Cant solve this?

Question

(1/3 x^2 * 9 -99/x)

Can some1 plzz explain this in easy steps? i cant solve this qwestion. thanx!!

Sory i mean like this

(1/3 x^2) (9 -99/x)

Answer 1

distributive property:
(1/3 x^2)(9-99/x)
(1/3x^2)(9) - (1/3x^2)(99/x)
3x^2-33x

Answer 2

Without brackets, question can have many meanings.
Will take a guess at:-
[ (1 / 3) x²) ] [ ( 9 - 99 / x ] as written by you!
[ (1/3) x² ](9) (1 - 9 / x)
3x² (1 - 9 / x)
3x² (x - 9) / x
3x (x - 9)
What worries me is that you may mean:-
[ 1 / (3x²) ] [ 9 - 99 / x] which would be a completely different question.

Moral behind story
------------------------
Use brackets!!!!!!

Answer 3

(1/3x^2)(9-99/x)
(x^2/3)(9/1-99/x)....lowest common multple(l.c.m.) for the 2nd bracket
(x^2/3)[(9x-99]/x)....multiply both brackets
(9x^2-99x)/3.....factorise the numerator
[9(x^2-11)/3]
after dividing by '3', you get
3(x^2-11)

Answer 4

1/3x^2*9-99/x
=-1/3x^2*90/x
=-90/3x^3
=-30/x^3

Answer 5

2x + 1 = 5x/3 multiply each side by 3 3(2x + 1) = 3(5x/3) 6x + 3 = 5x 6x - 5x = -3 x = -3

Answer 6

(1/3x^2)(9-99/x)
=(9/3x^2)(1-11/x)
=(3/x2)(x-11)/x
=3(x-11)/x^3 ans

Answer 7

Since there is no equal sign there is nothing to solve. You should think carefully about how to ask a question as your question makes no sense.

Answer 8

1/3x*x*(-90/x)
x gets cancelled
1/3x*(-90)
-30x