Let 
denote the rocket's position, velocity, and acceleration vectors at time 
.
We're given its initial position
and velocity
Immediately after launch, the rocket is subject to gravity, so its acceleration is
where 
.
a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,
(the integral of 0 is a constant, but it ultimately doesn't matter in this case)
and
b. The rocket stays in the air for as long as it takes until 
, where 
is the -component of the position vector.
The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):
c. The rocket reaches its maximum height when its vertical velocity (the -component) is 0, at which point we have
Scientific notation is
n times 10^x
where 1≤n<10 and x is the number of places the decimal place moved (if it moved to the right, x is negative, if it moved to the left, x is positive)
602, 200,000,000,000,000,000,000.0
we see we need to move decimal to the left 23 spaces to here
6.022 times 10^23
answer is
Answer:
Step-by-step explanation:
1. y=2x-3
2. y=2x+3
3. y=-2x+3
4. y=-2x-3
Answer:
Look that the coordinates and trace those points to a corresponding number on the x or y axis. Then write the first number in the coordinate in the x value and the second number for the y value, repeat the process for the rest of the points.
Step-by-step explanation:
Number 7 you divide 6.48 by 4
