How can you use parentheses and brackets in [math]\LaTeX[/math]?
For parentheses and brackets, you can write[math]1^\text{st}[/math] Bracket:[math]\LaTeX[/math] code: \left( \frac{x}{y} \right) Output: [math]\left( \frac{x}{y} \right)[/math][math]2^\text{nd}[/math] Bracket:[math]\LaTeX[/math] code: \left\{ \frac{x}{y} \right\} Output:[math]\left\{ \frac{x}{y} \right\}[/math][math]3^\text{rd}[/math] Bracket:[math]\LaTeX[/math] code: \left[ \frac{x}{y} \right] Output: [math]\left[ \frac{x}{y} \right][/math]Besides,[math]\LaTeX[/math] code: \left< \frac{x}{y} \right> Output: [math]\left< \frac{x}{y} \right>[/math]** this < and > can be replaced by \langle and \rangle respectively.Those \left and \right commands are used to adjust the sizes automatically. But if you want to adjust their size explicitly, you follow this codes in this table below:[math]\begin{array}{|c|c} \LaTeX \,\text{Code} & \text{Output} \\ \hline \text{\big( \Big( \bigg( \Bigg(}\, - \text{\Bigg) \bigg) \Big) \big)} & \big( \Big( \bigg( \Bigg( - \Bigg) \bigg) \Big) \big) & \\ \text{\big\{ \Big\{ \bigg\{ \Bigg\{}\, - \text{\Bigg\} \bigg\} \Big\} \big\}} & \big\{ \Big\{ \bigg\{ \Bigg\{ - \Bigg\} \bigg\} \Big\} \big\} & \\ \text{\big[ \Big[ \bigg[ \Bigg[}\, - \text{\Bigg] \bigg] \Big] \big]} & \big[ \Big[ \bigg[ \Bigg[ - \Bigg] \bigg] \Big] \big] & \\ \text{\big< \Big< \bigg< \Bigg<}\, - \text{\Bigg> \bigg> \Big> \big>} & \big< \Big< \bigg< \Bigg< - \Bigg> \bigg> \Big> \big> & \\ \end{array}[/math]Moreover,\tiny ) \Tiny \}\scriptsize ]\small )\normalsize \} \large ]\Large )\LARGE \}\huge ]\Huge )For more [math]\LaTeX[/math] you can visit this links:An introduction to beautiful math on QuoraShareLaTeX, Online LaTeX EditorGood luck.
Removing Brackets Help Me!?
4(2x-5)= 4*2x - 4* 5 = 8x -20 and 5(x+6) = 5*x + 5*6 = 5x +30 now adding both 8x-20+5x+30 will give u 13x + 10
Remove brackets and simplify, is this correct?
yes, remove brackets and simplify, but you have to distribute the negative sign, this is because the - in front of parentheses is a -1, so you multiply through by -1... it should look like this: (2x + y) - (x - 4y) = x - 3y 2x + y - x + 4y = x - 3y and solve (3a - b) - (2a - 3b) = a - 4y 3a - b - 2a + 3b = a - 4y and solve
What algorithm I can use to remove all brackets from math expression and still have valid math expression?
Reverse Polish Notation. RPN. ...named for Jan Łukasiewicz.Accumulate values in one direction (push each onto a stack) while applying operations in the other. (discarding the values and replacing them with the result) For example:Algebra: (24 + 6) / (3 * sqrt(25)) ...becomes the sequence...RPN: 24 6 + 3 25 sqrt * /Where each value gets pushed onto the stack and each operator processes and discards its arguments and replaces them with the result of the operation.So the above RPN is evaluated as...push the value 24 onto a stackpush the value 6 onto that stackadd the top two values (24 and 6) and replace them on the stack with the result (30)push the value 3 onto the stackpush the value 25 onto the stacktake the sqrt of the top value (25) and replace it on the stack with that on the stack (5)multiply the top two values (3 and 5) and replace them on the stack with the result (15)divide the top value on the stack (15) into the value beneath it (30), replacing them on the stack with the result (2)Now, all that's left on the stack is the answer. (2)You can use RPN to eliminate all grouping operators. In fact that is how a compiler deals with the evaluation of math expressions. If you want to code this up, search for Recursive Descent Parser - a very simple algorithm that makes this easy. Okay, the OP has now updated his question with an example that appears to show that he/she is not looking just for a solution that eliminates the brackets/parenthesis but, instead, is looking for an expansion of terms. If that is indeed the question, it looks like Richard Morris has provided a good solution.
How would you add parentheses and brackets to make this sentence true: 45 ÷ 2 x 4.7 - 4.4 x 6 = 54?
Hehe...one more... A printing error in a math book removed the brackets and parentheses from the original expression of (7 x 3.4) – [(2.8 x 5) – (4.3 x 2)] + 16. Give the order of the operations the student solving this problem would have to use to evaluate the expression with the printing error, and find the value of the incorrect expression and correct expression.
5a - 4 (3a - 1) Multiply Out Brackets?
I'm trying to complete 5a - 4 (3a - 1) by multiplying out the brackets. The answer is -7a + 4 according to the book however I'm really struggling to see how I would get there. I can see that the +4 could come from the multiplication of -4 and -1 however am I missing something with 5a and 3a? Is the book wrong? Thanks.
How can I express this in math?
Think about what "consecutive" means. (An example of "five consecutive even integers" is 6, 8, 10, 12, 14. Another example is 12, 14, 16, 18, 20. An example of five consecutive odd integers, incidentally, is 134073, 134075, 134077, 134079, 134081. Another example is -3, -1, 1, 3, 5. You'd be able to approach either case similarly.)If you were asked to express the sum of those first three consecutive integers, for example, one way you could do that is [math]n + n+2 + n+4[/math]. (Additionally, you could write that as [math]\sum_{k=0}^2 (n+2k)[/math], but that may be too abstruse, depending on the level you're working at.) But, as you can probably guess, you can simplify this expression. If you understand this, then you should be able to do the same for all five of the integers in question.
Grade 9 math help?? algebra?
Thats not correct. =(4x^2+2xy-7y^2) - (-5x^2+7xy-9y^2) - (4x^2+3xy+2y^2) = 4x^2+2xy-7y^2 + 5x^2 - 7xy + 9y^2 - 4x^2 - 3xy - 2y^2 = 4x^2-4x^2+5x^2+2xy-7xy-3xy-7y^2-2y^2+ 9y^2 = 5x^2 - 8xy -9y^2+9y^2 = 5x^2 - 8xy
If [math]2x = (y^2 - 13 = 36)[/math], then what is [math]x[/math]?
Remove brackets as they are not required, there is no order of precedence being changed by using them.2x = y^2–13 = 362x = 36n.b. it also equals y^2–13 but you don’t even need this if just solving for x because the equation states all three equal the same outcome, it’s like saying 2x = y^-13 = 36 = 2*18 = 37–1 = 72/2……they all amount to the same value 36.X = 18to solve y and so you can test the whole equationy^2–13 = 36y^2 = 49y = 7plugging the values x=18 and y = 7 back in we get2*18 = (7^2 - 13 = 36)36 = (49 - 13 = 36)36 = (36 = 36)again remove those brackets as not required, we are not doing any calculation on the right that requires them such as multiplication after addition36 = 36Therefore the valuesx = 18y = 7must be trueSome of the answers work on the basis that solving Y first gives:2x = 7because y=7but that is not true2x = y^2 - 132x = 7^2 - 132x = 49 - 132x = 36x = 18If you tried to plug 3.5 into the equation where the answer given is 2x = y and y is 7 you would get the following:2x = (y^2 - 13 = 36)2*3.5 = (7^2 - 13 = 36)7 = (49–13=36)7 = 36 = 36which is wrong.That’s my take on it. Happy to be shown if it is incorrect.