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Mathematics

1 answer:

Mariana [72]3 years ago

6 0

$\boxed{a_1=-4}\\\\a_2=a_1\cdot\,?\\\\-24=-4\cdot\,?\\\\\boxed{?=6}$

Answer B.

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Which equation could be used to find the length of the hypotenuse

Aloiza [94]

Answer:

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Step-by-step explanation:

6 0

2 years ago

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

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:

$S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx$

- 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:

$S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi$