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
Answer:
(-∞,+∞) interval notation
it is not clear if the square root on the x only or the four too.
the answer above is if the equation is :
√x +4/x -2
-2(5y-20) - 5y = -5
-10y + 40 - 5y = -5
-15y + 40 = -5
-15y = -45, y = 3
x = 5(3) - 20
x = 15 - 20, x = -5
Solution: x = -5, y = 3... or (-5,3)
Hopefully this helps, and sorry I didn't do 8 I'm limited on time
Hey,
I could actually answer the question now, so I am going to put it here as well so the people don't have to read the comments above.
<span>Use the Pythagorean Theorem to find the hypotenuse:<span>a² + b² = c²
5² + 5² = c²
25 + 25 = c²
50 = c²
√50=√(c^2 )
7.07 = c
Answer: 7
Cheers,
Izzy</span></span>
