Answer:
x = 34 degrees
Step-by-step explanation:
Each solid line and triangle add up to 180, so start from the left and slowly do your math.
180 - 112 = 68
180 - 68 - 43 = 69
180 - 69 = 111
180 - 111 - 35 = 34
x and the last angle are the same.
X + y = 15
Rearrange this to isolate for x.
x = 15 - y
x + z = 20
Rearrange this to isolate for x also.
x = 20 - z
Since they are both equal to x, that means they are equal to each other (x = x)
Set them equal to each other.
15 - y = 20 - z
Now, isolate for y.
y = 15 - 20 + z
Now, use y + z = 25 and isolate for y.
y = 25 - z
Since they are both equal to y, we set them equal to each other.
15 - 20 + z = 25 - z
Solve for z
z + z = 25 - 15 + 20
2z = 30
2z/2 = 30/z
z = 15
Now you can plug the value for z into the other equations to find y and x.
x + z = 20
x +15 = 20
x = 20-15
x = 5
y + z = 25
y + 15 = 25
y = 25 - 15
y = 10
Answer:
To add percentages, you have to make the percentage into a decimal with "1" as the whole number if it's lower than 100%.
So, $60 x 1.60 =
$96.00
The original price of the pair of shoes were: $96.00
Answer:
x 
Step-by-step explanation:
x
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.
