This is so provided that the velocity changes continuously in which case we can apply the mean value theorem.
<span>Velocity (v) is the derivative of displacement (x) : </span>
<span>v = dx/dt </span>
<span>Monk 1 arrives after a time t* and Monk 2 too. </span>
<span>Name v1(t) and v2(t) their respective velocities throughout the trajectory. </span>
<span>Then we know that both average velocities were equal : </span>
<span>avg1 = avg2 </span>
<span>and avg = integral ( v(t) , t:0->t*) / t* </span>
<span>so </span>
<span>integral (v1(t), t:0->t*) = integral (v2(t), t:0->t*) </span>
<span>which is the same of saying that the covered distances after t* seconds are the same </span>
<span>=> integral (v1(t) - v2(t) , t:0->t*) = 0 </span>
<span>Thus, name v#(t) = v1(t) - v2(t) , then we obtain </span>
<span>=> integral ( v#(t) , t:0->t*) = 0 </span>
<span>Name the analytical integral of v#(t) = V(t) , then we have </span>
<span>=> V(t*) - V(0) = 0 </span>
<span>=> V(t*) = V(0) </span>
<span>So there exist a c in [0, t*] so that </span>
<span>V'(c) = (V(t*) - V(0)) / (t* - 0) (mean value theorem) </span>
<span>We know that V(0) = V(t*) = 0 (covered distances equal at the start and finish), so we get </span>
<span>V'(c) = v#(c) = v1(c) - v2(c) = 0 </span>
<span>=> v1(c) = v2(c) </span>
<span>So there exist a point c in [0, t*] so that the velocity of monk 1 equals that of monk 2. </span>
X+y=24 so we can say
x=24-y making 3x+5y=100 become
3(24-y)+5y=100
72-3y+5y=100
72+2y=100
2y=28
y=14, since x=24-y
x=10
So there are 10 3-point questions and 14 5-point questions.
Y=(x+1)(x-1) this is factored out
Answer:
1/2
Step-by-step explanation:
from point to point, you rise by one and run to the right by two
therefore, slope = 1/2
