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.
You didn't include the formula.
Given that there is no data about the mass, I will suppose that the formula is that of the simple pendulum (which is only valid if the mass is negligible).
Any way my idea is to teach you how to use the formula and you can apply the procedure to the real formula that the problem incorporates.
Simple pendulum formula:
Period = 2π √(L/g)
Square both sides
Period^2 = (2π)^2 L/g
L = [Period / 2π)^2 * g
Period = 3.1 s
2π ≈ 6.28
g ≈ 10 m/s^2
L = [3.1s/6.28]^2 * 10m/s^2 =2.43 m
Hope this helps you!!!!
<span><em>Ps: Please mark brainliest!!!! I am only a few away from ranking up, it would help a lot, and I will make a shoutout for you!! </em></span>
Answer:
Options 1,2,3 are correct statements
The first values in an ordered pair ( the coordinates) are the domain : ( -3,-2,-1,0,1)
The second values are the range : (-1,1,2,3,4,5)
One input should have only one output for a function.
But -1 input has 2 outputs: -1 and 3.
Thus, it is a relation and not a function.
Hope it helps :)
Mark it Brainliest pls:)
The answer would be b. THis is becuse the second choice has less than or equal to and just less than signs. This is what is shown on the graph. The open circles are the less than sides, and the closed circles mean less than or equal to. Hope this helps.
8n = -3m + 1
n = -2, 2,4
first add -2, 2 and 4 which = 4
then do 8(4) = -3m +1 which makes 24 = -3m +1 so subtract 1 on both sides then you have 23 = -3m divde -3 on both sides which equal -7.66 or 8
