Answer:
V1 = 60 km/h
V2 = 40 Km/h
Step-by-step explanation:
The speed of an object is defined as
Speed = distance / time
Let
V1 be the speed of the faster car
V2 be the speed of the other car
t1 the time it took for the first car to arrive
t2 the time it took for the second car to arrive
d1 the distance traveled by first car
d2 the distance traveled by second car
We know thanks to the problem that
V1 = V2 + 20 Km/h
t1 = t2 - 1 hour
d1 = d2 = 120 Km
d1 = V1 * t1
d2 = V2* t2
V1 * t1 = V2* t2
V1* t1 = (V1 -20)*(t1 +1)
The system of equations
(V1 -20)*(t1 +1) = 120
V1 * t1 = 120
120 + (120/t1) -20*t1 = 140
(120/t1) -20*t1 = 20
Which gives,
t1 = 2
This means
V1 = 60 km/h
V2 = V1 - 20 Km/h = 40 Km/h
Answer:
Probability = 0.12025
Step-by-step explanation:
P (Am) = 1/50 = 0.02 {Magazine ad}
P (At) = 1/8 = 0.125 {Television ad}
P (Am ∩ At ) = 1/100 = 0.01 {Both ads}
P (Am U At) = P (Am) + P (At) - P (Am ∩ At )
= 0.02 + 0.125 - 0.01
P (Am U At) = 0.135 {Person sees either ad}
P (Am' ∩ At') = 1 - P (Am U At)
P (Am' ∩ At') = 1 - 0.135 = 0.865 {Person sees none ad}
Prob (Purchase) = Prob (Purchase with ad) + Prob (purchase without ad)
P (P/ A) = 1/4 = 0.25 , P (P / A') = 1/10 = 0.1
P (P) = (0.25) (0.135) + (0.1) (0.865)
= 0.03375 + 0.0865
0.12025
Answer: 43
Step-by-step explanation:
Let point (x, y) be any point on the graph, than the distance between (x, y) and the focus (3, 6) is sqrt((x - 3)^2 + (y - 6)^2) and the distance between (x, y) and the directrix, y = 4 is |y - 4|
Thus sqrt((x - 3)^2 + (y - 6)^2) = |y - 4|
(x - 3)^2 + (y - 6)^2 = (y - 4)^2
x^2 - 6x + 9 + y^2 - 12y + 36 = y^2 - 8y + 16
x^2 - 6x + 29 = -8y + 12y = 4y
(x - 3)^2 + 20 = 4y
y = 1/4(x - 3)^2 + 5
Required answer is f(x) = one fourth (x - 3)^2 + 5
