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1 year ago

Entre Celia y Quique suman 14€. Si Celia tuviera 1 €, tendría el doble de dinero que Quique.¿ Cuanto dinero tiene cada uno?

Mathematics

2 answers:

denpristay [2]1 year ago

7 0

Answer:

La cantidad de dinero

Celia tiene = x = 9.33 €

Quique tiene = y = 4.67 €

Step-by-step explanation:

Representemos la cantidad de dinero

Celia tiene = x

Quique tiene = y

Entre Celia y Quique suman 14 €.

x + y = 14

x = 14 - y

Si Celia tuviera 1 euro, tendría el doble de dinero que Quique.

Por lo tanto,

x = 2 años

Nosotros sustituimos

2y + y = 14

3 años = 14

y = 14/3

y = 4.67 €

x = 14 € - y

x = 14 € - 4.67 €

x = 9.33 €

Por lo tanto,

La cantidad de dinero

Celia tiene = x = 9.33 €

Quique tiene = y = 4.67 €

slega [8]1 year ago

3 0

Answer:

Celia tiene = x = 9.33 €

Quique tiene = y = 4.67 €

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

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