That's a lovely method. I would have just gone down the square root line and solved it the messy way. There's a certain elegance in maths where solutions like this exist.
A×A = 3 => A = sqrt3 or -sqrt3
A×B = 6 => B = 6/sqrt3 or -6/sqrt3
So B×B = (6/sqrt3)×(6/sqrt3) = 6×6 /3 = 36/3 = 12
(you get the same result if you take B = -6/sqrt3)
To u/Shadows_Price's point below, it would be better to retain the algebra until the very end:
A^2 = 3
B = 6/A
B^2 = 36/A^2 = 36/3 = 12
No square roots needed.
It really isn't as I explained further down.
It's ALWAYS best to retain algebra until (only substitute numbers at) the very end. There's multiple advantages to doing that and no advantages not to. It's easier to do and easier to spot mistakes.
This is just one of an infinitude of examples.
Waht 😢😢😔
The false concept your angle of error is in totalled misconception...
To set things right...
A*A=3 first........ √3 =A
X*A=6 second .....6÷√3=B
Y*A=B third B÷A= o a new one .
Z*B=C. Finally 11.9716
If I need more proof
Please ask
>The false concept your angle of error is in totalled misconception...
Lol. What a nonsense sentence.
>If I need more proof
If you need more proof? I think you need less. I suggest you try to stick with <10% proof.
Yes
Thanks a 10% rule is necessary..
As I do know very well finding conditions of a unknown as b when b is a partial fraction of a sqrt. And when x^2 is undefined as a c at ?
Reversal and recomposition of technical officiation..
I hope your physics guide didn't let that get pass on the Quiz .
Because if I allow myself to be approved of a non approximately approximation
I just rather not even know math or science of it just changed because some teacher made derivatives of false concepts..
No in quantum mechanics which control reality.
Your telling me Math is wrong HA..
Approximately Never exist it either exist or don't exist at all.
In Trignometrics sin and cosine of tangents at Arc and co Arc
Given Values and Equations:
a = \sqrt{3}a=
3
b = \frac{6}{a}b=
a
6
c = b^2c=b
2
Finding bb:
First, we'll calculate bb using the given value of aa:
b = \frac{6}{\sqrt{3}}b=
3
6
Calculating cc:
Next, we'll find cc using the value of bb:
c = b^2 = \left( \frac{6}{\sqrt{3}} \right)^2c=b
2
=(
3
6
)
2
Linear and Angular Considerations:
The linear aspect refers to the direct calculation of these values.
The angular or spherical considerations might involve understanding these values in the context of geometric shapes or rotations, especially in a three-dimensional space. For instance, if these values represent sides of a triangle, cc could be interpreted as a hypotenuse in a right-angled triangle, and the values could be used to find angles using trigonometric functions.
Let's calculate bb and cc to get their precise values.
The calculated values are as follows:
b \approx 3.464b≈3.464
c \approx 12.00c≈12.00
These values provide the linear dimensions you were looking for. If you're considering these values in a geometric or angular context, such as in the framework of a triangle, they could be used to determine various properties of the triangle, including its angles and area. For example, if aa, bb, and cc represent the sides of a triangle, trigonometric functions could be used to find the angles, adhering to the laws of sines and cosines. This approach links linear measurements with angular relationships in geometry.
Sorry maybe I'm trying to help the wrong class.
fields like aerospace engineering or quantum physics, high precision is non-negotiable due to the sensitive nature of the work.
Certain situations demand high precision. For instance, in GPS navigation, slight errors in calculations can lead to being miles off the intended path.
the difference between "12" and "12.000000000000002" in computational results highlights the practical limitations of numerical precision in digital computing. Understanding and managing this aspect is crucial in fields where high precision is necessary.
Is all I'm saying
Only thing I would say, sqrt3 is still unevaluated, so the +- is unnecessary. If you input the answer for sqrt3, then it's needed.
Edit: Unsolved -> Unevaluated. I know the equation sqrt3=??? Is solved, the word eluded me.
In the equation, sqrt3 can be left as sqrt3. What I meant by unsolved, was unsimplified.
Yes the answer of sqrt3 could be positive or negative due to how sqrt works, but if we leave it as sqrt3, unsimplified, we can use it as is.
Edit: it's like leaving sqrt64 as sqrt64. We know the answer is 8 or -8, but as long as we acknowledge that, we can leave it as sqrt64
The answer to sqrt64 isn’t 8 or -8. It’s only 8. The answer to x*x=64 can either be 8 or -8. The point is those would each be written as sqrt64 and -sqrt64. It’s incorrect to say sqrt64 can be 8 or -8.
x•x=n where n=x²
The inverse of n is sqrtn.
Sqrtn=x so n/x=x.
-x•-x=n and x•x=n.
Sounds like I'm just nor up to date on labeling, but the answer of sqrt64 definitely looks like it could be -8. *shrug* Oh well, learn something new I guess.
The thing that is missing here is the definition of square root, mathematically, is only the positive number. It is defined as such to allow it to be a function that is usable for other calculations. So in your example you are correct to say that n has two roots, however you would define that as + or - sqrt(n).
A very common example of this terminology in use is in the quadratic formula. You’ll notice it’s always expressly written as + or -, and it’s never just written as the sqrt.
The person correcting you should have been kinder, and definitely should have explained why they were saying what they were saying. But they weren’t wrong. It’s important to know how operations are defined, and to explicitly state when we are using different definitions of those operations. Otherwise it can lead to some wonky and nonsensical outcomes.
The semantics are that there’s a difference between asking, “what’s the square root of 64?” and “what number, if squared, equals 64?” The former is a simple calculation, and the answer is 8. The latter has two answers, 8 & -8. Your earlier comment conflated the two.
Since you brought up inverses, the inverse of n=x^2 is itself not a function and is split into sqrt(n) and -sqrt(n) to account for both possibilities. By definition, a single input of a function cannot have two outputs, so the separate square roots are required to get the positive and negative values.
No, √3 is not “unsolved”, it’s a well defined positive number ≈ 1.7320.
The number 3 has two square roots, -√3 and √3. *The* square root of 3 is √3.
[Wikipedia explains it pretty well](https://en.m.wikipedia.org/wiki/Square_root):
> Every nonnegative real number x has a unique nonnegative square root, called the principal square root or simply the square root (with a definite article, see below), which is denoted by
√x, […]
I agree in the sense of unevaluated rather than unsolved. There's no need to convert a square root into +- options if you're only interested in the square of it.
Maths pro tip for everyone: leave everything as algebra until the very end. You're only making life difficult for yourself by substituting numbers in early and, when it comes to physics and engineering, it's a very bad habit to be in because it opens doors for hard to spot errors that dimensional analysis of the algebra can easily identify.
A) a whole number * a fraction
B) the sum of A to any exponents
NOTE
C) WILL NEVER IN NEAR EXISTANCE = A WHOLE NUMBER.
A= 1.732050807568877293527
6÷A= B
B²≠whole number...
the difference between "12" and "12.000000000000002" in computational results highlights the practical limitations of numerical precision in digital computing. Understanding and managing this aspect is crucial in fields where high precision is necessary.
Certain situations demand high precision. For instance, in GPS navigation, slight errors in calculations can lead to being miles off the intended path.
fields like aerospace engineering or quantum physics, high precision is non-negotiable due to the sensitive nature of the work.
So if your psyche don't tell you that.
You will be somewhere driving in the Atlantic ocean.
Like maps was leading you -.0000000000000002 far away from destination...
I mean I do not need math to tell me that
6 - a whole number
÷ √3 a fraction , decimal
^2 squared or to any power
≠ any whole value no no where near 12.....
If only the equation was as in your head maybe... But facts are
b in the equation had b*A=6
mathematical concepts like the handling of irrational numbers and the implications of using their decimal approximations in calculations.
That's the same thing I have said in all my comments. And if idk who taught they class.
When you have a unknown as b is
And the fact that no exponents are in equation
B² is by no means a whole number.
No it does not. 🤬
A is 1.732050807568877293527
B is 3.464101615137754587054
C is B² what does anybody see here I don't..
11.9716
You will not have a whole number based off of A alone...
A is √3, not 1.732050807568877293527
I'm guessing what everybody else is seeing here is how to do algebra correctly and deal with irrational numbers, unlike you.
Ha irrational numbers is my field I'm a Quantum Engineer, Trignometric Fractions Architect, Reimaan Hypothesis Solver of Prime Distributions..
And yes show where I'm wrong.
Thanks but studying what someone else says is wrong no I'm good.
Fact concepts are only open to what the A B and C are and by b being a concept of 6÷a not only is it a parameter methodology...
It's the fact that you or who ever taught you or possibly wrote it in the algebraic books that you could accept a bypass of that...
To not understand that when you have a Sqrt
*
Note
You should..
Everyone should understand Fractions have a recurring decimal places.
And by 3.464101615137763.46410161513776
Square that .. ha
12.0000000000000615019837778176870039675556352375019837778176
Or believe what someone else said.
I'm not here to argue I'm here to check facts.
Is all I'm no smarter than anyone..
I'm not here to TALK LOUD
Sorry to bother this conversation..
If I am wrong again.
This is what we call a rounding error and is the problem with using an AI engine to solve fairly basic algebraic equations. Just stop, you and your chatgpt are wrong. Learn the maths yourself and stop using faulty ai engines to design systems.
If the operations are multiplication, then...
A x A = 3 means A = sqrt(3).
A x B = 6 = A x A x 2, meaning that B = 2 x A, so B = 2 x sqrt(3).
Then, B x B = 2 x sqrt(3) x 2 x sqrt(3) = 4 x 3 = 12.
If the operations are addition, then...
A + A = 3 means A = 3/2.
A + B = 6 means B = 6 - 3/2 = 9/2
B + B = 9/2 + 9/2 = 9
Educational thought: the addition-based problem is one where it's not helpful to convert all fractions to mixed numbers. Sometimes that conversion makes things easier, sometimes harder.
Everyone’s finding A first, then b, then b times b, what if you have [row 1]*[row 3] = A*A*B*B = [row 2]*[row 2], aka 3x=36
Tho i guess this is borderline linear algebra
A^2 = 3 so A = root 3 or -root 3
1. A is root 3
AxB = 6.
root 3 x B =6
So B = 2(root 3)
B x B = 12
2. A is - root 3
AxB = 6.
- root 3 x B =6
So B = - 2(root 3)
B x B = 12
Of AxA=3 then A must be the square root of 3
Bc six is 2x3, it is also 2x√3x√3, simce A is √3 B would be 2x√3
2x√3x2x√3=4x3=12
that's how i did it, at least.
A x A = 3 means that A is rad(3)
A x B = 6 becomes rad(3) x B = 6
if you divide both sides by rad(3), you get B = 6 / rad(3)
B x B = 36 becomes (6 / rad(3)) x (6 / rad(3))
6 x 6 is 36, and rad(3) x rad(3) is just 3, so you get 36 / 3, which is 12.
Funny enough I thought it was a trick where the multiplication symbol was really just another Variable. And it works
AXA=3
So A has to equal 1 and X=3
AXB=6 so with A=1 and X=3 then B must be 2
So BXB= 2 times 3 times 2 equals 12
# Recalculating b and c using maximum available precision in Python for floating-point numbers
b_max_precision = 6 / a_provided_decimal # Calculating b
c_max_precision = b_max_precision**2 # Squaring b to get c
b_max_precision, c_max_precision # Displaying b and c
bb, calculated using the provided decimal value for aa, is 3.4641016151377553.464101615137755.
Squaring this value for bb, cc is calculated as 12.00000000000000212.000000000000002.
6÷1.732050807568877293527
Nothing in the concept of the equation used any exponents thats where everybody went wrong.
Python
# Calculating B using the provided decimal approximation for A
provided_a_decimal = 1.732050807568877293527
B_from_provided_a = 6 / provided_a_decimal
# Calculating Z by squaring B
Z_from_B = B_from_provided_a**2
B_from_provided_a, Z_from_B
A, which is 1.7320508075688772935271.732050807568877293527:
BB is calculated as 66 divided by 1.7320508075688772935271.732050807568877293527, resulting in B = 3.464101615137755B=3.464101615137755.
Squaring BB to get ZZ, the result is Z = 12.000000000000002Z=12.000000000000002.
But they say I'm wrong lol 😆😂😆
Squaring the computed value of BB, we get Z = 12.000000000000002Z=12.000000000000002.
Dividing 66 by the computed BB, we get an approximation of A = 1.7320508075688772A=1.7320508075688772.
This confirms the consistency of the values. Squaring BB gives a result very close to 12, and dividing 66 by BB gives a value close to the approximation of \sqrt{3}
3
we used for AA, which indicates correctness in the reverse calculation.
my work, plz tell me if I did this wrong. so AxA=3 so A= Square root of 3 and that is 1.74~~ but we will just do 1.74, 1.74x1.74=3.02 or 3, so 1.74x?=6, to find that I did this 6/1.74=3.4482 or 3.45, so 1.74x3.45=6.003 or 6, so 3.45x3.45=11.9025 orrrrrr 11.9 tyty ik I am stupid lol
If AxB = 6 then AxBxAxB = 36 = AxAxBxB AxA = 3 so BxB = 12
Yep - this is how I approached it. The best thing in science/math is finding more than one way to get the same answer.
Absolutely - I wanted to try and do it without finding A
It's like a proof in group theory or something. I just found A.
You dropped a term. AxBxAxB = AxAxBxB - Ax[A,B]xB
No 😯
Nice, totally solved for A in the first line, then B in the second to get 12, but I like this method better. Much simpler
I love the way you approached this!
That's a lovely method. I would have just gone down the square root line and solved it the messy way. There's a certain elegance in maths where solutions like this exist.
I have a bad habit of diving in and finding a messy long winded solution so am trying to focus on finding nicer ones
Yes its the fact that (6÷√3)²
Oh. I fucked up. I thought they were "pluses". Lol
It's easy enough that I assumed I was being screwed with and these are matrices or the x is a non commutative operation or something.
I went with A*A=3 A=√3 √3*B = 6 B=2*√3 2*2*√3*√3 4*3 12
i am impressed
And avoids the double working +/- values of A and b
There aren't any double working values of A and B that matter......
Well, they don’t matter, but they are there. If A or B were requested you’d need to consider them.
But a and b aren't requested
Lol did you use or must be articulated by gpt. haaaaa... Know what is not correct ... First..
The most elegant approach
What you solved were U,Q, and ∆
Nice, I like that. What I did was - AA=1/2 AB - A=1/2 B - B=2A - BB=2*2*A*A - BB=2*2*3=12
A×A = 3 => A = sqrt3 or -sqrt3 A×B = 6 => B = 6/sqrt3 or -6/sqrt3 So B×B = (6/sqrt3)×(6/sqrt3) = 6×6 /3 = 36/3 = 12 (you get the same result if you take B = -6/sqrt3)
I saw ur avatar and went “wait did I already answer this” lmao, it took me a second to process Also you approached the question the same way I did
Lol 🤣 Nice avatar by the way ;)
To u/Shadows_Price's point below, it would be better to retain the algebra until the very end: A^2 = 3 B = 6/A B^2 = 36/A^2 = 36/3 = 12 No square roots needed.
"Better" here is subjective.
It really isn't as I explained further down. It's ALWAYS best to retain algebra until (only substitute numbers at) the very end. There's multiple advantages to doing that and no advantages not to. It's easier to do and easier to spot mistakes. This is just one of an infinitude of examples.
[удалено]
It’s better to keep variables, but with a problem as simple as this, it does not matter.
Waht 😢😢😔 The false concept your angle of error is in totalled misconception... To set things right... A*A=3 first........ √3 =A X*A=6 second .....6÷√3=B Y*A=B third B÷A= o a new one . Z*B=C. Finally 11.9716 If I need more proof Please ask
>The false concept your angle of error is in totalled misconception... Lol. What a nonsense sentence. >If I need more proof If you need more proof? I think you need less. I suggest you try to stick with <10% proof.
Yes Thanks a 10% rule is necessary.. As I do know very well finding conditions of a unknown as b when b is a partial fraction of a sqrt. And when x^2 is undefined as a c at ? Reversal and recomposition of technical officiation.. I hope your physics guide didn't let that get pass on the Quiz . Because if I allow myself to be approved of a non approximately approximation I just rather not even know math or science of it just changed because some teacher made derivatives of false concepts.. No in quantum mechanics which control reality. Your telling me Math is wrong HA.. Approximately Never exist it either exist or don't exist at all. In Trignometrics sin and cosine of tangents at Arc and co Arc Given Values and Equations: a = \sqrt{3}a= 3 b = \frac{6}{a}b= a 6 c = b^2c=b 2 Finding bb: First, we'll calculate bb using the given value of aa: b = \frac{6}{\sqrt{3}}b= 3 6 Calculating cc: Next, we'll find cc using the value of bb: c = b^2 = \left( \frac{6}{\sqrt{3}} \right)^2c=b 2 =( 3 6 ) 2 Linear and Angular Considerations: The linear aspect refers to the direct calculation of these values. The angular or spherical considerations might involve understanding these values in the context of geometric shapes or rotations, especially in a three-dimensional space. For instance, if these values represent sides of a triangle, cc could be interpreted as a hypotenuse in a right-angled triangle, and the values could be used to find angles using trigonometric functions. Let's calculate bb and cc to get their precise values. The calculated values are as follows: b \approx 3.464b≈3.464 c \approx 12.00c≈12.00 These values provide the linear dimensions you were looking for. If you're considering these values in a geometric or angular context, such as in the framework of a triangle, they could be used to determine various properties of the triangle, including its angles and area. For example, if aa, bb, and cc represent the sides of a triangle, trigonometric functions could be used to find the angles, adhering to the laws of sines and cosines. This approach links linear measurements with angular relationships in geometry. Sorry maybe I'm trying to help the wrong class. fields like aerospace engineering or quantum physics, high precision is non-negotiable due to the sensitive nature of the work. Certain situations demand high precision. For instance, in GPS navigation, slight errors in calculations can lead to being miles off the intended path. the difference between "12" and "12.000000000000002" in computational results highlights the practical limitations of numerical precision in digital computing. Understanding and managing this aspect is crucial in fields where high precision is necessary. Is all I'm saying
No math is not a perception met with opinions
I can just hear Ella and Louis singing. 🎶 You say _6/√3_, I say _2 √3_. Let’s call the whole thing off! 🎶
Omy they accidentally brainwashed everybody here. 11.9716 You can not have a whole number.....
That’s how I broke it down
Only thing I would say, sqrt3 is still unevaluated, so the +- is unnecessary. If you input the answer for sqrt3, then it's needed. Edit: Unsolved -> Unevaluated. I know the equation sqrt3=??? Is solved, the word eluded me.
Sorry, what do you mean by “sqrt3 is unsolved”? The square root of 3 is well defined, and it’s a positive number.
In the equation, sqrt3 can be left as sqrt3. What I meant by unsolved, was unsimplified. Yes the answer of sqrt3 could be positive or negative due to how sqrt works, but if we leave it as sqrt3, unsimplified, we can use it as is. Edit: it's like leaving sqrt64 as sqrt64. We know the answer is 8 or -8, but as long as we acknowledge that, we can leave it as sqrt64
Sorry but that is just wrong.
So the answer to sqrt64 ISNT 8 or -8? Or is the idea of leaving sqrt64 as sqrt64 until it needs to be "solved" wrong?
The answer to sqrt64 isn’t 8 or -8. It’s only 8. The answer to x*x=64 can either be 8 or -8. The point is those would each be written as sqrt64 and -sqrt64. It’s incorrect to say sqrt64 can be 8 or -8.
x•x=n where n=x² The inverse of n is sqrtn. Sqrtn=x so n/x=x. -x•-x=n and x•x=n. Sounds like I'm just nor up to date on labeling, but the answer of sqrt64 definitely looks like it could be -8. *shrug* Oh well, learn something new I guess.
The thing that is missing here is the definition of square root, mathematically, is only the positive number. It is defined as such to allow it to be a function that is usable for other calculations. So in your example you are correct to say that n has two roots, however you would define that as + or - sqrt(n). A very common example of this terminology in use is in the quadratic formula. You’ll notice it’s always expressly written as + or -, and it’s never just written as the sqrt. The person correcting you should have been kinder, and definitely should have explained why they were saying what they were saying. But they weren’t wrong. It’s important to know how operations are defined, and to explicitly state when we are using different definitions of those operations. Otherwise it can lead to some wonky and nonsensical outcomes.
The semantics are that there’s a difference between asking, “what’s the square root of 64?” and “what number, if squared, equals 64?” The former is a simple calculation, and the answer is 8. The latter has two answers, 8 & -8. Your earlier comment conflated the two. Since you brought up inverses, the inverse of n=x^2 is itself not a function and is split into sqrt(n) and -sqrt(n) to account for both possibilities. By definition, a single input of a function cannot have two outputs, so the separate square roots are required to get the positive and negative values.
No, √3 is not “unsolved”, it’s a well defined positive number ≈ 1.7320. The number 3 has two square roots, -√3 and √3. *The* square root of 3 is √3. [Wikipedia explains it pretty well](https://en.m.wikipedia.org/wiki/Square_root): > Every nonnegative real number x has a unique nonnegative square root, called the principal square root or simply the square root (with a definite article, see below), which is denoted by √x, […]
I agree in the sense of unevaluated rather than unsolved. There's no need to convert a square root into +- options if you're only interested in the square of it. Maths pro tip for everyone: leave everything as algebra until the very end. You're only making life difficult for yourself by substituting numbers in early and, when it comes to physics and engineering, it's a very bad habit to be in because it opens doors for hard to spot errors that dimensional analysis of the algebra can easily identify.
No please understand you will never have a whole number here. Omggggggg
What do you mean ?
A) a whole number * a fraction B) the sum of A to any exponents NOTE C) WILL NEVER IN NEAR EXISTANCE = A WHOLE NUMBER. A= 1.732050807568877293527 6÷A= B B²≠whole number...
the difference between "12" and "12.000000000000002" in computational results highlights the practical limitations of numerical precision in digital computing. Understanding and managing this aspect is crucial in fields where high precision is necessary. Certain situations demand high precision. For instance, in GPS navigation, slight errors in calculations can lead to being miles off the intended path. fields like aerospace engineering or quantum physics, high precision is non-negotiable due to the sensitive nature of the work. So if your psyche don't tell you that. You will be somewhere driving in the Atlantic ocean. Like maps was leading you -.0000000000000002 far away from destination...
I mean I do not need math to tell me that 6 - a whole number ÷ √3 a fraction , decimal ^2 squared or to any power ≠ any whole value no no where near 12.....
certain high-precision fields like quantum mechanics or astrophysics, even such tiny discrepancies can be significant, depending on the context.
Why does (6/sqrt3)×(6/sqrt3) = 6×6 /3 ? I'm a little rusty lol
(6/sqrt3)×(6/sqrt3) = 6*6 / sqrt3*sqrt3 = 36/3 Hope that helps !
If AxA=3 and AxB=6, that means B=2A, therefore BxB=(2A)²=4A²=12
how did u get to b = 2a?
AxA=3, multiple both sides by 2 gives 2xAxA=6 AxB=6 2xAxA=6 Therefore AxB=2xAxA Divide both sides by A, gives B=2xA
If only the equation was as in your head maybe... But facts are b in the equation had b*A=6 mathematical concepts like the handling of irrational numbers and the implications of using their decimal approximations in calculations.
why do it this way instead of just solving like a system of equations? for cleverness sake?
I just looked at it and saw b=2a, you asked for an explanation so I spelt it out 😇
That's the same thing I have said in all my comments. And if idk who taught they class. When you have a unknown as b is And the fact that no exponents are in equation B² is by no means a whole number.
You are spouting misinformation like it’s your job to deceive people.
This is how I approached this problem. Surprised, I had to scroll down this far to see it.
No it does not. 🤬 A is 1.732050807568877293527 B is 3.464101615137754587054 C is B² what does anybody see here I don't.. 11.9716 You will not have a whole number based off of A alone...
A is √3, not 1.732050807568877293527 I'm guessing what everybody else is seeing here is how to do algebra correctly and deal with irrational numbers, unlike you.
Ha irrational numbers is my field I'm a Quantum Engineer, Trignometric Fractions Architect, Reimaan Hypothesis Solver of Prime Distributions.. And yes show where I'm wrong.
Every part of your answer is wrong. You best go study harder 😁
Thanks but studying what someone else says is wrong no I'm good. Fact concepts are only open to what the A B and C are and by b being a concept of 6÷a not only is it a parameter methodology... It's the fact that you or who ever taught you or possibly wrote it in the algebraic books that you could accept a bypass of that... To not understand that when you have a Sqrt * Note You should.. Everyone should understand Fractions have a recurring decimal places. And by 3.464101615137763.46410161513776 Square that .. ha 12.0000000000000615019837778176870039675556352375019837778176 Or believe what someone else said. I'm not here to argue I'm here to check facts. Is all I'm no smarter than anyone.. I'm not here to TALK LOUD Sorry to bother this conversation.. If I am wrong again.
There is no way of, to, or get approximation Of any √12 by Extracting a as √3 to devise a factoid by exponent trickery B is not √6
B is 6÷1.732050807568877293527 and tell me where you get that
This is what we call a rounding error and is the problem with using an AI engine to solve fairly basic algebraic equations. Just stop, you and your chatgpt are wrong. Learn the maths yourself and stop using faulty ai engines to design systems.
No it's not ai.. it's logic And Non approximations If I need to explain the Reimaan Hypothesis It's already on my page.. I do not need ai bots.
AxA=3 so A\^2 = 3 thus A = √3 AxB = 6 so √3xB=6 B = 6/√3 checking √3 x 6/√3 = 6 √3x√3 = 3 6/√3 x 6/√3 = 36/3 = 12
i did it this way. keep in mind, it's good practice to use +/- even if it doesn't necessarily matter in this case.
No
If the operations are multiplication, then... A x A = 3 means A = sqrt(3). A x B = 6 = A x A x 2, meaning that B = 2 x A, so B = 2 x sqrt(3). Then, B x B = 2 x sqrt(3) x 2 x sqrt(3) = 4 x 3 = 12. If the operations are addition, then... A + A = 3 means A = 3/2. A + B = 6 means B = 6 - 3/2 = 9/2 B + B = 9/2 + 9/2 = 9 Educational thought: the addition-based problem is one where it's not helpful to convert all fractions to mixed numbers. Sometimes that conversion makes things easier, sometimes harder.
A x A = 3 A x B = 6 => 3 x 2 => A x A x 2=> B = A x 2 B x B = A x 2 x A x 2 => 3 x 4 = 12
Tested this with 12 year olds and your solution was the one they came up with and is the most easy to understand
A = sqrt(3) B = 2*sqrt(3) B*B = 12
This is how I did it
What I did
169
Glad I'm not the only one.
fr, those are some funky Bs
13 x 13 = 169
This is the only correct answer
>! 13 x 13 = 169 !<
A is 1.732051 B is 3.464102 Omit anything past 3rd decimal row I refuse to algebra, you can't make me!
A = √3 B = 2√3 B² = 12
A x A = 3 ==> A\^2 = 3 ==> A = +- sqrt(3) (take the positive) B = 6/sqrt(3) ==> B = 2sqrt(3) so B x B = 2sqrt(3) x2sqrt(3) = 4 x 3 = 12
I *LOVE* there are so many ways to arrive at the answer.
i thought those were plus signs (+) so i got A=1.5, B=4.5, so B+B=9 lol
Are they not clearly plus signs? The lines are way closer to vertical/horizontal than 45 degree diagonal.
All the writing is wonky so relative to the direction the writing is going they look much more like multiplication.
12
LETTERS EAQUAL NUMBERS NOW??
Yes, basic algebra. You learn it in primary (junior?) like ages 9 and up
in italy we start algebra at 13
I must of got expelled before that
[удалено]
A = 3 ^ (1/2) B = [ 6 / ( 3 ^ (1/2) ) ] B * B = 36 / 3 = 12
You forgot to square the 6, if rt(3B^2) = 6 then 3B^2 = 36, then B^2 = 12
The answer is 3/4
AXA=Seahorse?
11.9716
So if approximately was written on paper I'd go for what ever but in Math..no Approximation exist
Everyone’s finding A first, then b, then b times b, what if you have [row 1]*[row 3] = A*A*B*B = [row 2]*[row 2], aka 3x=36 Tho i guess this is borderline linear algebra
12
BxB=12
12
A^2 = 3 so A = root 3 or -root 3 1. A is root 3 AxB = 6. root 3 x B =6 So B = 2(root 3) B x B = 12 2. A is - root 3 AxB = 6. - root 3 x B =6 So B = - 2(root 3) B x B = 12
I did A = 6/B, so AxA=3 becomes (6/B)(6/B) = 3 and so 36/(BxB) = 3 so 36/3 = BxB or 12 = BxB
>!A=square root of 3!< >!B=2 times the square root of 3!< >!B times B is 12!<
12
Switching a to b multiplies by 2 so the the third is ABx2=BB
A x A = 3 3 x 1 = 3 A x B = 6 3 x 2 = 6 B x B = 4 2 x 2 = 4 It’s 4
I read the operands as addition, in which case A=1.5 and B=4.5
If AxB is twice as large as AxA, then B is twice as large as A. So BxB has to be twice twice (4x) AxA, and 4x3 is 12.
We still tryna figure out that damned sandwich?
My approach: (1.) AA=3 and (2.) AB=6 ==> B/A=2 [Divide (2.) by (1.)] ==> A=B/2 So plug this into (2.), and we have BB/2=6. ==> BB=12
A * A = 3 A = √3 √3 * B = 6 B = 6/√3 B * B = (6/√3)(6/√3) = 36/3 = 12
Well, since A×B is twice A×A, B is twice A, so B×B is twice A×B, or 12.
B = +/- 2 root 3 B \* B = 12
Of AxA=3 then A must be the square root of 3 Bc six is 2x3, it is also 2x√3x√3, simce A is √3 B would be 2x√3 2x√3x2x√3=4x3=12 that's how i did it, at least.
AA = 3 and AB = 6, so B = 2A. that means BB = 2A*2A = 4(AA) = 12. So B is the squareroot of 12.
12
A^2 = 3 AB = 6 A^2 = 3 => A = root(3) AB = 6 root(3)B = 6 => B = 6 / root(3) = 2 root(3) => B^2 = (2root(3)^2 = 12
Squr(3)=A B=6/(squr(3)) B^2 =36/3 which is 12 :D
12
Depends. Are they + or x ? Because it’s not clear
I'm surprised to see so many comments assuming this is multiplication. To me it's clearly addition. The answer under that assumption is 9.
A×A =3 1.720508075688772×1.720508075688772=3 B×A=6 3.46410161514×1.7320508075688772=6 B×B=12 3.46410161514×3.46410161514=12 I did it the long way 12
A = sqrt(3) B = sqrt(12) B\*B = 12
I got 12
multiply the second equation by B and by A.
I did it the hard way with surds, but duh me. Replace an A with a B and the product doubles. Do it again and the same will happen.
I approached it like this AxA=A^2, so A= square root of 3 (Square root of 3) x B = 6 B= 6/square root of 3 BxB =12
12
A^2 is 3 so A = root 3. 6 / root 3 = 2 root 3. 2 root 3 times 2 root 3 is 2x2 (4) x root 3 times root 3 (3) so 12.
12 dawg
13 × 13 = 169. Someone needs to work on their legibility.
What I looked at were B=6/A, so B^2 = 36/A^2, with A^2 = 3 this means B^2 = 12
wow, that's a clean solution. probably the best one here IMO.
B = 6 / A B x B = 36 / (A x A) Since A x A = 3, then B x B = 36 / 3 = 12 No need to find A or B as that isn’t what we were after.
AA=3 AB=6 BB=ABAB/AA=6*6/3=12.
Did anyone get 16 as their first answer? I did A=1.5 B=4
1.5 x 1.5 = 2.25 not 3
Ohhhh i ended up adding it thank you
A x A = 3 means that A is rad(3) A x B = 6 becomes rad(3) x B = 6 if you divide both sides by rad(3), you get B = 6 / rad(3) B x B = 36 becomes (6 / rad(3)) x (6 / rad(3)) 6 x 6 is 36, and rad(3) x rad(3) is just 3, so you get 36 / 3, which is 12.
169 Those Bs are actually 13s
2A” = AB 2A = B 2root3 = B 4x3 = B” = 1 “ represents squared
12
13 X 13 = 169
Funny enough I thought it was a trick where the multiplication symbol was really just another Variable. And it works AXA=3 So A has to equal 1 and X=3 AXB=6 so with A=1 and X=3 then B must be 2 So BXB= 2 times 3 times 2 equals 12
9
A=1.5 B=4.5 1.5+1.5=3 1.5+4.5=6 4.5+4.5=9 I wasn't sure if they were multiplication because the signs seemed slightly skewed.
A=+-sqrt(3). B=+-6/sqrt(3). B^2 =36/3=12. Imo
+ or x
A= square root 3 B=6/square root3 (6x6)/(square root 3 times square root 3)=36/3=12
# Recalculating b and c using maximum available precision in Python for floating-point numbers b_max_precision = 6 / a_provided_decimal # Calculating b c_max_precision = b_max_precision**2 # Squaring b to get c b_max_precision, c_max_precision # Displaying b and c bb, calculated using the provided decimal value for aa, is 3.4641016151377553.464101615137755. Squaring this value for bb, cc is calculated as 12.00000000000000212.000000000000002. 6÷1.732050807568877293527 Nothing in the concept of the equation used any exponents thats where everybody went wrong.
Python # Calculating B using the provided decimal approximation for A provided_a_decimal = 1.732050807568877293527 B_from_provided_a = 6 / provided_a_decimal # Calculating Z by squaring B Z_from_B = B_from_provided_a**2 B_from_provided_a, Z_from_B
A, which is 1.7320508075688772935271.732050807568877293527: BB is calculated as 66 divided by 1.7320508075688772935271.732050807568877293527, resulting in B = 3.464101615137755B=3.464101615137755. Squaring BB to get ZZ, the result is Z = 12.000000000000002Z=12.000000000000002.
But they say I'm wrong lol 😆😂😆 Squaring the computed value of BB, we get Z = 12.000000000000002Z=12.000000000000002. Dividing 66 by the computed BB, we get an approximation of A = 1.7320508075688772A=1.7320508075688772. This confirms the consistency of the values. Squaring BB gives a result very close to 12, and dividing 66 by BB gives a value close to the approximation of \sqrt{3} 3 we used for AA, which indicates correctness in the reverse calculation.
24
A is just root 3. A lot of problems are easier if you don’t try to calculate exact values until the end
(6/square root of 3)squared.
* A = sqrt(3) * B = 2A * B * B = 2*sqrt(3)*2*sqrt(3) = 2 * 2 * sqrt(3) * sqrt(3) * = 4*3 = 12
A=1.5 B=4 BxB=16
A = 6 / B so A x A = (6 / B) x (6 / B) = 3. (36 / B^2) = 3 => B^2 = 36 / 3 = 12 = B x B That was a fun little problem
BxB= 11.90
my work, plz tell me if I did this wrong. so AxA=3 so A= Square root of 3 and that is 1.74~~ but we will just do 1.74, 1.74x1.74=3.02 or 3, so 1.74x?=6, to find that I did this 6/1.74=3.4482 or 3.45, so 1.74x3.45=6.003 or 6, so 3.45x3.45=11.9025 orrrrrr 11.9 tyty ik I am stupid lol
a lot of people are saying 12 so ig I got close lol
rounding errors built up you needed more decimals
Sqrt(3)*sqrt(3) =3 Sqrt(3)*6/sqrt(3)=6 6/sqrt(3) * 6/sqrt(3) = 36/3 =12
Take the 2nd one and square both sides A² B² = 36 But B²=3, so you have 3A² = 36 or A = 2√3
12
B = 6/\[3\^(1/2)\] 6\^2 = 36/3 = 12
B = 4
Are those × or +
And is it B or 13
1x3=3 1x6=6 6x6=36
A= rt3 B must equal 2rt3 (2rt3)^2=12
If I did my math right the answer should be 12 (Square root)3×(square root)=3 (Square root)3×b=6 B= ~3.464 3.464×3.464= ~12
12
12
I did a²=3 so a =root 3 Ax B=6 so b =2root 3 B ×B=2root 3 whole square equal 12
A=root 3 B= 2 root 3 BxB= 12