Explanation:
When the inequality symbol is replaced by an equal sign, the resulting linear equation is the boundary of the solution space of the inequality. Whether that boundary is included in the solution region or not depends on the inequality symbol.
The boundary line is included if the symbol includes the "or equal to" condition (≤ or ≥). An included boundary line is graphed as a solid line.
When the inequality symbol does not include the "or equal to" condition (< or >), the boundary line is not included in the solution space, and it is graphed as a dashed line.
Once the boundary line is graphed, the half-plane that makes up the solution space is shaded. The shaded half-plane will be to the right or above the boundary line if the inequality can be structured to be of one of these forms:
- x > ... or x ≥ ... ⇒ shading is to the right of the boundary
- y > ... or y ≥ ... ⇒ shading is above the boundary
Otherwise, the shaded solution space will be below or to the left of the boundary line.
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Just as a system of linear equations may have no solution, so that may be the case for inequalities. If the boundary lines are parallel and the solution spaces do not overlap, then there is no solution.
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The attached graph shows an example of graphed inequalities. The solutions for this system are in the doubly-shaded area to the left of the point where the lines intersect. We have purposely shown both kinds of inequalities (one "or equal to" and one not) with shading both above and below the boundary lines.
Answer:

Step-by-step explanation:
1. Distribute the X across the equation: 
2.
3.
3.
Answer:
a) No. t < 0 is not part of the useful domain of the function
b) 2.0 seconds
Step-by-step explanation:
a) A graph of the function is shown below. It shows t-intercepts at t=-0.25 and t=2.0. We presume that t is measured forward from some event such as the ball being thrown or hit. The model's predicted ball location has no meaning prior to that event, when values of t are negative.
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b) It is convenient to use a graphing calculator to find the t-intercepts. Or, the equation can be solved for h=0 any of several ways algebraically. One is by factoring.
h = 0 = -16t² +28t +8 . . . . . . . . . . . . the ball hits the ground when h = 0
0 = -4(4t² -7t -2) = -4(4t +1)(t -2)
This has t-intercepts where the factors are zero, at t=-1/4 and t=2.
The ball will hit the ground after 2 seconds.
Answer:
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Step-by-step explanation:
Part A:
Circumference = πd
Circle A: d = 7
7π = 21.98
π = 21.98/7 = 3.14
Circle B: d = 6
6π = 18.84
π = 3.14
Part B:
Area = πr^2
Circle A: r = 3.5
π3.5^2 = 38.465
12.25π = 38.465
π = 3.14
Circle B: r = 3
π3^2 = 28.26
9π = 28.26
π = 3.14
Part C: The value of π is the same in circles A and B