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Vitek1552 [10]
2 years ago
10

How many triangles are formed by the angles and sides (45°, 108°, 7 cm)--unique triangle, more than one triangle, or no triangle

?

Mathematics
1 answer:
Kruka [31]2 years ago
5 0

Answer:

no triangle triangle is formed

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luda_lava [24]

Answer:

It is a right angled triangle.

so

tan32 = p/b

or, tan32 = x/5

or, tan32×5 = x

so, x = 3.124

for 5 no. the answer is a. 3.124

For y,

cos32 = b/h

or, cos32 = 5/y

or, y = 5/cos32

so, y = 5.895

for 6 no. 5.895 is closest to 5.994, so the answer is c. 5.994.

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What is a real life example of trigonometric ratios
Brut [27]

Answer:

Trigonometry can be used to measure the height of a building or mountains

Step-by-step explanation:

if you know the distance from where you observe the building and the angle of elevation you can easily find the height of the building. Similarly, if you have the value of one side and the angle of depression from the top of the building you can find and another side in the triangle, all you need to know is one side and angle of the triangle.

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PLEASE HELP!!!!
Lunna [17]

NOTES:

Given a quadratic function in standard format (y = ax² + bx + c), the direction of the parabola is as follows:

  • if "a" is positive, then opens UP
  • if "a" is negative, then opens DOWN

Given a quadratic function in standard format (y = ax² + bx + c), the vertex can be found as follows:

  • the Axis Of Symmetry (x-value) is: x = \frac{-b}{2a}  
  • y-value is found by plugging in the AOS for "x" in the equation

****************************************************************************************

1) y = x² + 11x + 24

  • a = +1 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-11}{2(1)}  = -\frac{11}{2}
  • y = (-\frac{11}{2})^{2} + 11(-\frac{11}{2} ) + 24 = -\frac{25}{4}
  • vertex (-\frac{11}{2}, -\frac{25}{4}) is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-3, 0) and (-8, 0), and y-intercept (0, 24)

******************************************************************************************

2) y = -x² - 6x - 8

  • a = -1 so the parabola opens DOWN
  • x = \frac{-b}{2a} = \frac{-(-6)}{2(-1)}  = \frac{6}{-2} = -3
  • y = -(-3)² - 6(-3) - 8 = -9 + 18 - 8 = 1
  • vertex (-3, 1) is in Quadrant 2 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-4, 0), and y-intercept (0, -8)

******************************************************************************************

3) y = x² - 2x + 3

  • a = 1 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-(-2)}{2(1)}  = \frac{2}{2} = 1
  • y = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2
  • vertex (1, 2) is in Quadrant 1 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 3), and its mirror image (2, 3). <em>There are no x-intercepts</em>

******************************************************************************************

4) y = x² + 4x + 4

  • a = 1 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-4}{2(1)}  = -2
  • y = (-2)² + 4(-2) + 4 = 4 - 8 + 4 = 0
  • vertex (-2, 0) is in Quadrant 2 and is on the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 4), and its mirror image (-4, 4). <em>The x-intercept is the vertex.</em>

Compared to the other four graphs, this is most likely the equation for the rain gauge!

******************************************************************************************

5) y = 3x² + 21x + 30

  • a = +3 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-21}{2(3)}  = -\frac{7}{2}
  • y = 3(-\frac{7}{2})^{2} + 21(-\frac{7}{2} ) + 30 = -\frac{27}{4}
  • vertex (-\frac{7}{2}, -\frac{27}{4}) is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-5, 0), and y-intercept (0, 30)

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4 0
3 years ago
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Step-by-step explanation:


8 0
3 years ago
A figure has vertices at L(3,3), M(2,4), and N(−1,0). After a transformation, the image of the figure has vertices at L′(2,−5),
alexandr402 [8]

The transformation was a reflection over the line x = -0.5

And the image and preimage can be seen in the graph below.

<h3>How to identify the preimage and image?</h3>

First, the original vertices are:

L(3, 3)

M(2, 4)

N(-1, 0)

3 vertices means that our figure is a triangle.

Now, the transformed vertices are:

L'(2, -5)

M'(3, -4)

N'(-1, 1)

And the graph of both sets can be seen below.

Now we can try to identify the transformation.

Now, notice that all the points (L, M, N, L', N', M') are equidistant to the horizontal line x = -0.5

This is a clear implication that the transformation was a reflection over the line x = -0.5

If you want to learn more about transformations:

brainly.com/question/4289712

#SPJ1

3 0
1 year ago
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