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Rudiy27
2 years ago
11

In the following questions, look at different ways to represent the relation given by the equation y = x2 - 1. The table below s

hows some values for the given equation. Find the values of a and b.
Mathematics
1 answer:
Verdich [7]2 years ago
4 0

Question: In the following questions, look at different ways to represent the relation given by the equation y = x2 - 1. The table below shows some values for the given equation. Find the values of a and b.

Answer:

a= 3

b= 0

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Solve 3x + 2 = 15 for x using the change of base formula log base b of y equals log y over log b. −1.594 0.465 2.406 4.465
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<u>Answer:</u>

The value in 3x + 2 = 15 for x using the change of base formula is 0.465 approximately and second option is correct one.

<u>Solution:</u>

Given, expression is 3^{(x+2)}=15

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Now, let us first apply logarithm for the given expression.

Then given expression turns into as, x+2=\log _{3} 15

By using change of base formula,

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x = 2.4649 – 2  = 0.4649

Hence, the value of x is 0.465 approximately and second option is correct one.

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What is the greatest common factor of x2y and xy2 ?
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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
Greeley [361]

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

8 0
2 years ago
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