6x^4 = (3x^3)(____) how can i find the missing factor?

Question

also -35x^2y = (-5xy)(____)?

Answer 1

6x^4=3x^3A
A=6x^4/3x^3
A=2x

-35x^2y=-5xyA
A=-35x^2y/-5xy
A=7x

Answer 2

it's 2x
you simply divide 6x^4 by 3x^3

6x^4/3x^3 =
first deal with the powers, when dividing powers you subtract from the total so you are left with x^1
then deal with the numbers
6/3=2
You're answer is 2x^1 or just simply 2x

as for the second question that's a little trickier seeing as how the "x" is to the power of the "y"

i don't really want to get into details but the easiest way to go about this is to use logarithms... if you haven't been taught logarithms then don't bother because you're probably supposed to do it some other time consuming way but if yo have been taught them then that would be what to use



EDIT: also don't listen to ntsoaE answer for the second question because it is most definently wrong... same with Kemmy, they both thought that it was -35x2y not -35x^2y

Answer 3

6x^4=3x^3 * -------

6x^4/3x^3=6/3=2 x^4.x^3=x

6x^4/3x^3=2x

-35x^2y=-5xy *--------

-35x^2y/-5xy=-35/-5=7 x^2/x=x y/y=1

-35x^2y/-5xy= 7x*

Answer 4

The first is 2x. It's like asking, what number times (3x^3) gives you 6x^4, which is: 2x. Follow the same logic for the other problem ;p

Answer 5

6x^4 = (3x^3)(2x)

-35x^2y = (-5xy)(7x)

Answer 6

6x^4=(3x^3)(___)
Divideboth sides by 3X^3
6X^4/3X^3=3X^3(___)/3X^3
2X=(___)
Divide both sides by -5XY
-35X^2Y/-5XY=-5XY(___)/-5XY
7X=(___)

Answer 7

BY WATCHING WRESTLING!!!!!!!!!!!