It can be >:@ or >:X
There is no actual symbol. Generally people only put an 'x'
4704= 2 x 2352= 2 x 2 x 1176= 2 x 2 x 2 x 588= 2 x 2 x 2 x 2 x 294= 2 x 2 x 2 x 2 x 2 x 147= 2 x 2 x 2 x 2 x 2 x 3 x 49= 2 x 2 x 2 x 2 x 2 x 3 x 7 x 7672= 2 x 336= 2 x 2 x 168= 2 x 2 x 2 x 84= 2 x 2 x 2 x 2 x 42= 2 x 2 x 2 x 2 x 2 x 21= 2 x 2 x 2 x 2 x 2 x 3 x 7The prime factors they have in common are 2 x 2 x 2 x 2 x 2 x 3 x 7, so the greatest common factor is 672.
2 x 2 2 x 2 = 16 2 x 2 x 5 = 20 2 x 2 x 2 x 2 x 5 = 80, the LCM
2 x 2 x 2 x 2 x 2 x 3 = 96 2 x 2 x 2 x 2 x 3 x 3 = 144 2 x 2 x 3 x 3 x 5 = 180
The symbol commonly used to denote factorial is the exclamation point ( ! ). E.g. 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 x! = x*(x-1)*(x-2)*(x-3)*(x-4)* ... *3*2*1 , where x is an integer.
It is the symbol of infinity which means never ending.
The carrot symbol is used to express exponents. For instance x squared would be written on a computer as x^2. The symbol signifies that the number has been raised into superscript.
Fragmentation hazard
the greastest common factor with monomails the GCF of 24x3y2 and 18x2y 24x3y2=2 2*2*3*x*x*x*y*y 18x2y= 2* 3*3*x*x* y the answer is 2*3*x*x*y or 6x2y the * mean multiplication symbol
The less than symbol is < Example: x<2 x is less than two.
3x+2 x is a variable. A variable is a symbol (x, y, etc...) that does not have an assigned value.
so say y = 2^x dy/dx = ln2.2^x (. = multiplication symbol, ^ = to-the-power-of symbol) The general formula is (where 'a' is a constant, x is what you are differentiating with respect to and y is f(x)) y = a^x then dy/dx = lna.a^x Go ask a math teacher or look up exponential function differentiation on the internet for why.
X is greater than or equal to 2. The symbol for "greater than or equal to" is a "greater than" sign over a horizontal dash.
Use the caret symbol (^) to denote powers and parentheses to clarify what you are asking:I'll assume you meant: [ 3*x^3 - x^2 ] + [(x-2)/(x^2)]Find a common denominator (x^2), and then [ 3*x^3 - x^2 ] becomes:[( 3*x^3 - x^2)*x^2 ]/(x^2) = [ (3*x^5 - x^4)/(x^2) ]then: [ (3*x^5 - x^4)/(x^2) ] + [(x-2)/(x^2)] = (3*x^5 - x^4 + x - 2)/(x^2)
You can ask your math teacher
2(x-4) > 1/13