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dsp73
3 years ago
11

What is the slope of the line in the graph?

Mathematics
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

slope =  \frac{(y - y)}{x -x}  \\  =  \frac{1 - 0}{0 + 1} =1 \\ you \: take \: two \: points \: a(0)(1) \\ b( - 1)(0) \\ then \: the \: slope \: is \: 1

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3 years ago
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A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume tha
Mariulka [41]

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

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- 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

- The derivative of the function f'(x) is as follows:

                            f'(x) = \frac{21-x}{\sqrt{42x-x^2} }

- The square of derivative of f(x) is:

                            f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }

- 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

- The Volume of the pain coating is:

                           V = S*t

                           V = 210*π*3/2000

                          V = 63π / 200 m^3

8 0
3 years ago
Read 2 more answers
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