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2 years ago

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The linear functions f(x) and g(x) are represented on the graph, where g(x) is a transformation of f(x): A graph with two linear

functions; f of x passes through 5, 0 and 10, 10, and g of x passes through negative 3, 0 and 2, 10. Part A: Describe two types of transformations that can be used to transform f(x) to g(x). (2 points) Part B: Solve for k in each type of transformation. (4 points) Part C: Write an equation for each type of transformation that can be used to transform f(x) to g(x). (4 points)

1 answer:

Julli [10]2 years ago

5 0

Answer:

First, let's find the equations for our lines:

A linear relationship can be written as:

y = a*x + b

where a is the slope and b is the y-axis intercept.

For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:

a = (y2 - y1)/(x2 - x1).

f(x) passes through (5,0) and (10, 10), then the slope is:

a = (10 - 0)/(10 - 5) = 10/5 = 2.

then we have:

y = 2*x + b

And when x = 5, we have y = 0.

0 = 2*5 + b

0 = 10 + b

b = -10

Then the equation for f(x) is:

y = f(x) = 2*x - 10.

Now for g(x) we have the points:

(3, 0) and (2, 10)

a = (10 - 0)/(2 - 3) = -10

y = -10*x + b

0 = -10*3 + b

b = 30.

y = g(x) = -10*x + 30.

A) Ok, the transformations:

Transformation 1 or T1.

f(x) = 2*x - 10

g(x) = -10*x + 30.

Then, we start with f(x):

First, we can move f(x) up 4 units and get:

f'(x) = 2*X - 6

Now we can dilate f(x) with a scale factor of -5 from the origin, now we get:

f''(x) = -5*f'(x) = -10*x + 30.

And this is g(x).

Transformation 2 or T2.

Move f(x) up 10 units, so now we have:

f'(x) = 2*x

Do a reflection over the x-axis, so the sign of y changes, and now we get:

f''(x) = -2*x

Do a dilation of scale factor 5

f'''(x) = 5*-2*x = -10*x

Now do a vertical translation of 30 units up.

f''''(x) = -10*x + 30 = g(x).

These are two transformations that start with f(x) and end with g(x).

B) Ok, as i was writting the transformations i already solved them, so this part is already done.

C) the equation for the transformations are:

T1) g(x) = -5*(f(x) + 4)

T2) g(x) = -(f(x) + 10)*5 + 30

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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:

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

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If we square both sides of the equation we got:

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

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

Replacing we got:

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

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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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