Y= kx (k being a constant)
10= k × 4 k=5/2
y= 5x/2 The answer is C.
Answer:
x
Step-by-step explanation:
;\
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:
x = 28
Step-by-step explanation:
Given that lines AB and CD are straight lines that intersects at O, it follows that the pair of opposite vertical angles formed are congruent.
Thus,
<AOD = <BOC
<AOD = 152°
<BOC = 3x + x + (x + 12) (angle addition postulate)
<BOC = 5x + 12
Since <AOD = <BOC, therefore,
152° = 5x + 12 (substitution)
152 - 12 = 5x (subtraction property of equality)
140 = 5x
140/5 = x (division property of equality)
28 = x
x = 28
Answer:
-1.3
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-mercury