Step-by-step explanation:
b2 =5
.....................
I think it measures an angle of 180 degrees
Slope point form :
To put in slope point form, label any of the points as either X1,y1 and X and y, then plug in those values into the following equation form.
Y - y1 = m(X-X1)
But before, we must solve for the m value or slope.
M = y2-y1/x2-X1
M = 5/2 - -1/2 / -1/2 - 3/2.
M = 5/2 + 1/2 / -(1/2+3/2)
M = 6/2 / -(4/2)
M = 3/-2
Now we can place it in slope point and also in standard form of a line.
Y-y1 = m(X -X1)
Y - -1/2 = -3/2(X - 3/2)
Y + 1/2 = -3/2(X - 3/2)
This is in slope point form.
Y + 1/2 = -3/2x + 9/4
Y + 1/2 - 1/2 = -3/2x + 9/4 - 1/2
1/2 = 2/4
Y = -3/2x + 7/4
-3/2x = -6/4x
Y = -6/4x + 7/4
Y • 4 = 4( -6/4 X + 7/4)
4y = -6x + 7
4y + 6x = -6x + 6x +7
6x + 4y = 7
This is in standard form. If you have any questions of the steps just ask.
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.