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Nadusha1986 [10]
2 years ago
13

Which is the radian measure of central angle KOL? StartFraction 3 pi Over 8 EndFraction radians StartFraction 3 pi Over 4 EndFra

ction radians StartFraction 7 pi Over 8 EndFraction radians StartFraction 13 pi Over 5 EndFraction radians
Mathematics
2 answers:
Anton [14]2 years ago
6 0

Answer: B

Step-by-step explanation:

bagirrra123 [75]2 years ago
5 0

The above question is incomplete.

Complete Question

Minor arc KL measures 135°. Circle O is shown. Line segments K O and O L are radii. Which is the radian measure of central angle KOL?

a) StartFraction 3 pi Over 8 EndFraction radians

b) StartFraction 3 pi Over 4 EndFraction radians

c) StartFraction 7 pi Over 8 EndFraction radians

d) StartFraction 13 pi Over 5 EndFraction radians

Answer:

b) StartFraction 3 pi Over 4 EndFraction radians

Step-by-step explanation:

Minor arc KL measures 135°. Circle O is shown. Line segments K O and O L are radii.

Note that:

π radians = 180°

Therefore,

180° = π radians

135° = x

Cross Multiply

180° × x = 135° × π radians

x = 135° × π radians/180

x = 3π/4 radians

Therefore, the radian measure of central angle KOL is 3π/4 radians. Option b) is the correct option.

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Find the approximate area of the regions bounded by the curves y = x/(√x2+ 1) and y = x^4−x. (You may use the points of intersec
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The approximate area of the region bounded by the curves f(x) = x / √(x² + 1) and g(x) = x⁴ - x is approximately 0.806.

<h3>How to determine the approximate area of the regions bounded by the curves</h3>

In this problem we must use definite integrals to determine the area of the region bounded by the curves. Based on all the information given by the graph attached below, the area can be defined in accordance with this formula:

A = A₁ + A₂                                                                (1)

A₁ = ∫ [g(x) - f(x)] dx, for x ∈ [- 0.786, 0]                   (2)

A₂ = ∫ [f(x) - g(x)] dx, for x ∈ [0, 1.151]                       (3)

g(x) = x⁴ - x                                                               (4)

f(x) = x / √(x² + 1)                                                      (5)

Then, we proceed to find the integrals:

∫ g(x) dx = ∫ x⁴ dx - ∫ x dx = (1 / 5) · x⁵ - (1 / 2) · x²                          (6)

∫ f(x) dx = ∫ [x / √(x² + 1)] dx = (1 / 2) ∫ [2 · x / √(x² + 1)] dx = (1 / 2) ∫ [du / √u] = √u = √(x² + 1)                                                                                  (7)

And the complete expression for the integral is:

A = A₁ + A₂                                                                                      (1b)

A₁ = (1 / 5) · x⁵ - (1 / 2) · x² - √(x² + 1), for x ∈ [- 0.786, 0]               (2b)

A₂ = √(x² + 1) - (1 / 5) · x⁵ + (1 / 2) · x², for x ∈ [0, 1.151]                  (3b)

A₁ = 0.023

A₂ = 0.783

A = 0.023 + 0.783

A = 0.806

The approximate area of the region bounded by the curves f(x) = x / √(x² + 1) and g(x) = x⁴ - x is approximately 0.806.

To learn more on definite integral: brainly.com/question/14279102

#SPJ1

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