A. Factor the numerator as a difference of squares:
![\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto9%7D%5Cfrac%7Bx-9%7D%7B%5Csqrt%20x-3%7D%3D%5Clim_%7Bx%5Cto9%7D%5Cfrac%7B%28%5Csqrt%20x-3%29%28%5Csqrt%20x%2B3%29%7D%7B%5Csqrt%20x-3%7D%3D%5Clim_%7Bx%5Cto9%7D%28%5Csqrt%20x%2B3%29%3D6)
c. As
![x\to\infty](https://tex.z-dn.net/?f=x%5Cto%5Cinfty)
, the contribution of the terms of degree less than 2 becomes negligible, which means we can write
![\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B4x%5E2-4x-8%7D%7Bx%5E2-9%7D%3D%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac%7B4x%5E2%7D%7Bx%5E2%7D%3D%5Clim_%7Bx%5Cto%5Cinfty%7D4%3D4)
e. Let's first rewrite the root terms with rational exponents:
![\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto1%7D%5Cfrac%7B%5Csqrt%5B3%5Dx-x%7D%7B%5Csqrt%20x-x%7D%3D%5Clim_%7Bx%5Cto1%7D%5Cfrac%7Bx%5E%7B1%2F3%7D-x%7D%7Bx%5E%7B1%2F2%7D-x%7D)
Next we rationalize the numerator and denominator. We do so by recalling
![(a-b)(a+b)=a^2-b^2](https://tex.z-dn.net/?f=%28a-b%29%28a%2Bb%29%3Da%5E2-b%5E2)
![(a-b)(a^2+ab+b^2)=a^3-b^3](https://tex.z-dn.net/?f=%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29%3Da%5E3-b%5E3)
In particular,
![(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3](https://tex.z-dn.net/?f=%28x%5E%7B1%2F3%7D-x%29%28x%5E%7B2%2F3%7D%2Bx%5E%7B4%2F3%7D%2Bx%5E2%29%3Dx-x%5E3)
![(x^{1/2}-x)(x^{1/2}+x)=x-x^2](https://tex.z-dn.net/?f=%28x%5E%7B1%2F2%7D-x%29%28x%5E%7B1%2F2%7D%2Bx%29%3Dx-x%5E2)
so we have
![\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto1%7D%5Cfrac%7Bx%5E%7B1%2F3%7D-x%7D%7Bx%5E%7B1%2F2%7D-x%7D%5Ccdot%5Cfrac%7Bx%5E%7B2%2F3%7D%2Bx%5E%7B4%2F3%7D%2Bx%5E2%7D%7Bx%5E%7B2%2F3%7D%2Bx%5E%7B4%2F3%7D%2Bx%5E2%7D%5Ccdot%5Cfrac%7Bx%5E%7B1%2F2%7D%2Bx%7D%7Bx%5E%7B1%2F2%7D%2Bx%7D%3D%5Clim_%7Bx%5Cto1%7D%5Cfrac%7Bx-x%5E3%7D%7Bx-x%5E2%7D%5Ccdot%5Cfrac%7Bx%5E%7B1%2F2%7D%2Bx%7D%7Bx%5E%7B2%2F3%7D%2Bx%5E%7B4%2F3%7D%2Bx%5E2%7D)
For
![x\neq0](https://tex.z-dn.net/?f=x%5Cneq0)
and
![x\neq1](https://tex.z-dn.net/?f=x%5Cneq1)
, we can simplify the first term:
![\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x](https://tex.z-dn.net/?f=%5Cdfrac%7Bx-x%5E3%7D%7Bx-x%5E2%7D%3D%5Cdfrac%7Bx%281-x%5E2%29%7D%7Bx%281-x%29%7D%3D%5Cdfrac%7Bx%281-x%29%281%2Bx%29%7D%7Bx%281-x%29%7D%3D1%2Bx)
So our limit becomes
The answer is C because you have three points p to Q, Q to R the two different segments then the whole thing is P to R which is also a segment.
Answer:
192 cubes
Step-by-step explanation:
So all we need to do is:
1) Find the area of 1 cube.
2) Divide the large prism volume with this.
SO 1)
the volume of cube = Length * width * height = (1/4)^3 = 1/64
SO 2)
the volume of prism / the volume of cube = 3/(1/64) = 192
Answer: 192 cubes
Answer:
1= A
2= C
Step-by-step explanation:
I think this is right I'm not sure but I think it is.
hope this helps :)