Answer:
we conclude that:
If 4x - 6≠4, then 2x–5≠5 is the contrapositive of a conditional statement if 2x -5=5, then 4x-6=14.
Step-by-step explanation:
We know that the contrapositive of a conditional statement of the form "If p then q" is termed as "If ~q then ~p".
In other words, it is symbolically represented as:
' ~q ~p is the contrapositive of p q '
For example, the contrapositive of "If it is a rainy day, then they suspend the match" is "If they do not suspend the match, then it won't be a rainy day."
Given
p: 2x -5=5
q: 4x-6=14
As the contrapositive of a conditional statement of the form "If p then q" is termed as "If ~q then ~p
Thus, we conclude that:
If 4x - 6≠4, then 2x–5≠5 is the contrapositive of a conditional statement if 2x -5=5, then 4x-6=14.
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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.
16 length height bc it’s a triangle so the triangle will be 17 feet tall and long
