Part A:
The first thing you should do is to graph both lines. Notice that one is of positive slope and another of negative slope.
The solution of the inequation system is given by the shaded region. That is, all the points that are in the shaded region satisfy the system of inequations.
Part B:
the point (-2, -2) is NOT included in the solution area (it is not in the shaded region.
Mathematically it is demonstrated by substituting inequalities and seeing that they are not satisfied
inequality 1
-2 <4 (-2) - 2
-2 <-10 (false)
inequality two
-2> = - (5/2) (- 2) - 2
-2> = 3 (false).

6 0

2 years ago

Find the two smallest possible solutions to part 1a

bixtya [17]

<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.

7 0

2 years ago

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