Solve the inequality, -4 less than 4(6y-12)-2y
1 answer:
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Answer:
Points of intersection of these graphs are (-2, -3) and (0.6, 4.8).
Step-by-step explanation:
Equation of the circle → (x - 2)² + y² = 25 ---------(1)
→ (x - 2)² + (y - 0)² = 5²
By comparing this equation with the standard equation of the circle,
(x - a)² + (y - b)²= r²
Here (a, b) is the center and r is the radius of the circle.
Therefore, center of the circle is (2, 0) and radius = 5 units
Second equation is a linear equation → y = 3x + 3 -------(2)
x-intercept of the equation → x = -1
y-intercept of the equation → y = 3
By graphing these equations we can get the point of intersections.
Solving these equations algebraically,
Substitute the value of y from equation (2) in the equation (1),
(x - 2)² + (3x + 3)² = 25
x² - 4x + 4 + 9x² + 18x + 9 = 25
10x² + 14x - 12 = 0
5x² + 7x - 6 = 0
x =
x =
x =
x = -2, 0.6
From equation (2),
y = -3, 4.8
Therefore, points of intersection of these graphs are (-2, -3) and (0.6, 4.8).
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.
