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IgorLugansk [536]
3 years ago
11

El horario y el minutero de un reloj miden respectivamente 0.7 y 1.2cm. Determinar la distancia entre los extremos de dichas man

ecillas a las 13:30 horas

Mathematics
1 answer:
Fed [463]3 years ago
8 0

Answer:

1.5887 cm

Step-by-step explanation:

If the time is 13:30, The hour pointer will be between the numbers 1 and 2.

Each number of the 12 number clock is separated by 30 degrees, so the hour pointer is at 45 degrees.

The minute pointer is at 30 minutes, so it is in number 6, being therefore at angle 180.

The angle formed by the pointers will be 180 - 45 = 135 degrees.

We can draw a triangle with both pointers and this angle (image attached)

So, we can calculate the distance between the tip of the pointers (let's call it "d") using law of cosines:

d^2 = 0.7^2 + 1.2^2 - 0.7*1.2*cos(135)

d^2 = 0.49 + 1.44 - (-0.5940) =  2.524

d = 1.5887 cm

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Factors are the numbers you multiply together to get another number. when you find the factors of two or more numbers, and then find some of the numbers (factors) are the same (common) you have common factors.

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I need help with this problem! Please Help<br> t=-2.74 + sqrt(2.74^(2)-4*-4.9*(-10)) -: 2*-4.9
Ulleksa [173]

Answer:

Step-by-step explanation:

t=(-2.74 + sqrt(2.74^(2)-4*-4.9*(-10))) / -2*-4.9

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3 years ago
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Answer:

Depending on how the input of each function defined,

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Step-by-step explanation:

A function between two sets (domain and range) should

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The second choice can't be a function since the element a from the domain is mapped to more than one element in the range.

Keep in mind that a function should be defined for all elements in its domain. For the first relation to be a function, its domain needs to be \lbrace a,\, 6, \, C\rbrace. Similarly, the domain for the third and fourth relations should be \lbrace 1,\, 2, \, 3\rbrace and \lbrace a,\, b, \, c\rbrace

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