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dolphi86 [110]
3 years ago
5

What is the mode of the following numbers? 2, 5, 11, 7, 5

Mathematics
2 answers:
Romashka-Z-Leto [24]3 years ago
5 0
5 is the mode because wit occurred the most often
WINSTONCH [101]3 years ago
5 0

The mode of this set of numbers is 5.

The mode of a set of numbers, is the most common/frequent number within the set. Since there are two 5's but only one of each other number. Thus making 5 the mode.

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Write an equation for the function
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Answer:

  m(u) = -0.25(u +2)² +1    or    -0.25u² -u

Step-by-step explanation:

The equation is fairly easily written in vertex form, as the vertex point is on a grid line intersection at (-2, 1). The parabola opens downward, so the scale factor is negative.

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What is the sum of the first six terms of the geometric series?
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1. Derive the half-angle formulas from the double
lilavasa [31]

1) cos (θ / 2) = √[(1 + cos θ) / 2], sin (θ / 2) = √[(1 - cos θ) / 2], tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) (x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°). The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) The <em>linear</em> function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

<h3>How to apply trigonometry on deriving formulas and transforming points</h3>

1) The following <em>trigonometric</em> formulae are used to derive the <em>half-angle</em> formulas:

sin² θ / 2 + cos² θ / 2 = 1                      (1)

cos θ = cos² (θ / 2) - sin² (θ / 2)           (2)

First, we derive the formula for the sine of a <em>half</em> angle:

cos θ = 2 · cos² (θ / 2) - 1

cos² (θ / 2) = (1 + cos θ) / 2

cos (θ / 2) = √[(1 + cos θ) / 2]

Second, we derive the formula for the cosine of a <em>half</em> angle:

cos θ = 1 - 2 · sin² (θ / 2)

2 · sin² (θ / 2) = 1 - cos θ

sin² (θ / 2) = (1 - cos θ) / 2

sin (θ / 2) = √[(1 - cos θ) / 2]

Third, we derive the formula for the tangent of a <em>half</em> angle:

tan (θ / 2) = sin (θ / 2) / cos (θ / 2)

tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) The formulae for the conversion of coordinates in <em>rectangular</em> form to <em>polar</em> form are obtained by <em>trigonometric</em> functions:

(x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) Let be the point (x, y) = (2, 3), the coordinates in <em>polar</em> form are:

r = √(2² + 3²)

r = √13

θ = atan(3 / 2)

θ ≈ 56.309°

The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°).

Let be the point (r, θ) = (4, 30°), the coordinates in <em>rectangular</em> form are:

(x, y) = (4 · cos 30°, 4 · sin 30°)

(x, y) = (2√3, 2)

The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) Let be the <em>linear</em> function y = 5 · x - 8, we proceed to use the following <em>substitution</em> formulas: x = r · cos θ, y = r · sin θ

r · sin θ = 5 · r · cos θ - 8

r · sin θ - 5 · r · cos θ = - 8

r · (sin θ - 5 · cos θ) = - 8

r = - 8 / (sin θ - 5 · cos θ)

The <em>linear</em> function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

To learn more on trigonometric expressions: brainly.com/question/14746686

#SPJ1

4 0
2 years ago
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