domain is what x values you are allowed to put into an equation.In this case there is no restrictions. x= all real numbers.range is the output once you put these numbers in the equation.h(x)=>all real numbers > 0 because the absolute value can not be negative.
if you have the equation, choose one of the pairs and put one of them in the position of x and solve. if you get the leftover value, the one you put in the equation is the x, independant variable, the other is y
In expressions such as "x-y", both "x" and "y" can have any value. The value of "x-y" will depend on what the value of "x" and the value of "y" are.
Let one of the numbers be x and other be y. According to question: xy = -3 Put any value of x then value of y = -3/x. Let us say x = 1 then y = -3/1 = -3. And if x = 3 then y = -3/3 = -1. So for any value of x(except 0) we get a value of y. So there are infinitely many solutions of the given equation.
There exists infinitely many solutions to the problem. Let the required number be x: By Euclid's Division Lemma we have x = 9q + 1(Since dividend is 9 and remainder is 1) Put any value of q(q must be natural number), say 5. On putting q = 5, we get x = 9*5 + 1 = 46. Put any value of q and you will get different value of x.
In algebra, variables are represented by letters such as x. A variable could be any number. That number is the "value" of the variable. In an expression, you can choose a number to put in for x, and simplify to get a number which is the value of the expression. In an equation, you can solve for the value of x, which will be the value of x which makes the equation true.
domain is what x values you are allowed to put into an equation.In this case there is no restrictions. x= all real numbers.range is the output once you put these numbers in the equation.h(x)=>all real numbers > 0 because the absolute value can not be negative.
i think you are missing the word point in the question, and if so, then yes. the domain of a function describes what you can put into it, and since your putting x values into the function, if there is a point that exists at a certain x value, then that x is included in the domain.
Boxing a value x is to put it in a class type rather than its primitive one. Ex. int x = 3; Integer Ix = x; // boxing x
A constant function is one like f(x)=2 so no matter what value of x I put in, the output is 2. If I have f(x)=x, the identity, this is not constant. I put in 3 for example and get out 3, but if I put in 4 I get out 4. f(x)=x^2 is another non constant. I put in a number and the output is the number squared so this is NOT constant.
if you have the equation, choose one of the pairs and put one of them in the position of x and solve. if you get the leftover value, the one you put in the equation is the x, independant variable, the other is y
In expressions such as "x-y", both "x" and "y" can have any value. The value of "x-y" will depend on what the value of "x" and the value of "y" are.
Let one of the numbers be x and other be y. According to question: xy = -3 Put any value of x then value of y = -3/x. Let us say x = 1 then y = -3/1 = -3. And if x = 3 then y = -3/3 = -1. So for any value of x(except 0) we get a value of y. So there are infinitely many solutions of the given equation.
There exists infinitely many solutions to the problem. Let the required number be x: By Euclid's Division Lemma we have x = 9q + 1(Since dividend is 9 and remainder is 1) Put any value of q(q must be natural number), say 5. On putting q = 5, we get x = 9*5 + 1 = 46. Put any value of q and you will get different value of x.
when you times the number by its value. for example if the number is 2. then to square it you would put 2 x 2. and for 8, it will be 8 x 8.
the value of x is 3
If the reflection is over the x value, the x-value does not change.