There's literally no reason to write the problem that way without parentheses. Problems are meant to model real world analogues for analysis so if you know the situation and write the problem that way anyway then you are perpetuating a useless notation.
For crying out loud, it is not 9, it is 1. 1 = 6/6 = 6/(2+4) = 6/2(1+2). Quoting PEMDAS does NOT allow you to do this: 6/2(1+2) = 6/2 * (1+2) = 3*(1+2) = 3*3 = 9. That 2(1+2) = (2+4) . It's like the 2 is still part of the parentheses and comes under the P and not the M. What was just taking out a factor of 2 is being treated as a separate operand in the multiplication operation, but then you switch it and you make it an operand of the other operator, the /. which is not a commutative operator, it is an operator in which the order matters. 2/4 != 4/2. Which is why you can't do that, please stop.
6'÷2(1'+2')=1 and is in fact a Distance divided by circumference equation of 1 revolution. D÷2πr=D÷2π(h+t). 6/2(1+2)=9 in exactly the same way as any other standard formula with a fractional coefficient. Kinetic energy, triangular areas, spherical volumes, etc. 6/2(1+2)=6(1/2)(1+2) and is all a single product of factors.
@clivehill4470 Well no. 6/2(1+2)≡(6/2)(1+2)=9 Just exactly like 1/2Mv², kinetic energy, triangular areas, spherical volume and a host of other standard formulas with fractional coefficients. 6÷2(1+2)≡6÷[2×(1+2)]=1 Because the coefficient of the parenthetical expression is 2, the composite quantity of 2(1+2) has a value of 6. 6÷6=1 There is an issue with 6/(2×(1+2)), called the Neutral Element property of multiplication in that all variables and symbols of grouping have coefficients. Even if that coefficient is the multiplicative identity of 1. So 6/1(2×(1+2))≠6÷[2×(1+2)] 6×6≠6÷6
@@Vibe77Guy Wow, you guys are persistent. You could write Mv²/2, then it would be correct, and clear that you don't actually mean what you wrote, 1/(2Mv²). I could maybe, slightly, see your point if you were trying to type this into a calculator, where you have to press your * key each time, but in the original there are parentheses. So there, you are failing to respect that 2(1+2)=(2+4) and you are 100% wrong when you do that.
@clivehill4470 I did not write 1/(2Mv²) I wrote 1/2 Mv² interpreted as (1/2)(Mv²) in almost all reference pages of the physical sciences textbooks. You could write 4/3πr³ as 4πr³/3 as well. No one does, but since multiplication is commutative, the order of the factors 4(1/3)(π)(r³) really is immaterial. Just as the order of the factors 2(1+2)=(1+2)2 has no effect on the correct results.
@@Vibe77Guy When you wrote 1/2Mv² it is equivalent to 1/(2Mv²). What you quote from textbooks is really going to be ½Mv² which is a whole different story. Yes, multiplication is commutative, but division is not, and that is a very important point. It really matters if Mv² is multiplied by the 1 and then divided by the 2 or if it is multiplied by the 2 and then 1 is divided by that. The way you are writing it puts an implied multiplication between the 2 and the Mv² and makes the formula wrong. The way you want to do things, how do I know you don't mean v²/2M ?
There's literally no reason to write the problem that way without parentheses.
Problems are meant to model real world analogues for analysis so if you know the situation and write the problem that way anyway then you are perpetuating a useless notation.
For crying out loud, it is not 9, it is 1. 1 = 6/6 = 6/(2+4) = 6/2(1+2). Quoting PEMDAS does NOT allow you to do this: 6/2(1+2) = 6/2 * (1+2) = 3*(1+2) = 3*3 = 9. That 2(1+2) = (2+4) . It's like the 2 is still part of the parentheses and comes under the P and not the M. What was just taking out a factor of 2 is being treated as a separate operand in the multiplication operation, but then you switch it and you make it an operand of the other operator, the /. which is not a commutative operator, it is an operator in which the order matters. 2/4 != 4/2. Which is why you can't do that, please stop.
Get a load of this guy
what is bro yapping about 🤯
maybe 1 is the friend we made along the way 😋
Bro, you are not the teacher!
@@raedomos He his correct. The teacher is wrong and it is getting tiring seeing this coming up every now and then.
6'÷2(1'+2')=1 and is in fact a Distance divided by circumference equation of 1 revolution.
D÷2πr=D÷2π(h+t).
6/2(1+2)=9 in exactly the same way as any other standard formula with a fractional coefficient. Kinetic energy, triangular areas, spherical volumes, etc.
6/2(1+2)=6(1/2)(1+2) and is all a single product of factors.
Absolutely not. 6/2(1+2)=6/(2(1+2)), 6(1/2)(1+2)=6(1+2)/2.
@clivehill4470
Well no.
6/2(1+2)≡(6/2)(1+2)=9
Just exactly like 1/2Mv², kinetic energy, triangular areas, spherical volume and a host of other standard formulas with fractional coefficients.
6÷2(1+2)≡6÷[2×(1+2)]=1
Because the coefficient of the parenthetical expression is 2, the composite quantity of 2(1+2) has a value of 6. 6÷6=1
There is an issue with 6/(2×(1+2)), called the Neutral Element property of multiplication in that all variables and symbols of grouping have coefficients. Even if that coefficient is the multiplicative identity of 1.
So 6/1(2×(1+2))≠6÷[2×(1+2)]
6×6≠6÷6
@@Vibe77Guy Wow, you guys are persistent. You could write Mv²/2, then it would be correct, and clear that you don't actually mean what you wrote, 1/(2Mv²). I could maybe, slightly, see your point if you were trying to type this into a calculator, where you have to press your * key each time, but in the original there are parentheses. So there, you are failing to respect that 2(1+2)=(2+4) and you are 100% wrong when you do that.
@clivehill4470
I did not write 1/(2Mv²)
I wrote 1/2 Mv² interpreted as
(1/2)(Mv²) in almost all reference pages of the physical sciences textbooks.
You could write 4/3πr³ as 4πr³/3 as well.
No one does, but since multiplication is commutative, the order of the factors
4(1/3)(π)(r³) really is immaterial.
Just as the order of the factors
2(1+2)=(1+2)2 has no effect on the correct results.
@@Vibe77Guy When you wrote 1/2Mv² it is equivalent to 1/(2Mv²). What you quote from textbooks is really going to be ½Mv² which is a whole different story. Yes, multiplication is commutative, but division is not, and that is a very important point. It really matters if Mv² is multiplied by the 1 and then divided by the 2 or if it is multiplied by the 2 and then 1 is divided by that. The way you are writing it puts an implied multiplication between the 2 and the Mv² and makes the formula wrong. The way you want to do things, how do I know you don't mean v²/2M ?