0
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.
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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.
What does x 0 and y o mean?
x0 and y0 usually denote the original condition or value for the variables x and y or may represent the first values in an array of values. Some final condition my be represented by x1 or y1 or x12 or y12 depending on how many conditions you want the values of x and y for.
Does the equation x plus y equals 41 define y as a function of x?
Both. If you look at it like this: y=41-x you say x is independent and y is dependent of x (i.e. y is a function of x) For x=41-y you say y is independent and x is dependent of y (i.e. x is a function of y)
Does the equation y equals 2x plus 1 represent a function?
The [ 2x + 1 ] represents a function of 'y' .
Is the linear equation X equals -14 a function?
No, this is not a function. The graph would have a vertical line at x=-14. Since there are more than one y value for every given x value, the equation does not represent a function. The slope of the equation also does not exist.
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.
What is y0 divided by y?
It is simply y0/y.
Does the graph x-y2 equals 1 represent x as a function of y?
X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.
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 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.
What does x 0 and y o mean?
x0 and y0 usually denote the original condition or value for the variables x and y or may represent the first values in an array of values. Some final condition my be represented by x1 or y1 or x12 or y12 depending on how many conditions you want the values of x and y for.
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
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.
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()
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.
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
What is the equation of a circle whose centre is at -1 -7 and has a radius of 10 on the Cartesian plane?
The equation of a circle with centre (x0, y0) and radius r is given by: (x - x0)² + (y - y0)² = r² For the circle with centre (-1, -7) and radius 10 this gives: (x - -1) + (y - -7)² = 10² → (x + 1)² + (y + 7)² = 100 This can be expanded and rearranged to give: x² +2x + y² + 14y - 50 = 0
