525cm bro because i just took it right now
Answer:
Answer:
t ≅ 5.09 min
Step-by-step explanation:
we have that if in 4000 L/sol there is 132 kg salt and the pumping speed is 12L/s, we must find how much of salt is pumping per second and then find the amount of salt remaining
12L/s*132kg salt/4000L = 0.396 Kg salt/s, this means that 0.396 kg per second comes out , It should be found that the amount of salt must be drained so that only 11 kg of salt remain
132kg salt - 11 kg salt = 121 kg salt, so
121Kg salt*s/0.396Kg salt ≅ 305.55 s ⇒ 305.55s*min/60s ≅ 5.09 min
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.
Answer:
Step-by-step explanation:
√3(√6+√15)
3√2+3√5
2√3+3√5
3√7
3√2+9√5
