Answer:
Step-by-step explanation:
The identities you need here are:
and 
You also need to know that
x = rcosθ and
y = rsinθ
to get this done.
We have
r = 6 sin θ
Let's first multiply both sides by r (you'll always begin these this way; you'll see why in a second):
r² = 6r sin θ
Now let's replace r² with what it's equal to:
x² + y² = 6r sin θ
Now let's replace r sin θ with what it's equal to:
x² + y² = 6y
That looks like the beginnings of a circle. Let's get everything on one side because I have a feeling we will be completing the square on this:

Complete the square on the y-terms by taking half its linear term, squaring it and adding it to both sides.
The y linear term is 6. Half of 6 is 3, and 3 squared is 9, so we add 9 in on both sides:

In the process of completing the square, we created within that set of parenthesis a perfect square binomial:

And there's your circle! Third choice down is the one you want.
Fun, huh?
A. <span>isolating the x² term and finding the square root of both sides </span>
1st number = n
2nd number = n+1
3rd number = n+2
sum of the squares of 3 consecutive numbers is 116
n² + (n+1)² + (n+2)² = 116
n² + (n+1)(n+1) + (n+2)(n+2) = 116
n² + [n(n+1)+1(n+1)] + [n(n+2)+2(n+2)] = 116
n² + n² + n + n + 1 + n² + 2n + 2n + 4 = 116
n² + n² + n² + n + n + 2n + 2n + 1 + 4 = 116
3n² + 6n + 5 = 116 Last option.
Step-by-step explanation:
s = -4.9t² + 49t
The vertex of the parabola is at t = -b/(2a).
t = -49 / (2 × -4.9)
t = 5
The rocket reaches its maximum height after 5 seconds.