Which of the following equations are equations of circles? Choose all that apply
1 answer:
Answer:
Step-by-step explanation:
The work form of a circle is
The standard form of a circle is
This standard form is a circle if A and C are the same number, there is a plus sign in between the x-squared and the y-squared terms, and the F value, when moved to the other side of the equals sign, is not negative. That F value represents the radius which is a measure of distance, and distances cannot be negative.
Looking at the first choice in your list, on the right side of the equals sign we have a negative 4. That means that that equation is not a circle.
Looking at the second choice in your list, we have a 2x^2 and a y^2, where A = 2 and C = 0. This is not a circle, either, since A and C have to be the same number in a circle, and they are not.
Looking at the third choice in your list, we have x^2 and y^2 with the same A and C value of 1, and if we move the F value of -5 over, we have a radius of 5, and there is a + sign in between the x^2 and the y^2 terms. Choice 3 is a circle.
Looking at the fourth choice, we have that A = 3 and C = 3, there is a + sign in between the x^2 and the y^2 terms, and if we move the -6 over to the other side, the radius is a positive number. Choice 4 is also a circle.
Looking at the last choice, there is a - sign in between the x^2 and the y^2 terms. Even though A and C are both the same number, 1, the - sign makes that possibly a hyperbola, but definitely NOT a circle.
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<h2>Answer:</h2>
The solution of the given equation is:
y=6
<h2>Step-by-step explanation:</h2>The expression is given by:
Now on solving for the given equation
since,
Hence, we get:
but y≠ 4
since, the denominator of the term in the left side of the given equality and the second term in the right side of the given equality will be zero and hence, the expression will be not defined.
Hence, the value of y is: 6
Answer:
Step-by-step explanation:
First, look at your terms. You have two terms and two variables in each. So they can be factored out because you can distribute them back in easily.
Focus one variable first, x. You cant factor an x^12 out of an x^4. You can, but you really don't want to, it'll turn your problem into a mess. So the only thing you can do is factor out the lower exponent, x^4.
We arent done yet because the terms still share the y variable. You cant factor a y^12 from a y^4. This is the same problem with the x variable. So take out the lower one, y^4.