Math questions, please and thanks?

1) -4x^2 + 16x^3y / 4x^2
step 1 - if your dividend is not monomial, separate the terms and copy put the divisor as their denominator

-4x^2/4x^2 + 16x^3y/4x^2

step 2 - divide the numerator by the denominator, i.e, divivde the constant in the numerator to the constant in the denominator; subtract the power of the variable in the numarator with the power of the variable in the denominator(subtract only the variables which are similar letters).

-1 + 4xy ◄--- answer

note: if there is a variable in either term which does not exist in the other term, the y in the second term for instance, just copy the variable and, if any, its power.

2. (3x)^ -2
step 1 - when the given exponent/power is a negative integer, invert the given term, i.e., put a #1 as your numarator then put the given term as your denominator but this time, your negative integer will be positive.

1/(3x)^ 2

step 2 - distribute the power to each constant/variable in the parenthesis.

1/3^2x^2

step 3 - raise the constant to the given power/exponent

1/(3*3)x^2

1/9x^2 ◄--- answer

3) I'm not sure about this one. Hope my answers to the first two questions help!

http://img442.imageshack.us/img442/7562/document1p0151hr3.jpg

For number 1:
First factor the numerator, thus: -4x^2 + 16x^3y into (4x^2)(-1 + 4xy)
so the whole equation becomes (4x^2)(-1 + 4xy)/(4x^2).
Now you can cancel (4x^2) from numerator and denominator leaving only (-1+4xy).

For number 2:
(3x)^ -2 is equivalent to (3^-2)(x^-2) which is equal to (1/3^2)(1/x^2).
So now we have (1/9)(1/x^2) = 1/(9x^2)

For number 3:
First distribute the 3 across the parenthesis
x - 3 < 6x - 3
now add 3 to both sides and subtract x
0 < 5x and dividing both sides by 5 gives 0 < x

2. 1/9x

3. 0 < x

1. 4x^2(-1 + 2xy)/4x^2 =
2xy - 1

2. 1/(9x^2)

3. x - 3 < 6x - 3
-5x < 0
x > 0

I do not have no clue.

Be careful here! Everyone has made mistakes in their answers on this one. Even M. Abuhela.

1.
-4x²+16x³y/4x²
= -4x²+4xy (cancel out a factor of 4x² from top and bottom)
= 4x(-x+y) (factor out 4x from each term)
= 4x(y-x) (reorder the terms to make it look nicer)

1b. You may have meant, instead, to write
(-4x²+16x³y)/4x² (like the whole thing is divided by 4x²)
In this case:
(-4x²+16x³y)/4x²
=(-4x²)/4x² + (16x³y)/4x² (split up the terms
=--1 + (16x³y)/4x² (cancel out 4x² top and bottom 1st term)
=--1 + (4x²)(4xy)/4x² (factor out 4x²)
=--1 + 4xy (cancel out 4x² top and bottom 2nd term)
=4xy - 1 (make it look pretty)
2.
3x^(-2)
=1/(3x)² (negative exponent means 1 over the whole thing)
=1/9x² (just square the bottom now, (3x)(3x) = 9x²)

3.
x-3 < 3(2x-1)
x-3 < 6x-3 (expand the brackets on the right by multiplying it out)
x < 6x (add 3 to both sides, notice it cancels the +3 on each side)
0 < 5x (subtract x from each side)
0 < x (divide each side by 5)
x > 0 (flip the direction of the inequality)

A great way to check your answers is on quickmath.com

Question 1
This type of question troubles me because I suspect that you mean , NOT AS GIVEN, but :-

(i) (- 4 x² + 16 x³ y) / 4x²

What you have in fact given is :-

(ii) -4 x² + (16 x³ y / 4 x² )

I will go for (i) ;-
- 1 + 4 xy

Question 2
1 / (3 x) ² = 1 / (9 x ²)

Question 3
x < 3 + 6x - 3

x < 6x

0 < 5x

x > 0

Best wishes and make 2009 the year of the brackets !

1)
(-4x^2 + 16x^3y)/4x^2
= 4x^2(-1 + 4xy)/4x^2
= -1 + 4xy
= 4xy - 1

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

3)
x - 3 < 3(2x - 1)
x - 3 < 3*2x - 3*1
x - 3 < 6x - 3
x - 6x < 3 - 3
-5x < 0
x > 0/-5
x > 0