From the figure, the vertices of the given triangle are
(-3,4), (1,3) and (5,5).
The given transformation is (x,y) -> (x+3, y-2).
Therefore the coordinates of the transformed triangle are
(-3+3,4-2), (1+3,3-2), (5+3,5-2) = (0,2), (4,1), (8,3)
The transformed triangle is shown in red color.
Answer:
The domain of a function is the set of all possible inputs for the function. For example, the domain of f(x)=x² is all real numbers, and the domain of g(x)=1/x is all real numbers except for x=0.
Area = length x width
Area = 150 x 75
Area = 11,250 square feet
Looks to be adjacent and virtually all of them are there
Answer:
V = 63π / 200 m^3
Step-by-step explanation:
Given:
- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:
y = √(42*x - x^2)
- The surface is coated with paint with uniform layer thickness t = 1.5 mm
Find:
The volume of paint needed
Solution:
- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:
![S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx](https://tex.z-dn.net/?f=S%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5Ea_b%20%7B%20%5Bf%28x%29%2A%5Csqrt%7B1%20%2B%20f%27%28x%29%5E2%7D%20%7D%5D%20%5C%2C%20dx)
- The derivative of the function f'(x) is as follows:
- The square of derivative of f(x) is:
- Now use the surface area formula:
![S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi](https://tex.z-dn.net/?f=S%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B42x-x%5E2%7D%20%2A%5Csqrt%7B1%20%2B%20%5Cfrac%7B%2821-x%29%5E2%7D%7B42x-x%5E2%20%7D%20%7D%5D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B42x-x%5E2%2B%2821-x%29%5E2%7D%20%7D%5D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B42x-x%5E2%2B441-42x%2Bx%5E2%7D%20%7D%5D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B441%7D%20%7D%5D%20%5C%2C%20dx%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%2021%7D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%2042%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20dx%7D%20%5C%2C%5C%5C%5C%5CS%20%3D%2042%2A%5Cpi%20%5B%206%20-%201%20%5D%5C%5C%5C%5CS%20%3D%2042%2A5%2A%5Cpi%20%5C%5C%5C%5CS%20%3D%20210%5Cpi)
- The Volume of the pain coating is:
V = S*t
V = 210*π*3/2000
V = 63π / 200 m^3
