Answer: The car's maximum speed
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I'm assuming the limit is supposed to be
![\displaystyle\lim_{x\to3}\frac{\sqrt{2x+3}-\sqrt{3x}}{x^2-3x}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto3%7D%5Cfrac%7B%5Csqrt%7B2x%2B3%7D-%5Csqrt%7B3x%7D%7D%7Bx%5E2-3x%7D)
Multiply the numerator by its conjugate, and do the same with the denominator:
![\left(\sqrt{2x+3}-\sqrt{3x}\right)\left(\sqrt{2x+3}+\sqrt{3x}\right)=\left(\sqrt{2x+3}\right)^2-\left(\sqrt{3x}\right)^2=-(x-3)](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%7B2x%2B3%7D-%5Csqrt%7B3x%7D%5Cright%29%5Cleft%28%5Csqrt%7B2x%2B3%7D%2B%5Csqrt%7B3x%7D%5Cright%29%3D%5Cleft%28%5Csqrt%7B2x%2B3%7D%5Cright%29%5E2-%5Cleft%28%5Csqrt%7B3x%7D%5Cright%29%5E2%3D-%28x-3%29)
so that in the limit, we have
![\displaystyle\lim_{x\to3}\frac{-(x-3)}{(x^2-3x)\left(\sqrt{2x+3}+\sqrt{3x}\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto3%7D%5Cfrac%7B-%28x-3%29%7D%7B%28x%5E2-3x%29%5Cleft%28%5Csqrt%7B2x%2B3%7D%2B%5Csqrt%7B3x%7D%5Cright%29%7D)
Factorize the first term in the denominator as
![x^2-3x=x(x-3)](https://tex.z-dn.net/?f=x%5E2-3x%3Dx%28x-3%29)
The
terms cancel, leaving you with
![\displaystyle\lim_{x\to3}\frac{-1}{x\left(\sqrt{2x+3}+\sqrt{3x}\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto3%7D%5Cfrac%7B-1%7D%7Bx%5Cleft%28%5Csqrt%7B2x%2B3%7D%2B%5Csqrt%7B3x%7D%5Cright%29%7D)
and the limand is continuous at
, so we can substitute it to find the limit has a value of -1/18.
3 kilometers is equal to 3000 meters
Distance of a ground is 300 meters sou need to run 10 rounds around thefield
2 because 8 divided by 4 is 2.
Answer:
you use disputation
Step-by-step explanation:
4 times 3x = 12x
4 times 6 = 24