Answer:
x = -5
y = 6
Step-by-step explanation:
3x + 7y = 27 --------------(I)
-3x + y = 21 -----------(II)
Add equation (I) & (II) and so x will be eliminated and we can find the value of y.
(I) 3x + 7y = 27
(II) <u> -3x + y = 21 </u> {add}
8y = 48
y = 48/8
y = 6
Plugin y = 6 in equation (I)
3x +7*6 = 27
3x + 42 = 27
3x = 27 - 42
3x = -15
x = -15/3
x = -5
The answer will be y^2/4 (I’ve written up properly on the photo as I can’t type it)
F it rose 10% that means that it is now worth 110% of what it was a year ago. So set up an equation: 297 = 110% of x297 = 1.1 x270 = x So it was worth $270 a year ago.
Angel D: 56 degrees that's the answer and your welcome
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.
