Answer: 2 bookcases.
Step-by-step explanation:
Let be "x" the number of bookcases that the custodian will be able to repaint completly with 5 gallons of paint.
According to the data given, the custodian needs 2 gallons of paint to repaint one bookcase. Since all the bookcases are equal, you can set up the following proportion:
Now you must solve for "x" in order to find its value:
Notice that the custodian will be able to repaint 2 bookcases completely, then you can say that:
3.3 ft --- 1 m
11.4 ft --- x m
x = 11.4/3.3 ≈ 3.45 m
Slope = (y2-y1)/(x2-x1)
The points: (2,0) and (0,3)
(3-0)/(0-2) = 3/-2 = -3/2
The solution is -3/2
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.
Answer: I think it's the first one.
Step-by-step explanation:
