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laceybabin1952
02.06.2020 •
Mathematics
Find the two smallest possible solutions to part 1a
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Ответ:
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.
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
Plugging either x value into the expression 12*sin(2pi/5*x)+10 should yield 16, which would confirm the two answers.
Ответ:
a
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
Ответ:
Step-by-step explanation:
question 4: a. x≤125
next question:
when you multiply or divide both sides by negative number, You need to change sign. in this case she should have written: x>-5