It takes 4.3 seconds for the rocket to return to earth.
The equation is:

where -9.8m/sec² is the acceleration due to gravity, v₀ is the initial velocity, and h₀ is the initial height. We will go from the assumption that the rocket is launched from the ground, so h₀=0, and we are told that the initial velocity, v₀, is 42. This gives us:

We will use the quadratic formula to solve this. The quadratic formula is:

Plugging in our information we have:

x=0 is when the rocket is launched; x=4.3 is when the rocket lands.
Answer:
The angle in the box is always 50 and that other acute angle is 46
Hope this helps
Step-by-step explanation:
Answer:
m=5
Step-by-step explanation:
-2m-10+14=-6
-2m+4=-6
-2m=-10
m=5
Answer:
c
Step-by-step explanation:
Here's how this works:
Get everything together into one fraction by finding the LCD and doing the math. The LCD is sin(x) cos(x). Multiplying that in to each term looks like this:
![[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?](https://tex.z-dn.net/?f=%5Bsin%28x%29cos%28x%29%5D%5Cfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%2B%5Bsin%28x%29cos%28x%29%5D%5Cfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%20%3D%3F)
In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

Put everything over the common denominator now:

Since
, we will make that substitution:

We could separate that fraction into 2:
×
and 
Therefore, the simplification is
sec(x)csc(x)
Good idea to look up the formula for "surface area of a sphere" and to write it down for later reference. It is A=4 pi r^2, where r is the radius of the sphere.
If the given sphere has a diameter of 7 cm, what is the radius of this sphere?
Write your answer inside the parentheses, below:
Sphere surface area = 4 pi (r)^2, or (in this case), A = 4(3.14)( )^2