It is true because
when you separate 2y^2 +9y -18, you are supposed to get the factors
(2y-3)(y+6), when you multiply this it gives you the first equation, we know
this because when we multiply the first term it gives us exactly 2y^2 and if we
multiply the last ones it gives you -18, now here is the tricky part, when we
multiply the middle part we get 12y and -3y, when we add these up we get 9y,
now is that the same middle as in the equation? Yes… then our factors are correct
making this true
Hope this helps
Answer:
Growth
Step-by-step explanation:
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.
9514 1404 393
Answer:
y = -1/2x +11/2
Step-by-step explanation:
The slope of the line is ...
m = (y2 -y1)/(x2 -x1)
m = (1 -2)/(9 -7) = -1/2
The y-intercept is ...
b = y -mx
b = 2 -(-1/2)(7) = 11/2
Then the slope-intercept equation is ...
y = -1/2x +11/2
_____
<em>Alternative solution</em>
A general form equation for the line can be ...
(y1 -y2)(x -x1) -(x1 -x2)(y -y1) = 0
(2 -1)(x -7) -(7 -9)(y -2) = 0
x-7 +2y -4 = 0
x +2y -11 = 0 . . . . . general form equation
x +2y = 11 . . . . . . . standard form equation
Note that we want the x-coefficient to be positive, so we chose the order of the points to make that be the case.
