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2 years ago

On a coordinate plane, a triangle has points (negative 5, 1), (2, 1), (2, negative 1).

Mathematics

2 answers:

daser333 [38]2 years ago

5 0

Answer:

Step-by-step explanation:

Given a triangle has points:

(-5,1),(2,1), (2,-1)

Let us label the points:

A(2,1),

B(-5,1) and

C(2,-1)

To find:

Distance between (−5, 1) and (2, −1) i.e. BC.

Horizontal leg AB and

Vertical leg, AC.

Solution:

Please refer to the attached diagram for the labeling of the points on xy coordinate plane.

We can simply use Distance formula here, to find the distance between two coordinates.

Distance formula :

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

For BC:

$x_2 = 2\\x_1 = -5\\y_2 = -1\\y_1 = 1$

$BC = \sqrt{(2--5)^2+(-1-1)^2} = \sqrt{63}$

Horizontal leg, AC:

$x_2 = -5\\x_1 = 2\\y_2 = 1\\y_1 = 1$

$AC = \sqrt{(2-(-5))^2+(1-1)^2} = 7$

Vertical Leg, AB:

$x_2 = 2\\x_1 = 2\\y_2 = -1\\y_1 = 1$

$AB = \sqrt{(2-2)^2+(-1-1)^2} = 2$

Annette [7]2 years ago

3 0

Answer:

The answer are 7, 2 and 53

Step-by-step explanation:

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Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

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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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