MATH HELP Thanks!!! :)?

Question

1. (9- 1/x^2) divided by (3x-1)

a. 8/ 3x-1
b. 9x +1 / x
c. 3x+ 1 / 3x-1
d, 3x+1 / x^2

2. It took Jan y hours to drive 200 km. If she had increased her speed by 10 km/h and driven for 2 h less, how far could she have gone?

a. 10y^2 + 180y - 400 / y
b. 10y^2 + 90y - 400
c. 10y - 200
d. 10y^2 - 200y + 180 / y

PLease and Thanks so much!!

Answer 1

1. (9 - 1/x^2) divided by 3x-1
= (9x^2 - 1) divided by x^2(3x-1)
= (3x-1)(3x+1)/x^2(3x-1)
= (3x+1)/x^2
Answer d

2. y hours 200 km therefore speed is 200/y
if speed is increased by 10 then new speed is 200/y + 10 =
(200 + 10y)/y
Driving 2 hours less means y -2
therefore distance is (y-2)(200 +10y)/y
= (10y^2 +180y - 400)/y
answer a

Answer 2

1. (9- 1/x^2) divided by (3x-1)

= (9x² - 1)/(x²*(3x-1))

= (3x+1)(3x-1)/(x²*(3x-1))

= 3x+1 / x²

Thus (d) 3x+1 / x^2

2. Original speed of Jane = 200/y km/hr

New speed = 200/y + 10 km/h

New time = y-2 hr

Distance travelled = Speed*time = [(200/y) + 10](y-2)
=(200+10y)(y-2)/y

=(10y² - 20y + 200 y - 400) /y

= (10y² +180y -400)/y

Thus (a) 10y^2 + 180y - 400 / y

Answer 3

9x^2-1 =(3x-1)*(3x+1) so simplify and you get

(9-1/x^2)/(3x-1) =( 9x^2-1)/((x^2*(3x-1))= (3x+1)/x^2

d)is the right answer
2) distance = velocity * time
d=v*y so y = d/v so y =200/v

the new velocity is v+10 and the time y-2 but

v= 200/y so the new distance would be

D = (200/y + 10)*(y-2) =(10y^2+180y -400)/y km

a) is the right answer

Answer 4

1. (d) 9- 1/x^2 mulitply by 1/(3x-1)
=(9x^2 -1)/x^2 multiply by 1/(3x-1)
=(9x^2 -1)/[ (x^2)(3x-1)]
=[(3x+1)(3x-1)]/[ (x^2)(3x-1)]
then cancel out the similar top and bottom part to get (d) 3x+1 / x^2

2. not so sure

Answer 5

1. d, 3x+1 / x^2

((9-1/x^2)) / (3x-1) =
((9x^2-1) / (x^2)) / (3x-1) =
((9x^2-1 / (x^2)) * (1/3x-1) =
(9x^2-1) / (3x^3 - x^2) =
(3x+1)(3x-1) / x^2(3x-1) =
3x+1/x^2

Answer 6

i hve no idea