Answer: The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.
Step-by-step explanation: this is the same paragraph The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.
well then, the volume of the nose cone will just be the sum of the volume of the cylinder below and the cone above.
since the diameter for both is 8, then their radius is half that, or 4.
![\bf \stackrel{\textit{volume of a cone}}{V=\cfrac{\pi r^2 h}{3}}~~ \begin{cases} r=radius\\ h=height\\ \cline{1-1} r=4\\ h=6 \end{cases}\implies V=\cfrac{\pi (4)^2(6)}{3}\implies V=32\pi \\\\\\ \stackrel{\textit{volume of a cylinder}}{V=\pi r^2 h}~~ \begin{cases} r=radius\\ h=height\\ \cline{1-1} r=4\\ h=6 \end{cases}\implies V=\pi (4)^2(6)\implies V=96\pi \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{volume of the nose cone}}{32\pi +96\pi \implies 128\pi }\qquad \approx \qquad 402.12](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bvolume%20of%20a%20cone%7D%7D%7BV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%20%5Ccline%7B1-1%7D%20r%3D4%5C%5C%20h%3D6%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20%284%29%5E2%286%29%7D%7B3%7D%5Cimplies%20V%3D32%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bvolume%20of%20a%20cylinder%7D%7D%7BV%3D%5Cpi%20r%5E2%20h%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%20%5Ccline%7B1-1%7D%20r%3D4%5C%5C%20h%3D6%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Cpi%20%284%29%5E2%286%29%5Cimplies%20V%3D96%5Cpi%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bvolume%20of%20the%20nose%20cone%7D%7D%7B32%5Cpi%20%2B96%5Cpi%20%5Cimplies%20128%5Cpi%20%7D%5Cqquad%20%5Capprox%20%5Cqquad%20402.12)
Answer:
1/10
10%
3/4
0.75
0.2
20%
Step-by-step explanation:
We shall use a negative sign for an elevation below sea level, and a positive sign for an elevation above sea level.
Given:
Elevation of city A = -96 ft (below sea level)
Elevation of city C = -24 ft (below sea level)
Let x = elevation of city D.
Part A.
Because the elevation of city D is 1/4 of the elevation of city C, therefore
x = (1/4)*(-24)
Part B.
Because 24/4 = 6, therefore
x = -6 ft.
Answer:
The elevation of city D is 6 feet below sea level.
The answer is negative because the elevation is below sea level.
127 is the answer bc if you subtract 45-20 that gives you 25 minutes and then in 25 minutes he read 30 pages so you can divide 30÷25 which gives you 1.2 so then you’re going to multiply 20 and 1.2 which gives u 25 and subtract 24 from 151
