I think they’re perpendicular lines
Interpreting your expression as
![\dfrac{3(1+2x)}{\sin(x)}](https://tex.z-dn.net/?f=%20%5Cdfrac%7B3%281%2B2x%29%7D%7B%5Csin%28x%29%7D%20)
when
approaches zero, the numerator approaches 3:
![3(1+2x) \to 3(1+2\cdot 0) = 3(1+0) = 3\cdot 1 = 3](https://tex.z-dn.net/?f=%203%281%2B2x%29%20%5Cto%203%281%2B2%5Ccdot%200%29%20%3D%203%281%2B0%29%20%3D%203%5Ccdot%201%20%3D%203%20)
The denominator approaches 0, because ![\sin(0)=0](https://tex.z-dn.net/?f=%20%5Csin%280%29%3D0%20)
Moreover, we have
![\displaystyle \lim_{x\to 0^-} \sin(x) = 0^-,\quad \displaystyle \lim_{x\to 0^+} \sin(x) = 0^+](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%200%5E-%7D%20%5Csin%28x%29%20%3D%200%5E-%2C%5Cquad%20%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%200%5E%2B%7D%20%5Csin%28x%29%20%3D%200%5E%2B%20)
So, the limit does not exist, because left and right limits are different:
![\displaystyle \lim_{x\to 0^-} \dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^-} = -\infty,\quad \displaystyle \lim_{x\to 0^+}\dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^+} = +\infty](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%200%5E-%7D%20%5Cdfrac%7B3%281%2B2x%29%7D%7B%5Csin%28x%29%7D%3D%20%5Cdfrac%7B3%7D%7B0%5E-%7D%20%3D%20-%5Cinfty%2C%5Cquad%20%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%200%5E%2B%7D%5Cdfrac%7B3%281%2B2x%29%7D%7B%5Csin%28x%29%7D%3D%20%5Cdfrac%7B3%7D%7B0%5E%2B%7D%20%3D%20%2B%5Cinfty%20)
Answer:
(2*60/30)*(24+6)
Step-by-step explanation:
(2*60/30)*(24+6)
(120/30)*(26)
4*26
104
Answer:
Step-by-step explanation:
The coordinate proof is a proof of a geometric theorem which uses "generalized" points on the Cartesian Plane to make an argument. ... The method usually involves assigning variables to the coordinates of one or more points, and then using these variables in the midpoint or distance formulas .