5<em>x</em>² - 7<em>x</em> + 2 = 0
5(<em>x</em>² - 7/5 <em>x</em>) + 2 = 0
5(<em>x</em>² - 7/5 <em>x</em> + 49/100 - 49/100) + 2 = 0
5(<em>x</em>² - 2 • 7/10 <em>x</em> + (7/10)²) - 49/20 + 2 = 0
5(<em>x</em> - 7/10)² - 9/20 = 0
5(<em>x</em> - 7/10)² = 9/20
(<em>x</em> - 7/10)² = 9/100
<em>x</em> - 7/10 = ± √(9/100)
<em>x</em> - 7/10 = ± 3/10
<em>x</em> = 7/10 ± 3/10
<em>x</em> = 10/10 = 1 or <em>x</em> = 4/10 = 2/5
Answer:
Step-by-step explanation:
(1). (8,4)
(2). (6,0)
(3). (3,5)
(4). (24,3)
(5). (12,17)
(6). (2,3)
(7). (4,1)
(8). (2,1)
Answer:
1
Step-by-step explanation:
If you convert it to a decimal both fractions equal to 1.5/1.5 which makes the answer 1
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.
