Answer:
It takes 4 one-sixth sized pieces to make the two-thirds. and we can see that there 1 \times 6 = 6 sixths in a whole.
Step-by-step explanation:
Which two quantities? Please upload a picture or type so we can help you with the best of our abilities!
The answer is B AC
because it's the 2 that connect the line between A and C
<u>Differentiate using the Quotient Rule</u> –

![\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\](https://tex.z-dn.net/?f=%5Cpink%7B%5Ctwoheadrightarrow%20%5Csf%20%5Cdfrac%7Bd%7D%7Bdx%7D%20%5Cbigg%5B%5Cdfrac%7Bf%28x%29%7D%7Bg%28x%29%7D%20%5Cbigg%5D%3D%20%5Cdfrac%7B%20g%28x%29%5C%3A%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbigg%5Bf%28x%29%5Cbigg%5D%20-f%28x%29%5Cdfrac%7Bd%7D%7Bdx%7D%5C%3A%5Cbigg%5Bg%28x%29%5Cbigg%5D%7D%7Bg%28x%29%5E2%7D%7D%5C%5C)
According to the given question, we have –
- f(x) = x^3+5x+2
- g(x) = x^2-1
Let's solve it!

![\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\](https://tex.z-dn.net/?f=%5Cgreen%7B%5Ctwoheadrightarrow%20%5Cbf%20%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbigg%5B%20%5Cdfrac%7Bx%5E3%2B5x%2B2%20%7D%7Bx%5E2-1%7D%5Cbigg%5D%7D%20%5C%5C)













