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gladu [14]
2 years ago
8

He repartido mi colección de canicas entre mis tres amigos. A thales le he dado 1/5 del total, a arquimedes un 1/3 del resto y p

or último a pitagoras le he regalado las 16 canicas que me quedaban
Mathematics
1 answer:
elena-14-01-66 [18.8K]2 years ago
4 0

Answer:

Ok, supongamos que en total hay N canicas.

1/5 de esas canicas se le dieron a Thales.

Entonces el resto ahora es:

1 - 1/5 = 4/5.

1/3 de esas canicas se le dieron a Arquimedes.

Entonces Arquimedes recibe:

(1/3)(4/5) = 4/15 de las canicas totales.

El resto, que eran 16, se le dieron a Pitagoras.

Ahora, las dos primeras entregas fueron:

1/5 + 4/15 = 3/15 + 4/15 = 7/15.

El resto seria:

1 - 7/15 = 8/15.

Y sabemos que el resto es igual a 16 canicas, entonces tenemos que:

(8/15)*N = 16.

De aca podemos calcular el numero total de canicas.

N = 16*(15/8) = 30

Esto quiere decir que inicialmente hay 30 canicas.

Thales recibe 30*(1/5) = 7 canicas.

Arquimedes recibe 30*(4/15) = 8 canicas.

Pitagoras recibe 16 canicas.

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\vec s_v\times\vec s_u=u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k

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\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_1^2(-u\cos v\,\vec\imath-u\sin v\,\vec\jmath+u^3\,\vec k)\cdot(u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k)\,\mathrm du\,\mathrm dv

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C. Phonograph were inferior in sound quality.

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(a) 6 units

(b) 4 units

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Step-by-step explanation:

The distance d from one point (x₁, y₁, z₁) to another point (x₂, y₂, z₂) is given by;

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Now from the question;

<em>(a) The distance from (4, -7, 6) to the xy-plane</em>

The xy-plane is the point where z is 0. i.e

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Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, -7, 0)</em>

d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]

d = √[(0)² + (0)² + (-6)²]

d = √(-6)²

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d = 6

Hence, the distance to the xy plane is 6 units

<em>(b) The distance from (4, -7, 6) to the yz-plane</em>

The yz-plane is the point where x is 0. i.e

yz-plane = (0, -7, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 6)</em>

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d = √(4)²

d = √16

d = 4

Hence, the distance to the yz plane is 4 units

<em>(c) The distance from (4, -7, 6) to the xz-plane</em>

The xz-plane is the point where y is 0. i.e

xz-plane = (4, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 6)</em>

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Hence, the distance to the xz plane is 7 units

<em>(d) The distance from (4, -7, 6) to the x axis</em>

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d = √[(49 + 36)]

d = √(85)

d = 9.22

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d = √[(16 + 36)]

d = √(52)

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<em>(f) The distance from (4, -7, 6) to the z axis</em>

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d = √[(16 + 49)]

d = √(65)

d = 8.06

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