Answer:12 months
Step-by-step explanation:
400+400=800 so 950-800=150 1800/150=12
Answer:
BC = 2.24 cm
AC = 1.99 cm
Step-by-step explanation:
well to start we have to know the relationships between angles, legs and the hypotenuse.
a: adjacent
o: opposite
h: hypotenuse
sin θ = o/h
cos θ= a/h
tan θ = o/a
θ = 40°
h = 3cm
a = AC
o = BC
first let's calculate BC
sin θ = o/h
o = sin θ * h
o = sin 40 * 3
o = 0.7451 * 3
o = 2.24 cm
BC = 2.24 cm
having one leg and the hypotenuse we can calculate the other leg with pitagoras
h^2 = c1^2 + c2^2
3^2 = 2.24^2 + AC^2
9 = 5.02 + AC^2
AC^2 = 9 - 5.02
AC = √ 3.98
AC = 1.99
Answer:
nothing?
Step-by-step explanation:
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>