PLEASE PLEASE PLEASE HELP. :(

Mathematics

1 answer:

mart [117]2 years ago

7 0

Integers are like -4,-3,-2,-1,0,1,2,3,4 etc

√49=7, integer

-3⁰=-(3⁰)=-(1)=-1, integer

1.2 times 10⁻²=1.2/100=0.012, not an integer

18/3=6, integer

answer is 3rd one

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Are the two triangles similar or not?

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Step-by-step explanation:

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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

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

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

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

$V=\int\limits^{1}_{-1} {(2-2x^2)^2} \, dx\\\\V=\int _{-1}^14-8x^2+4x^4dx\\\\V=\int _{-1}^14dx-\int _{-1}^18x^2dx+\int _{-1}^14x^4dx\\\\V=\left[4x\right]^1_{-1}-8\left[\frac{x^3}{3}\right]^1_{-1}+4\left[\frac{x^5}{5}\right]^1_{-1}\\\\V=8-\frac{16}{3}+\frac{8}{5}\\\\V=\frac{64}{15}$