SOLUTION
Given B is the midpoint of AC ABBC 1 and C is the midpoint of BD BCCD2 ie on adding 1 and 2 ART We get ABBCBCCD hence ABCD hence proved
Answer:
Distance from the airport = 894.43 km
Step-by-step explanation:
Displacement and Velocity
The velocity of an object assumed as constant in time can be computed as
Where is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as
The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as
The displacement of the plane in 2 hours is
Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are
The displacement in 1 hour is
The total displacement is the vector sum of both
The distance from the airport is the module of the displacement:
Answer and explanation:
Given : The position of an object moving along an x axis is given by where x is in meters and t in seconds.
To find : The position of the object at the following values of t :
a) At t= 1 s
b) At t= 2 s
c) At t= 3 s
d) At t= 4 s
(e) What is the object's displacement between t = 0 and t = 4 s?
At t=0, x(0)=0
At t=4, x(4)=14.24
The displacement is given by,
(f) What is its average velocity from t = 2 s to t = 4 s?
At t=2, x(2)=-1.76
At t=4, x(4)=14.24
The average velocity is given by,
x = -38/9 !!!
27(x+4)=−6
Step 1: Simplify both sides of the equation.
27(x+4)=−6
(27)(x)+(27)(4)=−6 (Distribute)
27x+108=−6
Step 2: Subtract 108 from both sides.
27x+108−108=−6−108
27x=−114
Step 3: Divide both sides by 27.
27x/27=−114/27
x=−389