the answer to your question is c
Answer:
332 ft²
Step-by-step explanation:
A rectangular garden measures 33 ft by 46 ft. Surrounding (and bordering) the garden is a path 2 ft. Find the area of this path. Be sure to include the correct unit in your answer.
Solution:
The area of the rectangular garden = length * breadth = 33 ft. * 46 ft. = 1518 ft²
Since the path surrounding the garden is 2ft, the length of the path with the garden = 46 + 2 + 2 = 50 ft
The width of the path with the garden = 33 + 2 + 2 = 37 ft.
Area of the path with the garden = length * breadth = 50 ft * 37 ft = 1850 ft².
Area of the path = 1850 ft² - 1518 ft.² = 332 ft²
Answer:
(4,4)
Step-by-step explanation:
hope this help
3 x 6 x 9 x 3 x 6 x 9 = ?
3 x 6 = 18
18 x 9 = 162
162 x 3 = 486
486 x 6 = 2,916
2,916 x 9 = 26,244
3 x 6 x 9 x 3 x 6 x 9 = 26,244
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.
