ANSWER ASAP!!! Find the area of the figure.

Mathematics

2 answers:

denis-greek [22]3 years ago

7 0

Area= (3 √(3) × s^2) ÷ s
s= side length
area=(3 √(3) × 16^2) ÷ 16

area= 83.1384387633

chubhunter [2.5K]3 years ago

3 0

Try 360. And tell me if it’s wrong

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Use the general slicing method to find the volume of the following solids. The solid whose base is the region bounded by the cur

Y_Kistochka [10]

Answer:

The volume is $V=\frac{64}{15}$

Step-by-step explanation:

The General Slicing Method is given by

<em>Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is</em>

$V=\int\limits^b_a {A(x)} \, dx$

Because a typical cross section perpendicular to the x-axis is a square disk (according with the graph below), the area of a cross section is

The key observation is that the width is the distance between the upper bounding curve $y = 2 - x^2$ and the lower bounding curve $y = x^2$

The width of each square is given by

$w=(2-x^2)-x^2=2-2x^2$

This means that the area of the square cross section at the point x is

$A(x)=(2-2x^2)^2$

The intersection points of the two bounding curves satisfy $2 - x^2=x^2$ , which has solutions x = ±1.

$2-x^2=x^2\\-2x^2=-2\\\frac{-2x^2}{-2}=\frac{-2}{-2}\\x^2=1\\\\x=\sqrt{1},\:x=-\sqrt{1}$

Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is

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