Formula for curvature for a well behaved curve y=f(x) is
K(x)= ![\frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B%7C%7By%7D%27%27%7C%7D%7B%5B1%2B%7By%7D%27%5E2%5D%5E%5Cfrac%7B3%7D%7B2%7D%7D)
The given curve is y=7
k(x)=![\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B7e%5E%7Bx%7D%7D%7B%5B%7B1%2B%287e%5E%7Bx%7D%29%5E2%7D%5D%5E%5Cfrac%7B3%7D%7B2%7D%7D)
![{k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}](https://tex.z-dn.net/?f=%7Bk%28x%29%7D%27%3D%5Cfrac%7B7%28e%5Ex%29%281%2B49e%5E%7B2x%7D%29%2849e%5E%7B2x%7D-%5Cfrac%7B1%7D%7B2%7D%29%7D%7B%5B1%2B49e%5E%7B2x%7D%5D%5E%7B3%7D%7D)
For Maxima or Minima
→
[not possible ∵there exists no value of x satisfying these equation]
→
Solving this we get
x= 
As you will evaluate 
<0 at x=
So this is the point of Maxima. we get y=7×1/√98=1/√2
(x,y)=[,1/√2]
k(x)=![\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}](https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%5Cinfty%20%7D%20%5Cfrac%7B7e%5E%7Bx%7D%7D%7B%5B%7B1%2B%287e%5E%7Bx%7D%29%5E2%7D%5D%5E%5Cfrac%7B3%7D%7B2%7D%7D)
k(x)=
k(x)=0
Answer:
( - x, - y )
Step-by-step explanation:
The starting is always Quadrant I.
270 degrees clockwise from Quadrant I is Quadrant III.
In Quadrant III, the points will be in the form ( - x, - y ).
Answer:
hi
Step-by-step explanation:
....................
You have to solve this algebraicly.
To do this set up the proportion and add x to each term.
You get 4x and 7x
Set this equal to 33
4x+7x=33
Combine like terms and get 11x=33
Divide by 11
x=3
Now go back to 4x and 7x
Multiply them by x and get 12(4×3=12) and 21(7×3=21)
One group is 12 and the other is 21.
In the first bag, picking a black marble would be impossible.
The second bag would be likely.
The third bag would be unlikely.
hope this helps!
