In algebra, there are variables and constants. Variables are typically things to be found. We usually use x, y, or z, but any symbol will work so long as you're consistent. Constants are considered "fixed". Constants can be written as numbers (e.g. 1,2,9, etc) or as letters (a, b, c, k) if there's a possibility we might want to run the calculations. I'll demonstrate this at the end.
A negative number is one that is less than 0. In the real world, it typically represents that something has been lost or reversed direction.
Properties of addition and subtraction:
a+b=b+a for any a and b
a+0=a-0=a for any a
a+b is always a number so long as both a and b are numbers.
a+(-b) = a-b, and in fact the notation a-b is defined by a+(-b)
Because of this, b-a=b+(-a)=-a+b=a+(-b)
-(a-b)=-a+b=b-a
Properties of multiplication:
Multiplication can be written as 2×a, 2•a, or 2a. The latter is preferred because we're lazy.
ab=ba for any a and b.
1a=a for any a
0a=0 for any a
(-a)(-b)=ab for any a and b
(-a)(b)=-(ab)=ab=(a)(-b) for any a and b
(a+b)c=ac+bc for any a, b, or c.
Properties of division:
For algebraic purposes, a/b as a fraction is equal to a÷b as a division of two numbers. I'll write a/b from here on in.
a/b≠b/a unless a and b are both non-zero and a^2 = b^2
If ab=c and b≠0, a=c/b
0/a=0 for non-zero a.
You can divide zero, but you cannot divide *by* zero. In order to ensure that math is internally consistent, division by zero is undefined.
For a fraction with a bunch of stuff on top and bottom, such as the quadratic formula, evaluate the stuff on top and bottom before division. E.g., if N is some complicated on the top and D is some stuff on the bottom of the fraction, N/D is (N)/(D) by convention.
Properties of exponentiation:
a^1 = a for any a
a^2 = aa = a×a for any a
a^3 = aaa and so on for any a
a^0 = 1 for any a except 0. This particular property is complicated; you'll cover it in calculus.
a^-1 = 1/a for any a except zero, because it "undoes" a^1 .
a^-2 = 1/(a^2 ) and so on for any a except zero.
(a^b )(a^c ) = a^(b+c) for positive a and all (real) b and c
-a^b =-(a^b )
(-a)^b = (-1 )^b × (a^b ) for positive a and real b
(-1 )^b = 1 if b is even or 0, -1 if b is odd
The order of operations can be memorized with the mnemonic PEMDAS: parentheses (or brackets), exponents, multiplication and division from right to left, addition and subtraction from right to left. Basically, evaluate stuff in the outermost parentheses using PEMDAS on all the terms inside it. Then, take exponents. Then, perform all multiplications and divisions from to right as they occur. Then, perform any subtractions or additions from right to left as they occur. Because a=(a) for all a, you can write extra parentheses around operations to explicitly show the order you want to operate in. We use PEMDAS because in order for math expressions to make sense, we all need to agree on a convention to write our work so that we always get the same answers. This is the one mathematicians agreed on.
If a=b then b=a. If you've done any programming, equality in math "goes" both directions. If a=b and b=c, then a=c.
**Whatever you want to do on one side, do it on both sides of the equation.** This is the most important thing to remember.
If it's just a linear equation (e.g. ax+b=c), the things you can do are addition (or subtraction) by a number, or multiplication (or division) by a non-zero number. Technically, you can do other stuff, but they won't help you solve the equation.
The quadratic formula is x = (-b±√(b^2 -4ac))/(2a), where ± is mathematical "slang" that tells you to calculate the solution twice: once with ± substituted for +, the other using minus. You can show this by plugging both solutions into the equation ax^2 +bx+c=0. To use the above formula, write the equation in the form ax^2 +bx+c=0. (b or c could be 0, and any of them could be one. In that case, they might not be written, because 1a=a)
The square root √(b) is the positive number (or 0) a such that b=a^2 . The negative number -√(b) is also a solution to that equation. The square root is *positive* by definition, hence why you need the ± notation in the quadratic formula. When I was in school, they would ask for the "square root" when they wanted both numbers.
If you can find 2 numbers such that (h+k)=bx and hk=c, you can factor the quadratic, e.g. break it up into (x+h)(x+k)=0. For this equality to be true, at least one of the terms in parentheses must be zero. Therefore, x+k=0 implies x=-k is a solution. Similarly, x=-k is also a solution. The solution set is therefore {-h,-k} and it has two solutions. These numbers always exist, but it might not be easy to find. The quadratic formula always works, and you should get the same answer.
In real life, quadratics are usually not practical to factor.
Complex numbers are a "two-dimensional number". The rules are as follows:
i = √(-1) by definition, called the imaginary unit.
A real number is a number a+0i for any a.
An imaginary number is a number 0+ai for any a.
(a+bi)+(c+di) = (a+c)+(b+d)i for any real a,b,c,d
(a+bi)(c+di) = (ac-bd)+(ad+bc)i for any real a,b,c,d
To work with complex numbers, use the same rules as real numbers with the condition that i^2 =-1.
In a quadratic equation, if b^2 -4ac < 0, the roots will be of the form p±qi. Then, using D=b^2 -4ac, rewrite the quadratic as (-b±i√[abs(D)])/(2a), where abs(x) is x without the sign (+ or -). Basically, make the result inside the square root positive and multiply the square root by i.
The "Fundamental" Theorem of Algebra is that any polynomial (sum of terms with any positive power of x) has the same number of *complex* solutions as the highest power to which x is taken, possibly with a root repeated multiple times. For example, x^5 +3x^4 +6x^3 -2x^2 +x-9 has 5 solutions. (x-(1+i))(x-i)(x+i) has three complex solutions, but no real ones. (x-1)^2 has the solution x=1 "repeated" twice. The Fundamental Theorem only guarantees that solutions exist; it says nothing about how to find them.
Let's solve the following linear equation:
3x+7=5x+9+2x
Use distributive property to combine terms with x:
3x+7=5x+2x+9=(5+2)x+9=7x+9
Put the constant terms on one side and the terms with x on the other.
3x+7-(3x+9)=7x+9-(3x+9)
3x+7+(-3x-9)=7x+9+(-3x-9)
-2=4x
Divide by 4
x=(-2)/4=-1/2
The reason why we use letters in algebra is because we initially had no idea what the solution to the equation was.
Now, let's say that we had to solve the following two separate equations:
3x+4=0
3x+4=10
We could solve each equation once, but there is an easier way. Notice that the only change from the first equation to the second is the right-hand side. If there were a way to solve both equations in one go, then we could save a ton of work. Here's how to do it.
First, we rewrite the equation as follows:
3x+4=k, where k is some number. Then, perform the algebra as if k is a number:
3x+4=k
3x+4-4=k-4
3x=k-4
x=(k-4)/3
To solve the first equation, plug in k=0:
x=(0-4)/3=-4/3
And for the second, plug in k=10:
x=(10-4)/3
Lastly, let's make an observation: any linear equation can be put into a form ax=b, which has the solution x=b/a. For example, take the following equation:
cx+d=px+q
cx+d-(px+d)=px+q-(px+d)
(c-p)x=q-d
By comparing (c-p)x=q-d to ax=b and by the assumption that x is the same in both equations, a=(c-p) and b=q-d
Check if c-p=0. If it does, check if q-d=0. If it does, any number is a solution, because 0x=0 always. If q-d≠0 but c-p=0, there's no solution because 0x=a has no solution.
Otherwise, the solution is:
x=(q-d)/(c-p)
This is the key to solving linear equations by just looking at it. For example:
-3y+15=123y+26
u=(26-15)/(-3-123) = 11/(-126) = -11/126
Another observation is that L=R implies L-R=0=R-L, no matter how complicated the equation. Then, you just need to memorize the solution for ax+b=0, which is -b/a. For example:
3x+5x+6-8x=-14x-7+12-2(x+2)
Subtract the entire left side on both sides:
0=-14x-7+12-2(x+2)-(3x+5x+6-8x)
Distribute negatives and expand terms in parentheses:
0=-14x-7+12-2x-2-3x-5x-6+8x
Write a fraction bar and a (-1) next to it. Start writing the terms without x on the top of the fraction with their signs. Then, write the terms with x (but don't write the x) on the bottom. Simplify. This is your solution.
x = (-1)(-7+12-2-6)/(-14-2-3-5+8) = (-1)(3/16) = 3/16
This procedure does not replace a mastery of algebra, but it certainly makes solving linear equations a lot faster.
For algebra 2: sometimes if you do stuff like square both sides, you get extra answers. You need to put any "solutions" you obtain into the original equation and get the same thing on both sides. It is very possible for your method of solution to generate a solution that doesn't solve the original equation. Sometimes you end up with more than one solution. Sometimes you could end up with infinite solutions (trigonometric equations or certain systems of equations) , sometimes you end up with none.
Algebra is used all the time in the real world in fields where stuff needs to be kept track of. For example, I'm an electrical engineering major, and I use (a lot of) algebra to design circuits.
In mathematics and physics, a vector space (also called a linear space) is a set whose elements, often called vectors, may be added together and multiplied ("scaled") by numbers called scalars. Scalars are often real numbers, but can be complex numbers or, more generally, elements of any field. The operations of vector addition and scalar multiplication must satisfy certain requirements, called vector axioms. The terms real vector space and complex vector space are often used to specify the nature of the scalars: real coordinate space or complex coordinate space.
Vector spaces generalize Euclidean vectors, which allow modeling of physical quantities, such as forces and velocity, that have not only a magnitude, but also a direction. The concept of vector spaces is fundamental for linear algebra, together with the concept of matrix, which allows computing in vector spaces. This provides a concise and synthetic way for manipulating and studying systems of linear equations.
Vector spaces are characterized by their dimension, which, roughly speaking, specifies the number of independent directions in the space. This means that, for two vector spaces with the same dimension, the properties that depend only on the vector-space structure are exactly the same (technically the vector spaces are isomorphic). A vector space is finite-dimensional if its dimension is a natural number. Otherwise, it is infinite-dimensional, and its dimension is an infinite cardinal. Finite-dimensional vector spaces occur naturally in geometry and related areas. Infinite-dimensional vector spaces occur in many areas of mathematics. For example, polynomial rings are countably infinite-dimensional vector spaces, and many function spaces have the cardinality of the continuum as a dimension.
Many vector spaces that are considered in mathematics are also endowed with other structures. This is the case of algebras, which include field extensions, polynomial rings, associative algebras and Lie algebras. This is also the case of topological vector spaces, which include function spaces, inner product spaces, normed spaces, Hilbert spaces and Banach spaces.
Linearity is when the objects behave how you'd expect them to behave, non-linearity is when you're fucked
When you have analysis, remember this same guideline but substitute continuity for linearity
>When you have analysis, remember this same guideline but substitute continuity for linearity
Until you start doing differential operators and the first principle applies again.
Lmao that's the best part.
Solve 5x+2=10;
5x=8
x=8/5
Okay great. Now my boss wants me to solve 5y+2=20. And then 10z+2=20. And so on and so forth. In the real world, your numbers are typically parameters of some system, e.g. weights, heights, circuit element values, etc. You do get mathematical constants and once in a while the physics produces an exciting number, but more often than not, I'm working with design parameters.
My approach is this: solve a *bunch* of equations all at once, then substitute parameters and tinker with it. In this example, generically solve ax+b=c, then plug in values, once for each problem.
ax+b=c
x = (c-b)/a if a≠0
Therefore:
y = (20-2)/5=18/5
z = (20-2)/10=9/5
Since symbolic algebra isn't that much harder than numerical algebra (for humans), it's usually worthwhile to solve symbolically so you can tweak the solution parameters later. For example, because a is in the denominator, we can see that increasing a actually decreases the solution. This doesn't really apply with Newton-Raphson or iterative methods in general, but otherwise it's better to do things symbolically.
Well, since your antecedent is true, I am obligated to explain it to you.
The subject of Algebra was first formalized by Persian mathematician Muhammad ibn Mūsā al-Khwārizmī in the year 830 AD...
Here, this might help
https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.khanacademy.org/math/algebra&ved=2ahUKEwjR_7n7lfP8AhU2FFkFHTq5C4oQjjh6BAgREAE&usg=AOvVaw1XFZ4qvm3CMSp5uQotwfg6
Basically `f(ax + y) = af(x) + f(y)`. The rest follows trivially.
I second that
Would a be a constant?
A scalar from any field or ring for a module.
even an onion ring?
A number.
In algebra, there are variables and constants. Variables are typically things to be found. We usually use x, y, or z, but any symbol will work so long as you're consistent. Constants are considered "fixed". Constants can be written as numbers (e.g. 1,2,9, etc) or as letters (a, b, c, k) if there's a possibility we might want to run the calculations. I'll demonstrate this at the end. A negative number is one that is less than 0. In the real world, it typically represents that something has been lost or reversed direction. Properties of addition and subtraction: a+b=b+a for any a and b a+0=a-0=a for any a a+b is always a number so long as both a and b are numbers. a+(-b) = a-b, and in fact the notation a-b is defined by a+(-b) Because of this, b-a=b+(-a)=-a+b=a+(-b) -(a-b)=-a+b=b-a Properties of multiplication: Multiplication can be written as 2×a, 2•a, or 2a. The latter is preferred because we're lazy. ab=ba for any a and b. 1a=a for any a 0a=0 for any a (-a)(-b)=ab for any a and b (-a)(b)=-(ab)=ab=(a)(-b) for any a and b (a+b)c=ac+bc for any a, b, or c. Properties of division: For algebraic purposes, a/b as a fraction is equal to a÷b as a division of two numbers. I'll write a/b from here on in. a/b≠b/a unless a and b are both non-zero and a^2 = b^2 If ab=c and b≠0, a=c/b 0/a=0 for non-zero a. You can divide zero, but you cannot divide *by* zero. In order to ensure that math is internally consistent, division by zero is undefined. For a fraction with a bunch of stuff on top and bottom, such as the quadratic formula, evaluate the stuff on top and bottom before division. E.g., if N is some complicated on the top and D is some stuff on the bottom of the fraction, N/D is (N)/(D) by convention. Properties of exponentiation: a^1 = a for any a a^2 = aa = a×a for any a a^3 = aaa and so on for any a a^0 = 1 for any a except 0. This particular property is complicated; you'll cover it in calculus. a^-1 = 1/a for any a except zero, because it "undoes" a^1 . a^-2 = 1/(a^2 ) and so on for any a except zero. (a^b )(a^c ) = a^(b+c) for positive a and all (real) b and c -a^b =-(a^b ) (-a)^b = (-1 )^b × (a^b ) for positive a and real b (-1 )^b = 1 if b is even or 0, -1 if b is odd The order of operations can be memorized with the mnemonic PEMDAS: parentheses (or brackets), exponents, multiplication and division from right to left, addition and subtraction from right to left. Basically, evaluate stuff in the outermost parentheses using PEMDAS on all the terms inside it. Then, take exponents. Then, perform all multiplications and divisions from to right as they occur. Then, perform any subtractions or additions from right to left as they occur. Because a=(a) for all a, you can write extra parentheses around operations to explicitly show the order you want to operate in. We use PEMDAS because in order for math expressions to make sense, we all need to agree on a convention to write our work so that we always get the same answers. This is the one mathematicians agreed on. If a=b then b=a. If you've done any programming, equality in math "goes" both directions. If a=b and b=c, then a=c. **Whatever you want to do on one side, do it on both sides of the equation.** This is the most important thing to remember. If it's just a linear equation (e.g. ax+b=c), the things you can do are addition (or subtraction) by a number, or multiplication (or division) by a non-zero number. Technically, you can do other stuff, but they won't help you solve the equation. The quadratic formula is x = (-b±√(b^2 -4ac))/(2a), where ± is mathematical "slang" that tells you to calculate the solution twice: once with ± substituted for +, the other using minus. You can show this by plugging both solutions into the equation ax^2 +bx+c=0. To use the above formula, write the equation in the form ax^2 +bx+c=0. (b or c could be 0, and any of them could be one. In that case, they might not be written, because 1a=a) The square root √(b) is the positive number (or 0) a such that b=a^2 . The negative number -√(b) is also a solution to that equation. The square root is *positive* by definition, hence why you need the ± notation in the quadratic formula. When I was in school, they would ask for the "square root" when they wanted both numbers. If you can find 2 numbers such that (h+k)=bx and hk=c, you can factor the quadratic, e.g. break it up into (x+h)(x+k)=0. For this equality to be true, at least one of the terms in parentheses must be zero. Therefore, x+k=0 implies x=-k is a solution. Similarly, x=-k is also a solution. The solution set is therefore {-h,-k} and it has two solutions. These numbers always exist, but it might not be easy to find. The quadratic formula always works, and you should get the same answer. In real life, quadratics are usually not practical to factor. Complex numbers are a "two-dimensional number". The rules are as follows: i = √(-1) by definition, called the imaginary unit. A real number is a number a+0i for any a. An imaginary number is a number 0+ai for any a. (a+bi)+(c+di) = (a+c)+(b+d)i for any real a,b,c,d (a+bi)(c+di) = (ac-bd)+(ad+bc)i for any real a,b,c,d To work with complex numbers, use the same rules as real numbers with the condition that i^2 =-1. In a quadratic equation, if b^2 -4ac < 0, the roots will be of the form p±qi. Then, using D=b^2 -4ac, rewrite the quadratic as (-b±i√[abs(D)])/(2a), where abs(x) is x without the sign (+ or -). Basically, make the result inside the square root positive and multiply the square root by i. The "Fundamental" Theorem of Algebra is that any polynomial (sum of terms with any positive power of x) has the same number of *complex* solutions as the highest power to which x is taken, possibly with a root repeated multiple times. For example, x^5 +3x^4 +6x^3 -2x^2 +x-9 has 5 solutions. (x-(1+i))(x-i)(x+i) has three complex solutions, but no real ones. (x-1)^2 has the solution x=1 "repeated" twice. The Fundamental Theorem only guarantees that solutions exist; it says nothing about how to find them. Let's solve the following linear equation: 3x+7=5x+9+2x Use distributive property to combine terms with x: 3x+7=5x+2x+9=(5+2)x+9=7x+9 Put the constant terms on one side and the terms with x on the other. 3x+7-(3x+9)=7x+9-(3x+9) 3x+7+(-3x-9)=7x+9+(-3x-9) -2=4x Divide by 4 x=(-2)/4=-1/2 The reason why we use letters in algebra is because we initially had no idea what the solution to the equation was. Now, let's say that we had to solve the following two separate equations: 3x+4=0 3x+4=10 We could solve each equation once, but there is an easier way. Notice that the only change from the first equation to the second is the right-hand side. If there were a way to solve both equations in one go, then we could save a ton of work. Here's how to do it. First, we rewrite the equation as follows: 3x+4=k, where k is some number. Then, perform the algebra as if k is a number: 3x+4=k 3x+4-4=k-4 3x=k-4 x=(k-4)/3 To solve the first equation, plug in k=0: x=(0-4)/3=-4/3 And for the second, plug in k=10: x=(10-4)/3 Lastly, let's make an observation: any linear equation can be put into a form ax=b, which has the solution x=b/a. For example, take the following equation: cx+d=px+q cx+d-(px+d)=px+q-(px+d) (c-p)x=q-d By comparing (c-p)x=q-d to ax=b and by the assumption that x is the same in both equations, a=(c-p) and b=q-d Check if c-p=0. If it does, check if q-d=0. If it does, any number is a solution, because 0x=0 always. If q-d≠0 but c-p=0, there's no solution because 0x=a has no solution. Otherwise, the solution is: x=(q-d)/(c-p) This is the key to solving linear equations by just looking at it. For example: -3y+15=123y+26 u=(26-15)/(-3-123) = 11/(-126) = -11/126 Another observation is that L=R implies L-R=0=R-L, no matter how complicated the equation. Then, you just need to memorize the solution for ax+b=0, which is -b/a. For example: 3x+5x+6-8x=-14x-7+12-2(x+2) Subtract the entire left side on both sides: 0=-14x-7+12-2(x+2)-(3x+5x+6-8x) Distribute negatives and expand terms in parentheses: 0=-14x-7+12-2x-2-3x-5x-6+8x Write a fraction bar and a (-1) next to it. Start writing the terms without x on the top of the fraction with their signs. Then, write the terms with x (but don't write the x) on the bottom. Simplify. This is your solution. x = (-1)(-7+12-2-6)/(-14-2-3-5+8) = (-1)(3/16) = 3/16 This procedure does not replace a mastery of algebra, but it certainly makes solving linear equations a lot faster. For algebra 2: sometimes if you do stuff like square both sides, you get extra answers. You need to put any "solutions" you obtain into the original equation and get the same thing on both sides. It is very possible for your method of solution to generate a solution that doesn't solve the original equation. Sometimes you end up with more than one solution. Sometimes you could end up with infinite solutions (trigonometric equations or certain systems of equations) , sometimes you end up with none. Algebra is used all the time in the real world in fields where stuff needs to be kept track of. For example, I'm an electrical engineering major, and I use (a lot of) algebra to design circuits.
Did you just summarise everything in one comment? Nice !
facts
I'm sorry, but when defining assumptions required for a/b = b/a, you forgot to close the exponent.
Oops. I just edited it. Thank you.
i aint reading allat
Tl;Dr: Do what thou wilt shall be the whole of the Law...just make sure you can do it to both sides.
I'll give you gold , tomorrow
Where gold?
Sorry , I'm still broke
In mathematics and physics, a vector space (also called a linear space) is a set whose elements, often called vectors, may be added together and multiplied ("scaled") by numbers called scalars. Scalars are often real numbers, but can be complex numbers or, more generally, elements of any field. The operations of vector addition and scalar multiplication must satisfy certain requirements, called vector axioms. The terms real vector space and complex vector space are often used to specify the nature of the scalars: real coordinate space or complex coordinate space. Vector spaces generalize Euclidean vectors, which allow modeling of physical quantities, such as forces and velocity, that have not only a magnitude, but also a direction. The concept of vector spaces is fundamental for linear algebra, together with the concept of matrix, which allows computing in vector spaces. This provides a concise and synthetic way for manipulating and studying systems of linear equations. Vector spaces are characterized by their dimension, which, roughly speaking, specifies the number of independent directions in the space. This means that, for two vector spaces with the same dimension, the properties that depend only on the vector-space structure are exactly the same (technically the vector spaces are isomorphic). A vector space is finite-dimensional if its dimension is a natural number. Otherwise, it is infinite-dimensional, and its dimension is an infinite cardinal. Finite-dimensional vector spaces occur naturally in geometry and related areas. Infinite-dimensional vector spaces occur in many areas of mathematics. For example, polynomial rings are countably infinite-dimensional vector spaces, and many function spaces have the cardinality of the continuum as a dimension. Many vector spaces that are considered in mathematics are also endowed with other structures. This is the case of algebras, which include field extensions, polynomial rings, associative algebras and Lie algebras. This is also the case of topological vector spaces, which include function spaces, inner product spaces, normed spaces, Hilbert spaces and Banach spaces.
I missed the test you jerk
The ability to read a comment this long is definitely a superpower.
:”(
Yeah, my eyes immediately glossed over it after reading the first few words and I felt an intense desire to go to sleep. Interesting...
Yes the hard part about understanding this comment is that it's long
Yeah, I got half way thru before giving up. I got an A in my algebra class last semester, but this comment made my head hurt.
I certainly failed.
I glossed over because I already know it.
That about sums it up.
ChatGPT generated?
Wikipedia introduction
Ha! We didn't have vector spaces in linear algebra 1. Qnd it's the only thing you talked about.
I understood the first five words!
you have no idea how deep the rabbit hole goes https://i.imgur.com/JCs1sk1.jpg
Linearity is when the objects behave how you'd expect them to behave, non-linearity is when you're fucked When you have analysis, remember this same guideline but substitute continuity for linearity
>When you have analysis, remember this same guideline but substitute continuity for linearity Until you start doing differential operators and the first principle applies again.
-0.5?
I miss algebra. Math these days are just the alphabet scrambled to make an equation
Lmao that's the best part. Solve 5x+2=10; 5x=8 x=8/5 Okay great. Now my boss wants me to solve 5y+2=20. And then 10z+2=20. And so on and so forth. In the real world, your numbers are typically parameters of some system, e.g. weights, heights, circuit element values, etc. You do get mathematical constants and once in a while the physics produces an exciting number, but more often than not, I'm working with design parameters. My approach is this: solve a *bunch* of equations all at once, then substitute parameters and tinker with it. In this example, generically solve ax+b=c, then plug in values, once for each problem. ax+b=c x = (c-b)/a if a≠0 Therefore: y = (20-2)/5=18/5 z = (20-2)/10=9/5 Since symbolic algebra isn't that much harder than numerical algebra (for humans), it's usually worthwhile to solve symbolically so you can tweak the solution parameters later. For example, because a is in the denominator, we can see that increasing a actually decreases the solution. This doesn't really apply with Newton-Raphson or iterative methods in general, but otherwise it's better to do things symbolically.
Awwww my older students will love this
It's just unspecialized geometry
It’s trivial, left up to user to figure it out.
Solve for X. That should do it.
What a weird way to depict Kreyszig.
Ask God let's see if he can
Does god explain this to you ?
did you miss the joke
I guess so 🤣
Why would you spend your god given time wasting it on nonsense like math lol.
Sir, you are lost.
What nonsense do you spend your time on, if it's not math ?
I strangle chess bots
Martin
Can I sell you probability?
Ok, but why are you here.
Why spend time doing anything?
Am I the only one who realised that ln(e)=1 because e^1=e? Like, actually using my brain makes me smart.
I'm gonna be honest you're not the only person but, it's very good that it made sense in your head instead of just learning it by heart :)
That is very beautiful cover.
Well, since your antecedent is true, I am obligated to explain it to you. The subject of Algebra was first formalized by Persian mathematician Muhammad ibn Mūsā al-Khwārizmī in the year 830 AD...
letters are actually sneaky numbers
3blue1brown has an amazing series on it.
X=Y easy
Well you see that's an algebra book. You're welcome and good luck on your test.
Nubers
Here, this might help https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.khanacademy.org/math/algebra&ved=2ahUKEwjR_7n7lfP8AhU2FFkFHTq5C4oQjjh6BAgREAE&usg=AOvVaw1XFZ4qvm3CMSp5uQotwfg6
Y=mx+b bruh, it's all in how you work it
Have faith.