Graph a... you go down on the y axis 3, and then using the rise over run method, you go up 3, over 4
Let's take the non-linear equation:
![{x}^{2} = \sqrt{3} y](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%3D%20%20%5Csqrt%7B3%7D%20y)
First Make The Equation To Be In Terms of Y
*Divide Both sides by
![\sqrt{3}](https://tex.z-dn.net/?f=%20%5Csqrt%7B3%7D%20)
![\frac{ {x}^{2} }{ \sqrt{3} } = \frac{ \sqrt{3} y}{ \sqrt{3} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7Bx%7D%5E%7B2%7D%20%7D%7B%20%5Csqrt%7B3%7D%20%7D%20%20%3D%20%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20y%7D%7B%20%5Csqrt%7B3%7D%20%7D%20)
*Simplify
![\frac{ {x}^{2} }{ \sqrt{3} } = y](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7Bx%7D%5E%7B2%7D%20%7D%7B%20%5Csqrt%7B3%7D%20%7D%20%20%3D%20y)
You shouldn't keep a radical in the denominator, however we leave it for the purpose of the reciprocal.
*Take The Reciprocal Of Both Sides
![\frac{ \sqrt{3} }{ {x}^{2} } = \frac{1}{y}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B%20%7Bx%7D%5E%7B2%7D%20%7D%20%3D%20%20%5Cfrac%7B1%7D%7By%7D%20%20)
A Linear Equation Takes The Format:
![y = mx + b](https://tex.z-dn.net/?f=y%20%3D%20mx%20%2B%20b)
So,
![\frac{1}{y} = y](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7By%7D%20%20%3D%20y)
![b = 0](https://tex.z-dn.net/?f=b%20%3D%200)
![m(slope) = \frac{ \sqrt{3} }{x}](https://tex.z-dn.net/?f=m%28slope%29%20%3D%20%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7Bx%7D%20)
![x = \frac{1}{x}](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B1%7D%7Bx%7D%20)
Resulting in :
12 if not just look it up on quiz lit and you’ll get it it’s on safari or use scan and solve
Answer:
They are both similar in the way that the formula is base multiplied by height. They are different because of the type of shape on the base.
Step-by-step explanation:
hope this helps