Can someone answer 33-37 for 20 points

33. A variable with a negative exponent would be 1/x^2 so it would look like this:

q^6/q^2  Now you can subtract or divide these exponents:

= q^4

34. Using the same method do this

3a x 2b^3/ b^5 a^2

= 15ab^3/ b^2 a^2

= 15b / a

36. 4g^5h^3 / 8g^2h

Use the same method above, you can divide 4/8 and you'll get 1/2

= 1h^2 / 2g^3

37. 6c^4d^2 / cd

= 6c^3d

answer:

step-by-step explanation:

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.