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There is no single solution for x2 + y2 = 13, as there are an infinite number of corresponding values that could be plugged in for x and y.
x2 + y2 = 13 is a function that describes a circle. The circle would have a center point of 0, 0, and a radius of √13.
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When using the slope formula does it have to be (y2-y1)(x2-x1) or can it be (y1-y2)(x1-x2)?
If you mean: (y2-y1)/(x2-x1) and (y1-y2)/(x1-x2) then either works out the same.
Which equations correctly represents a circle centered at the origin with a radius of 5?
x2 + y2 = 25radius of 10?x2 + y2 = 100
How do you find the slope when given two points?
Points: (x, y) and (x2, y2) Slope = y2-y divided by x2-x
How do you find a midpoint between two points on a graph?
Given two coordinates (x1,y1) and (x2,y2) The midpoint is ( ((x2+x1)/2) , ((y2+y1)/2) )
What is the equation for a semicircle?
x2+y2=r2, y>=0
Convert x2 plus y2 equals 2x?
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
How do you solve for slope from two points and a graph?
Points: (x1, y1) and (x2, y2) Slope: y1-y2/x1-x2
Find the centre and radius of the circle with equation x2 plus 2x plus y2 equals 12?
x2 + 2x + y2 = 12 (complete the square)(x2 + 2x + 1)+ y2 = 12 + 1(x + 1)2 + y2 = 13 (divide by 13 each element to both sides)(x + 1)2/13 + y2/13 = 1(x - -1)2/13 + (y - 0)2/13 = 1Thus, the center of the circle is (-1, 0) and its radius is 1.
How do you solve a slope problem?
Use the slope formula (y2-y1)/(x2-x1) and yes x2 is the x-value of your second x coordinate. Meanwhile y2 is the second y coordinate.
What is the expansion for x2 plus y2?
There is no expansion for x2 + y2
How do you solve the factoring different of two square?
X2 - Y2 = (X + Y)(X - Y)
The circle is centered at the origin and the length of its radius is 8 What is the circle's equation?
The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.
Explain how do you write an equation of a line in standard form when two points on the line are given?
Th e two points are X1,Y1 and X2, Y2 slope, m, is y2-y1 divided by x2-x1 then standard form is y = mx +b plug in y2 for y and x2 for x and solve b or plug in y1 for y and x1 for x and solve b
How do you solve this problem it just says solve x2 plus 3y equals 7 3x plus y2 equals 3?
x2 + 3y = 7 3x + y2 = 3 3y = x2 + 7 y2 = -3x + 3 y = x2/3 + 7/3 y = ± √(-3x + 3) If you draw the graphs of y = x2/3 + 7/3 and y = ± √(-3x + 3) in a graphing calculator, you will see that they don't intersect, so that the system of the given equations has not a solution.
How do you draw a square using line command?
Line (x1, y1, x2, y1); Line (x2, y1, x2, y2); Line (x2, y2, x1, y2); Line (x1, y2, x1, y1);
If xy equals x2 plus y2 plus 2xy?
If:xy = x2 + y2 + 2xyThen:x2 + xy + y2 = 0Do you want to solve it for x?x2 + xy + (y/2)2 = (y/2)2 - y2(x + y/2)2 = y2/4 - y2x + y/2 = ± √(-3y2/4)x = -y/2 ± y√(-3) / 2x = (-y ± yi√3) / 2
When using the slope formula does it have to be (y2-y1)(x2-x1) or can it be (y1-y2)(x1-x2)?
If you mean: (y2-y1)/(x2-x1) and (y1-y2)/(x1-x2) then either works out the same.
