Answer:
4,391
Step-by-step explanation:
next time use a caculator luv <3
A-length
b-width
perimeter of rectangular floor: 2a+2b=204
<span>The length is two times the width: a=2b
</span>2a + 2b = 204
2*2b + 2b = 204
6b = 204|:6
b = 34
a=2b=2*34=68
a=68->length
b=34->width
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:
Not reasonable. Not reasonable.
Step-by-step explanation:
1) 39x37=2183 We've been given this product.
Since 37 and 39 is greater than 35, then we can Round it up both to 40. Estimating that product means 40*40=1600. 2183 exceeds 1443. So the first product we've been given it's not a reasonable one.
39x37=2183 (Given)
Rounding up
40 x 40 = 1600
39 x 37 = 1443
Not Reasonable. Because 2183 exceeds the estimation (1600).
2) 27 x 8 = 2241
Since 27 is near to 30, as well as 8 is to 10. Let's round them both up:
30* 10=300
27 x 8 =216
Not Reasonable. Because 2241 exceeds the estimation (300).
Check out https://www.umass.edu/preferen/Game%20Theory%20Evolving/GTE%20Public/GTE%20Answers.pdf it will help!
