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2 years ago

Find the two smallest possible solutions to part 1a

Mathematics

2 answers:

bixtya [17]2 years ago

7 0

<h3>Answer: A. 5/12, 25/12</h3>

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Work Shown:

12*sin(2pi/5*x)+10 = 16

12*sin(2pi/5*x) = 16-10

12*sin(2pi/5*x) = 6

sin(2pi/5*x) = 6/12

sin(2pi/5*x) = 0.5

2pi/5*x = arcsin(0.5)

2pi/5*x = pi/6+2pi*n or 2pi/5*x = 5pi/6+2pi*n

2/5*x = 1/6+2*n or 2/5*x = 5/6+2*n

x = (5/2)*(1/6+2*n) or x = (5/2)*(5/6+2*n)

x = 5/12+5n or x = 25/12+5n

these equations form the set of all solutions. The n is any integer.

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The two smallest positive solutions occur when n = 0, so,

x = 5/12+5n or x = 25/12+5n

x = 5/12+5*0 or x = 25/12+5*0

x = 5/12 or x = 25/12

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Plugging either x value into the expression 12*sin(2pi/5*x)+10 should yield 16, which would confirm the two answers.

Mkey [24]2 years ago

4 0

Answer:

Step-by-step explanation:

12sin(⅖pi x) + 10 = 16

sin(⅖pi x) = 6/12

⅖pi x = sin^-1(½)

⅖pi x = pi/6, pi - pi/6

⅖pi x = pi/6, 5pi/6

x = 5/12, 25/12

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