Answer:
2x +3x =7-13=4+8
Step-by-step explanation:
5x= -6= 12
A transversal intersects two parallel lines, given that the lines are on the same plane.
Step-by-step explanation:
57,500,000,000 is in standard form
Check the picture below.
so, if we slap the bottom surplus to the empty area on the top, we can pretty much close off those squares above, that gives a solid area of at least a 4x5 square, the red part and the yellow part there, and that'd give us a perimeter of 4+4+5+5 = 20 at least.
we might also be able to augment the left-right sides together to close off those squares as well, but even if we did, our area will just be about a 4x6 or 24 square units.
A) is the area larger than 14 square units? ✔
B) is the area less than 35 square units? ✔
C) is the perimeter more than 16? ✔
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.
