The answer is D, tony would always be behind Robert.
For question number 1:The plot H = H(t) is the parabola and it reaches its maximum in the moment when exactly at midpoint between the roots t = 0 and t = 23. At that moment t = 23/2 or 11.5 seconds.
For question number 2:To find the maximal height, just simply substitute t = 11.5 into the quadratic equation. The answer would be 22.9.
For question number 3:H(t) = 0, or, which is the same as -16t^2 + 368t = 0.Factor the left side to get -16*t*(t - 23) = 0.t = 0, relates to the very start of the process, when the ash started its way up.The other root is t = 23 seconds, and it is precisely the time moment when the bit of ash will go back to the ground.
Answer:
The third one
Step-by-step explanation:
Its asking if its positive in the interval [-3, -2]
Which means you only look at the graph from -3, -2 on the x axis and we are looking see which one is positive in that space. And by positive it means going up in a positive direction. Only the third one does this.
First write it in vertex form :-
y= a(x - 2)^2 + 3 where a is some constant.
We can find the value of a by substituting the point (0.0) into the equation:-
0 = a((-2)^2 + 3
4a = -3
a = -3/4
so our equation becomes y = (-3/4)(x - 2)^2 + 3
