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Anton [14]
2 years ago
7

PLEASE HELP ME!!!!!!

Mathematics
1 answer:
Nata [24]2 years ago
8 0

9514 1404 393

Answer:

  vertex: (2, -3.2)

  axis of symmetry: x = 2

  zeros: x=0, x=4

  formula for axis of symmetry:

  • x = <x-coordinate of vertex>
  • x = -b/(2a) . . . . . . . where the quadratic is ax^2 +bx +c
  • x = average of zeros, (z1 +z2)/2

Step-by-step explanation:

a) The vertex x-coordinate is halfway between the zeros, so is x = (0+4)/2 = 2. It is where the graph has a turning point.

The vertex y-coordinate is the low point of this graph. It is not on a grid line, so we can only guess at its value. I choose to call it -3.2.

The vertex is (2, -3.2).

__

b) The axis of symmetry is the vertical line through the vertex. The constant in its equation is the x-coordinate of the vertex:

  x = 2

__

c) The zeros of the function are where the function crosses the x-axis. These are marked on the graph with red dots. The zeros are x=0 and x=4.

__

d) When the quadratic is written in standard form, ax^2 +bx +c, the equation of the axis of symmetry is ...

  x = -b/(2a) . . . . . perhaps this is the formula you're being asked for (?)

When there is no equation, the axis of symmetry can be found from the graph a couple of ways. One is to identify the x-coordinate of the vertex.

  x = <x-coordinate of the vertex>

Another is to average the zeros, since they are symmetrical about the axis of symmetry.

  x = average of zeros = (z1 +z2)/2

If the quadratic is written in vertex form, the vertex coordinate is the constant in the equation for the axis of symmetry.

  y = a(x -h)^2 +k . . . . . quadratic with vertex (h, k)

  x = h . . . . . . equation of axis of symmetry

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Options A,B and D. The correction about the area of the triangle are

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<h3>How to solve for the length of the Apothem using the Pythagoras theorem</h3>

Firstly this is an equilateral triangle

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<u>For B,</u>

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<h3>using tangent ratios </h3>

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This would also give us an approximate of 2.5

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Read more on triangles here:

brainly.com/question/18404158

#SPJ1

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