ASA, HL, and SAS.
AAS wouldn't work, and LA and LL don't exist
Hope this helps!
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?
2. They are both straight lines. The solution of a linear system is where the two lines intersect, and if they're both straight lines, then they can only cross once.
See the picture attached.
Answer:

Step-by-step explanation:
we have

Solve for h
That means ----> isolate the variable h
step 1
Multiply by 6 both sides

step 2
Divide by
both sides

If you go on my profile you will see a similar problem that I already answered today.
Lets start by using a formula (don't remember the name)
(y-y1)=M(x-x1)
M is slope and y is first y value and y1 is second y value same applies to X.
Substitute in the values.
(5-q) = 10(-6-(-7))
5-q = 10*(1)
5-q = 10
-q = 5
q = -5
Check:
Substitute in the value of "q" and use the formula to find slope:
(y1-y)/(x1-x)
Substitute
(-5-5)/(-7-(-6))
-10/-1
10
Your Answer:
The value of "q" is -5