Yes.
An equation that is not a function is called a relation. Functions are special types of relations where every input (or in other words each value in the domain) has exactly one output (or matches up with exactly one value in the range).
A relation would be where you plug in a number for x but instead of only getting one number out for y, you get more than one. Example:
y2=x
If you plug in 4 for x and solve for y by taking the square root, then y could equal either positive 2 or negative 2, since 22 is 4 and (-2)2 is also 4. In this case, x corresponds with two output values for y (2 and -2) which means that while this equation is a relation, it is not a function.
Domain here would refer to all numbers that make sense for x. In other words, what numbers can you plug in for x, and get an answer that is not imaginary or undefined. In the example above, I could not plug in negative numbers for x, because when I try to solve for y I would get an imaginary number. So we would say that the domain of that relation is x> or equal to 0.
The Range for a relation is all of the possible output values. So for all the values of x that you can plug in, what are all the possible values of y I could get out? If you look at it, since I'm only plugging in 0 for x or any other number larger than 0, that would imply that y can only be 0 or bigger as well. So the range here would be y > or equal to 0.
I hope that helps!
Zero is an acceptable value for a variable. As a general rule, if you solve an equation and get x equals zero, that's just what it is: zero. There is one solution. Example: 10x +1 = 5x +1 5x = 1 - 1 5x = 0 x = 0 There is one solution, zero. If you solve for x and get two solutions, such as 2 and 0, there are two answers. Later on, if you've studied domain and range (the numbers you can use for x so that the equation is still defined, (i.e. not 1/0, or the square root of negative 1), you may find that x = 0 does not always work. You need to check first, to see if zero is in the domain, before you say it's an answer.
DomainFind the domain of y = x + 4Domain simply means "what numbers can I use for X that give me an answer for Y?"Domain is looking for the acceptable X values that work in the equation y = x +4When looking for the domain, the rule of thumb is:* No fraction? * No radical? (sqrt, cube root, etc) * No problems! If you don't see a fraction, or a radical, or both in your equation, the domain will always be "All real numbers". This means you can pick any number you want, plug it in for X, and you'll get an answer for Y.If you do happen to have a fraction, radical (or both)...----Think about what "breaks" a fraction. What makes a fraction not work?If you try to divide by 0, you get an error, or undefined on your calculator.for example:Find the domain of y = 1/(x -1)1/x is a fraction, and we know having a 0 on the bottom would make it not work.So what we do is say the domain is "all real numbers, except" and then find out what numbers break the function and fill that in later.We set the bottom of the fraction = 0 to find out exactly which numbers will break it, and then solve for X.x - 1= 0x = 1So we know that is we plug in a 1, it will break the fraction.The domain of y = 1/(x - 1) is "All real numbers, except 1"---- If you have a radical expression in your function:example. y = sqrt(x+2)We need to know what breaks a radical expression, what number(s) won't work?Well, sqrt(0) is OK. it = 0.sqrt(1) is ok, it = 1.But what happens if I try sqrt(-1). Try it on your calculator.You get an error message, right?This is because you cannot take the root of a negative number. (at least, not yet)This means, negative numbers break a function with a radical. So similarly to how we found the numbers that broke a fraction, we'll set what's inside the radical less than 0. Since negative numbers are less than 0.x + 2 < 0 and solve for xx < -2So any number less than -2 will break my radical.Your answer would be, "All real numbers >= -2" (since anything less than -2 is broken, but -2 is still OK)To check:y = sqrt( -2 + 2) = sqrt (0) = 0 - OKy = sqrt( -3 + 2) = sqrt(-1) - breaksI chose a number smaller than -2 to check.---- If you get a problem where they use both fractions AND radicals, just use both techniques.Ex: y = 1/sqrt(x + 2)We know that having a 0 on the bottom of a fraction breaks it, but it OK for a radical to have a 0 in it. We need to combine both rules together.Take the inside of the radical and set it less than(what breaks the radical) or equal to 0 (what breaks the fraction).x + 2
y=x-5 is a linear equation that has been shifted down 5, the slope is still 1. So if you plug in points for x, you can find the position easily. When x=0, y=-5. When x=1, y=-4. And when x=-1, y=-6.
First of all, the word "equals" doesn't belong in an "inequality".Let's assume you actually meant to call it an "equation".Now, we have the nagging felling that you're not exactly sure what you mean by "answer",so we'll tell you: It means a pair of numbers that you can put in the places of 'x' and 'y'that make the whole equation a true statement.For example:Can 'x' be 1 and 'y' be zero ?If they are, you'd have 4(1) + 2(0) = 40. Is that a true statement ?Well, it says that 4 = 40, which is a big fat lie, so 'x' can't be 1 when 'y' is zero.Can 'x' be zero and 'y' be 20 ?If they are, then you have 4(0) + 2(20) = 40. Is that a true statement?Well, it says that 40 = 40, which is true, so 'x' can be zero when 'y' is 20,and that's a good 'answer'..But wait. You're not done. We want you to try another one:Can 'x' be 5 and 'y' be 10 ?If they are, then you have 4(5) + 2(10) = 40. Is that a true statement ?Well, it says that 20 + 20 = 40, and that's true, so 'x' can be 5 when 'y' is 10.That's another good answer.How many answers can this thing have ? We're sure that if you're still reading,you're about to be surprised.Your equation ... [ 4x + 2y = 40 ] ... is the equation of a straight line.Every point on the line is an 'answer' to the equation.How many points are there on a line ? An infinite number of them.Your equation has an infinite number of correct 'answers'.And also an infinite number of wrong answers ... all of the points that aren't on the line.
No. It is a noun: "You've still got your soup on the heat."And a verb: "You still need to heat up your soup."But not an adverb; that modifies a verb, and adjective, or another adverb.
A relation is when the domain in the ordered pair (x) is different from the domain in all other ordered pairs. The range (y) can be the same and it still be a function.
It is an equation. It could be an algebraic equation, or a trigonometric equation, a differential equation or whatever, but it is still an equation.
It is still an equation.
This music in in the public domain. Nothing from 1876 is still in copyright.
It is still an equation.
It is called an equation with no solutions in the domain under consideration. It is quite possible that an equation which does not have a solution in one domain has a simple solution in another - possibly broader - domain.For example,a junior school pupil may tell you that x + 3 = 2 has no solution. In the domain of positive integers it does not but perhaps a year later, they will tell you that the answer is x = -1.a little later they may tell you that 3*x = 2 has no solution in the domain of integers. True, but in the domain of rational numbers, the answer is x = 2/3.Next, x2 = 2 has no solution in rational numbers but x = sqrt(2), is a perfectly valid solution in the domain of real numbers (which includes irrational numbers).Still further, x2 = -9 has no solution in real numbers since we all know that 32 = +9 and (-3)2 = +9. However, in the domain of imaginary numbers, x = 3i.and so on.Also, there are equations which cannot be solved analytically, but do have solutions which can be found by numerical methods.
It is still called a quadratic equation!
The song itself is in the public domain, but certain arrangements, performances, and recordings may still be protected.
The song itself is in the public domain, but certain arrangements, performances, and recordings are still protected by copyright.
The song itself is in the public domain, but certain settings, performances, and recordings may still be protected.
The work itself is in the public domain, but certain transcriptions, adaptations, performances, and recordings may still be protected.
Since it was published in 1906, which was prior to 1923, the work itself is in the public domain. However certain editions of it may still be protected, at least to the extent they contain "creative and original work" not in the public domain editions.