30

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For geometry progression:

We calculate the common ratio (r)

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r = \frac{g(1)}{g(1-1)}

r = \frac{g(1)}{g(0)}

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From the table:

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Because the progression starts from 0, we make use of the following formula

g(n) = g(0) * r^{n}

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The explicit formula is: g(n) = 10 * 3^{n}

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2 years ago
Geometry - Similar
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Answer:

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Step-by-step explanation:

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