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lisov135 [29]
2 years ago
9

Which choice is equivalent to the quotient shown here for acceptable values of x?

Mathematics
2 answers:
snow_tiger [21]2 years ago
6 0

Answer:

Choice D

Step-by-step explanation:

The division of the two radicals can be re-written in the following format;

\frac{\sqrt{30(x-1)} }{\sqrt{5(x-1)^{2} } }

Using the properties of radicals division, the expression can further be written as;

\sqrt{\frac{30(x-1)}{5(x-1)^{2} } }

We simplify the terms under the radical sign to obtain;

\sqrt{\frac{6}{x-1} }

Choice D is thus the correct solution

Degger [83]2 years ago
3 0

Answer: OPTION D

Step-by-step explanation:

You need to remember this property:

\frac{\sqrt{x} }{\sqrt{y} }=\sqrt{\frac{x}{y} }

And remember that:

\frac{a}{a}=1

Then, the first step is rewrite the expression:

\frac{\sqrt{30(x-1)} }{\sqrt{5(x-1)^2}} =\sqrt{\frac{30(x-1)}{5(x-1)^2}} }

Now, to find the corresponding equivalent expression, you need to simplify the expression.

Therefore, the equivalent expression is the following:

\sqrt{\frac{6}{(x-1)}} }

Finally, you can observe that this matches with the option D.

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The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat
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Answer:

1) 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}.

2) 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

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

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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