Hello there!
6q + 9 ≤ 9
Start by subtracting 9 from both sides
6q + 9 - 9 ≤ 9 - 9
6q ≤ 0
Then divide both sides by 6
6q/6 ≤ 0/6
q ≤ 0
--------------
The second equation
6q + 9 ≥ -9
Start by subtracting 9 from both sides
6q + 9 - 9 ≥ -9 - 9
6q ≥ -18
Divide both sides by 6
6q/6 ≥ -18/6
q ≥ -3
Good luck with your studies!
When doing this, we would have to first straight them all out, and as we see above, they are just all over the place, and then, we would have to set them up as a sequence.
![\left[\begin{array}{ccc}5x2 + 5y2 - 20x + 30y + 40 = 0\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5x2%20%2B%205y2%20-%2020x%20%2B%2030y%20%2B%2040%20%3D%200%5Cend%7Barray%7D%5Cright%5D%20)
would be our first step in this problem mainly because it contains the most terms in this aspect.
Then we would then

, then,

.
These would only be our first 3 part of the sequence in this aspect.
The others would then be the following:

Thus, as we would have one more afterward, our last part of the sequence would then be the following.

I hope this was found helpful!
Answer:
30 2/3
Step-by-step explanation:
First multiply the two exponents
Second solve 2^6
Third multiply 64 times 5 = 320
Then multiply 16 times 3
Then add sums together 368
Lastly multiply 4 times 3 then divide 368 by 12
I hope this helps :)
The answer is B ! as vertex is (6,3) and axis of symmetry lies on x=6 !
if you need explanation, comment !