Your answer would be 2x^2=4x
also I have a question I need help with, would you mind?
Hey!
In order to simplify this equation, we'll first have to multiply both sides of the equation by v. This will give us v on its own.
<em>Original Equation :</em>
![\frac{-3}{v} = -6](https://tex.z-dn.net/?f=%20%5Cfrac%7B-3%7D%7Bv%7D%20%3D%20-6)
<em>New Equation {Added Multiply Both Sides by V} :</em>
![\frac{-3}{v} v=-6v](https://tex.z-dn.net/?f=%20%5Cfrac%7B-3%7D%7Bv%7D%20v%3D-6v)
<em>Solution {New Equation Solved} :</em>
![-3 = -6v](https://tex.z-dn.net/?f=-3%20%3D%20-6v)
Now we'll switch sides to get v on the left side of the equation which is generally where we always want the variables to be located in these types of equations.
<em>Old Equation :</em>
![-3 = -6v](https://tex.z-dn.net/?f=-3%20%3D%20-6v)
<em>New Equation {Switched} :</em>
![-6v=-3](https://tex.z-dn.net/?f=-6v%3D-3)
Now we'll divide both sides by v to get v on its own.
<em>Old Equation :</em>
![-6v = -3](https://tex.z-dn.net/?f=-6v%20%3D%20-3)
<em>New Equation {Added Divide Both Sides by V} :</em>
![\frac{-6v}{v} = \frac{-3}{v}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-6v%7D%7Bv%7D%20%3D%20%5Cfrac%7B-3%7D%7Bv%7D%20)
<em>Solution {New Equation Solved} :</em>
![v = \frac{1}{2}](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20)
<em>So, this means that in the equation
![\frac{-3}{v} =-6](https://tex.z-dn.net/?f=%20%5Cfrac%7B-3%7D%7Bv%7D%20%3D-6)
,</em>
![v = \frac{1}{2}](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20)
.
Hope this helps!
- Lindsey Frazier ♥
Answer:
you are already in the hundredths
Step-by-step explanation:
5.48
5= the ones place
4= the tenths place
8= the hundredths
Answer:
The answer is 94.8 m
Step-by-step explanation:
Area of a square is a²
So, 23.7² = 561.69 m²
The side of the square is equal to 23.7 m
Perimeter of a square is the side multiplied by 4
So we have that 23.7 x 4 = 94.8 m
Answer:
2x^3 - 2x^2 - 12x
Step-by-step explanation:
2x(x - 3)(x + 2)
= 2x ( x^2 + 2x -3x - 6)
= 2x (x^2 - x - 6)
= 2x^3 - 2x^2 - 12x (answer).