Answer:
The correct option is;
Increasing one fifth unit/sec
Step-by-step explanation:
The equation that gives the curve of the particle of the particle is y = 5·x² - 1
The rate of decrease of the y value dy/dt = 2 units per second
We have;
dy/dx = dy/dt × dt/dx
dy/dx = 10·x
dy/dt = 2 units/sec
dt/dx = (dy/dx)/(dy/dt)
dx/dt = dy/dt/(dy/dx) = 2 unit/sec/(10·x)
When x = 1
dx/dt = 2/(10·x) = 2 unit/sec/(10 × 1) = 1/5 unit/sec
dx/dt = 1/5 unit/sec
Therefore, x is increasing one fifth unit/sec.
Answer:
formula: y-y1 = m(x-x1)
Step-by-step explanation:
General:
1. find 2 points directly on the line and apply the formula y2-y1 over x2-x1 this answer equals m
2. then add it into the formula y-y1 = m(x-x1) remember x and y are variables so only put numbers in y1 and x1
3. solve the problem
Example:
line 1:
1. y2-y1 over x2-x1 is 4-6 over 6-4 = -1
2. y-6 = -1(x-(-4) or y-6 = -1(x+4)
3. y-6 = -1x-4
4. y = -1x-2
3. the gradient equals: y = -1x-2
Hope this helps :) Let me know if you need a better example
Answer:
The volume of the solid is: ![\mathbf{V = \pi [ \dfrac{8 \pi}{3} - 2\sqrt{3}]}](https://tex.z-dn.net/?f=%5Cmathbf%7BV%20%3D%20%5Cpi%20%5B%20%5Cdfrac%7B8%20%5Cpi%7D%7B3%7D%20-%202%5Csqrt%7B3%7D%5D%7D)
Step-by-step explanation:
GIven that :
This implies that the distance between the x-axis and the axis of the rotation = 2 units
The distance between the x-axis and the inner ring is r = (2+sec x) -2
Let R be the outer radius and r be the inner radius
By integration; the volume of the of the solid can be calculated as follows:
![V = \pi \int\limits^{\dfrac{\pi}{3}}_{\dfrac{-\pi}{3}} [(4-2)^2 - (2+ sec \ x -2)^2]dx \\ \\ \\ V = \pi \int\limits^{\dfrac{\pi}{3}}_{\dfrac{-\pi}{3}} [(2)^2 - (sec \ x )^2]dx \\ \\ \\ V = \pi \int\limits^{\dfrac{\pi}{3}}_{\dfrac{-\pi}{3}} [4 - sec^2 \ x ]dx](https://tex.z-dn.net/?f=V%20%3D%20%5Cpi%20%5Cint%5Climits%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%7D%20%5B%284-2%29%5E2%20-%20%282%2B%20sec%20%5C%20x%20-2%29%5E2%5Ddx%20%5C%5C%20%5C%5C%20%5C%5C%20V%20%3D%20%5Cpi%20%5Cint%5Climits%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%7D%20%5B%282%29%5E2%20-%20%28sec%20%5C%20x%20%29%5E2%5Ddx%20%5C%5C%20%5C%5C%20%5C%5C%20V%20%3D%20%5Cpi%20%5Cint%5Climits%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%7D%20%5B4%20-%20sec%5E2%20%5C%20x%20%5Ddx)
![V = \pi [4x - tan \ x]^{\dfrac{\pi}{3}}_{\dfrac{-\pi}{3}} \\ \\ \\ V = \pi [4(\dfrac{\pi}{3}) - tan (\dfrac{\pi}{3}) - 4(-\dfrac{\pi}{3})+ tan (-\dfrac{\pi}{3})] \\ \\ \\ V = \pi [4(\dfrac{\pi}{3}) - tan (\dfrac{\pi}{3}) + 4(\dfrac{\pi}{3})- tan (\dfrac{\pi}{3})] \\ \\ \\ V = \pi [8(\dfrac{\pi}{3}) - 2 \ tan (\dfrac{\pi}{3}) ]](https://tex.z-dn.net/?f=V%20%3D%20%5Cpi%20%5B4x%20-%20tan%20%5C%20%20x%5D%5E%7B%5Cdfrac%7B%5Cpi%7D%7B3%7D%7D_%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%7D%20%20%5C%5C%20%5C%5C%20%5C%5C%20V%20%3D%20%5Cpi%20%5B4%28%5Cdfrac%7B%5Cpi%7D%7B3%7D%29%20-%20tan%20%28%5Cdfrac%7B%5Cpi%7D%7B3%7D%29%20-%204%28-%5Cdfrac%7B%5Cpi%7D%7B3%7D%29%2B%20tan%20%28-%5Cdfrac%7B%5Cpi%7D%7B3%7D%29%5D%20%5C%5C%20%5C%5C%20%5C%5C%20V%20%3D%20%5Cpi%20%5B4%28%5Cdfrac%7B%5Cpi%7D%7B3%7D%29%20-%20tan%20%28%5Cdfrac%7B%5Cpi%7D%7B3%7D%29%20%2B%204%28%5Cdfrac%7B%5Cpi%7D%7B3%7D%29-%20tan%20%28%5Cdfrac%7B%5Cpi%7D%7B3%7D%29%5D%20%20%5C%5C%20%5C%5C%20%5C%5C%20V%20%3D%20%5Cpi%20%5B8%28%5Cdfrac%7B%5Cpi%7D%7B3%7D%29%20%20-%202%20%5C%20%20tan%20%28%5Cdfrac%7B%5Cpi%7D%7B3%7D%29%20%5D)
Answer:
6hours/1 day
Step-by-step explanation:
I REALLY hope this helps
Best of luck
