Check the picture below.
now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.
now, those are the values when the rocket is 30 feet up above.

2x+5y=21, x+y=3
y=3-x
2x+5y=21
2x+5(3-x) =21
2x+15-5x=21
2x-5x+15=21
-3x+15=21
-3x=21-15
-x=6/3
-x=2
x=-2
Answer:
d=7
Step-by-step explanation:
Answer:
x = 12
Step-by-step explanation:
sqrt(x+4) - 3 = 1
First get the sqrt all by itself on one side of the equation. Add 3 to both sides of the equation.
sqrt(x+4) = 1 + 3
sqrt(x+4) = 4
To "fix" the sqrt, that is, "undo it" and get rid of it, you have to SQUARE both sides of the equation.
(sqrt(x+4))^2 = 4^2
x + 4 = 16
subtract 4 to finish up.
x = 12
Check:
sqrt(12 + 4) - 3 = 1
sqrt16 - 3 = 1
4 - 3 = 1
1 = 1 Check!