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VladimirAG [237]
3 years ago
15

Which angles are corresponding angles check all that apply

Mathematics
2 answers:
Setler79 [48]3 years ago
8 0

Answer-

(1 and 3), (5 and 7), (6 and 8) are corresponding angles

<u>Solution-</u>

Corresponding Angles

-

When two parallel lines are crossed by another line (called the Transversal), the angles in matching corners are called Corresponding Angles.

e.g As shown in the attachment attached herewith, angle 1 and 2, 3 and 4 are Corresponding Angles.

∴ In the question, angle 1 and 3, 5 and 7, 6 and 8 are Corresponding Angles.

labwork [276]3 years ago
5 0
Corresponding angles are those in matching corners:
.. C. 1 and 3
.. D. 5 and 7
.. E. 6 and 8
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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.

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