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?
Answer:
∠1 ≅ ∠2
Step-by-step explanation:
The transitive property says that if ∠1 ≅ ∠3 and ∠1 ≅ ∠2, then ∠2 ≅ ∠3. The missing part of this statement in the paragraph is ∠1 ≅ ∠2.
The answer would be 5 cuz 5 x 40 = 200
Here we have a case of the least common multipl(lcm) of 6 and 20.
Prime numbers 2,3,5,7,11,13,17,19... (natural numbers greater than 1 that has no positive divisors other than 1 and itself) .
lcm(6,20)= 6 20 | 2
3 10 | 3
1 10 | 2
5 | 5
1
2*3*2*5=60 The first one to get both calendar and the animal toy will be 60th.
Explenation: First we look for the smallest prime number with wich 6 and 20 can be devided by. That is 2. Next is 3. Since 10 is not divisible by 3, we only copy it. Under the 6 we got 1, wich is our goal. Now we continue to devide 10 by prime numbers till we also get 1. We now multiple all divisors and we get the least common multiple.