Answer:
Step-by-step explanation:
If two polygons are similar, then corresponding sides are in proportion.
The corresponding sides:
4 → x
y → 15
3 → w
2 → 6
z → 9
therefore:
<em>cross multiply</em>
<em>divide both sides by 6</em>
let's recall the vertical line test, it's a function if when dropping a vertical line on the graph, it only touches it once on the way down.
Check the picture below.
Answer:
x = 7/15
Step-by-step explanation:
you can check by replacing the x with the number I gave you
Answer:
D. y = 5x + 29
Step-by-step explanation:
To write an equation of a line in slope-intercept form, we need to find the slope and the y-intercept.
The equation of a line is y=mx + b.
x and y are the coordinates of any point on a line.
m is the slope.
b is the y-intercept.
Lines are parallel to each other when they have the same slope. A line parallel to y = 5x + 2 would also have the slope 5.
m = 5
Since we have a point on the line, (-6, -1) and the slope, 5, there is only one missing variable, b, the y-intercept.
Substitute the known information into the equation and isolate b.
y = mx + b
-1 = 5(-6) + b <=Simplify by solving 5 X -6
-1 = -30 + b
-1 + 30 = -30 + 30 + b <= Add 30 to both sides to isolate b
29 = b
b = 29 <= Standard formatting puts the variable on the left side
Put the m and b values into the equation of a line to solve:
y = mx + b
y = 5x + 29
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.
