Keep in mind that we're framing it based on what the first sentence says, which is "If the cost of a competing factor of production, such as a machine that also could do the job, rises".
So if the cost of getting a machine part, various parts, or the entire machine cost rises, then demand for the machine will go down. This will make employers seek out substitutes. In this case, those substitutes would be human labor. As employers demand for labor goes up, the wages will rise assuming the supply of workers is held constant. If the supply of workers increased, then you could argue the wages could go down. So that's why I'm assuming the supply is held in check.
Answers: sin(2∅) = 120/169, cos(2∅) = 119/169, tan(2∅) = 120/119
<h3>What are trigonometric functions?</h3>
Trigonometric functions are used to establish the relationship between the sides and the angles of a right angle triangle.
Analysis:
If cos∅ = adjacent/hypotenuse = 12/13,
Then, opposite of the right angled triangle = 
= 5
sin∅ = 5/13, cos∅ = 12/13, tan∅ = 5/12
sin(2∅) = 2sin∅cos∅ = 2(5/13)(12/13) = 120/169
cos(2∅) = 
∅ - 
∅ = 
- 
= 144/169 - 25/169 = 119/169
tan(2∅) = sin(2∅) / cos(2∅) = 120/169 ÷ 119/169 = 120/119
Learn more about trigonometric functions: brainly.com/question/24349828
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Answer:
Step-by-step explanation:
To determine the time when the height is 40 meters above the ground, we would substitute 40 for h in the given equation. It becomes
1.2 + 4t0t - 5t² = 40
5t² - 40t + 40 - 1.2 = 0
5t² - 40t + 38.8 = 0
The general formula for solving quadratic equations is expressed as
x = [- b ± √(b² - 4ac)]/2a
From the equation given,
a = 5
b = - 40
c = 38.8
Therefore,
t = [- - 40 ± √(- 40² - 4 × 5 × 38.8)]/2 × 5
t = [40 ± √(1600 - 776)]/10
t = [40 ± √824]/10
t = (40 + 28.71)/10 or x = (40 - 28.71)/10
t = 6.871 seconds or t = 1.129 seconds
4 - (-12) = 16
difference = 16 degrees celcius.
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Answer:
Area segment = 3/2 π - (9/4)√3 units²
Step-by-step explanation:
∵ The hexagon is regular, then it is formed by 6 equilateral Δ
∵ Area segment = area sector - area Δ
∵ Area sector = (Ф/360) × πr²
∵ Ф = 60° ⇒ central angle of the sector
∵ r = 3
∴ Area sector = (60/360) × (3)² × π = 3/2 π
∵ Area equilateral Δ = 1/4 s²√3
∵ The length of the side of the Δ = 3
∴ Area Δ = 1/4 × (3)² √3 = (9/4)√3
∴ Area segment = 3/2 π - (9/4)√3 units²
