0
It is simply y0/y.
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What is the equation of sphere?
The equation of a sphere with radius r, centered at (x0 ,y0 ,z0 ) is (x-x0 )+(y-y0 )+(z-z0 )=r2
What is 4x to the third and y to the zero?
(4x)3 = 64x3 y0 = 1
What is the angle between the two lines x0 and y0?
x0 and y0 aren't lines. Do you mean x=0 and y=0? If so, they are the y axis and the x axis, respectively, and the answer is 90 degrees as noted above.
How to calculate year over year percentage?
Suppose the value of some quuantity Y, in year 0 was Y0. Suppose the following year, it was Y1. Then year over year percantage change is = 100*(Y1/Y0 -1).
What is x-y0 and 7x-3y24?
If you mean: x -y = 0 and 7x -3y = 24 then they are simultaneous equations whose solutions are x = 6 and y = 6
Equation for linear approximation?
The general equation for a linear approximation is f(x) ≈ f(x0) + f'(x0)(x-x0) where f(x0) is the value of the function at x0 and f'(x0) is the derivative at x0. This describes a tangent line used to approximate the function. In higher order functions, the same concept can be applied. f(x,y) ≈ f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0) where f(x0,y0) is the value of the function at (x0,y0), fx(x0,y0) is the partial derivative with respect to x at (x0,y0), and fy(x0,y0) is the partial derivative with respect to y at (x0,y0). This describes a tangent plane used to approximate a surface.
Does y0 represent y as a function of x?
y0(x) could represent a function of x but usually y(0) represents the function y that is evaluated at x = 0 and so is no longer a function of x but a constant.
What is the equation of sphere?
The equation of a sphere with radius r, centered at (x0 ,y0 ,z0 ) is (x-x0 )+(y-y0 )+(z-z0 )=r2
What is 4x to the third and y to the zero?
(4x)3 = 64x3 y0 = 1
Which circles lie completely within the fourth quadrant?
A circle with centre (x0, y0) and radius r has the equation of:(x -x0)² + (y - y0)² = r²By writing the equation of any circle in this form its centre and radius can be determined.To completely lie within a quadrant, the centre of the circle must be more than r away from the y- and x-axes:In the first quadrant if: x0 > r and y0 > rIn the second quadrant if: x0 < -r and y0 > rIn the third quadrant if: x0 < -r and y0 < -rIn the fourth quadrant if: x0 > r and y0 < -rIf either x0 or y0 (or both) is exactly r away from the y- or x-axis then the circle is on boundary between quadrants, and if either x0 or y0 (or both) is less than r away from the y- or x-axis, then the circle is in more than one boundary.f x0 < r from the y-axis then the circle is in quadrants I and II, or y0 < r from the x-axis then the circle is in quadrants III and IV; if both less than r away from their respective axes, the the circle is in all four quadrants.
What are slope-intercept and point-slope forms please show examples?
Slope-intercept is: y = mx + c where m = slope; and c = intercept. Example y = 3x + 5 slope = 3 intercept = 5 Point-slope is: y - y0 = m(x - x0) where m = slope; point (x0, y0) is a point on the line. Example y - 8 = 3(x - 1) slope = 3 point (1, 8) is on the line. The two are interchangeable: y - y0 = m(x - x0) → y - y0 = mx - mx0 → y = mx + (y0 - mx0) Which means that y0 - mx0 = c, the intercept. Example: y - 8 = 3(x - 1) → y = 3x + (8 - 3×1) → y = 3x + 5 → the two examples above are the same line in different forms.
If the slope of a line is already known what else is needed in order to create an equation?
Either a point or the y-intercept.If the y-intercept is known (call it b), and, calling the slope m, the equation is y = mx + b.If a sample point (x0, y0) is known, then the equation is y - y0 = m(x - x0).
What is the angle between the two lines x0 and y0?
x0 and y0 aren't lines. Do you mean x=0 and y=0? If so, they are the y axis and the x axis, respectively, and the answer is 90 degrees as noted above.
Which equation represents the line that passes through points (1 5) and (3 17)?
The general equation of a line through point (x0, y0) with gradient m is given by: y - y0 = m(x - x0) The gradient m between two points (x0, y0) and (yx1, y1) is given by: m = change_in_y/change_in_x = (y1 - y0)/(x1 - x0) → line through points (1, 5) and (3, 17) is given by: y - 5 = ((17 - 5)/(3 - 1))(x - 1) → y - 5 = (12/2)(x - 1) → y - 5 = 6(x - 1) → y - 5 = 6x - 6 → y = 6x - 1
Is y0 a linear equation?
Hard to tell because of problems with the browser. If the question is about "y = 0" the answer is YES.
Code in c to draw line?
// macros for simplicity #define MAX(x,y) (x>y?x:y) #define MIN(x,y) (x<y?x:y) /* ** drawLine ** ** Draw a line from vertex (x0,y0) to vertex (x1,y1) using ** the midpoint line algorithm, implemented using OpenGL. ** */ void drawLine( GLint x0, GLint y0, GLint x1, GLint y1 ) {GLint dE, dNE, x, y, d, dx, dy; // check if we need to switch the points if( x0 > x1 ) { x0 = x0 + x1; x1 = x0 - x1; x0 = x0 - x1; y0 = y0 + y1; y1 = y0 - y1; y0 = y0 - y1;} // calculate deltas dy = y1 - y0; dx = x1 - x0; // special cases if( dx -1 - diag down-right glBegin(GL_POINTS); for( x = x0, y = y0; x <= x1; x++, y-- ) { glVertex2i(x,y);} glEnd();}else { // general cases // midpoint algorithm if( abs(dy) < dx ) { // small slope dE = 2 * abs(dy); dNE = 2 * (abs(dy) - dx); d = dE - dx; glBegin(GL_POINTS); for( x = x0, y = y0; x <= x1; x++ ) { glVertex2i(x,y); if( d <= 0 ) { d+= dE;}else { y += (dy>0?1:-1); d += dNE;}}// for x = x0 to x1 glEnd();}else { // large slope dE = 2 * dx; dNE = 2 * (dx - abs(dy)); d = dE - abs(dy); glBegin(GL_POINTS); for( x = x0, y = y0; (y0 < y1 && y <= y1) (y0 > y1 && y >= y1); y+=(y0 < y1?1:-1) ) { glVertex2i(x,y); if( d <= 0) { x ++; d+= dE; }else { d += dNE; }}// for y = y0 to y1 glEnd();}} }// drawLine()
How to calculate year over year percentage?
Suppose the value of some quuantity Y, in year 0 was Y0. Suppose the following year, it was Y1. Then year over year percantage change is = 100*(Y1/Y0 -1).
