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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 quadrant do points lie if their abscissa is positive and ordinate is negative?
The fourth quadrant
Is this equation linear or nonlinear y0?
linear (A+)
What are the four quadrants of a cartesian graph?
Quadrant I ( + , + ) Quadrant II ( - , + ) Quadrant III ( - , - ) Quadrant IV ( + , - )
In which quadrant will the value of x be negative?
The value of x will be negative in the bottom left quadrant (quadrant 3) and the top left quadrant (quadrant 2).
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.
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 the net displacement of the particle between 0 seconds and 80 seconds?
That would besqrt[ (x80 - x0)2 + (y80 - y0)2 ) at an angle of tan-1 (y80 - y0) / (x80 - x0)or(x80 - x0) i + (y80 - y0) j
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 distance between the origin ( 0 0) and a point at ( -15 8)?
Use Pythagoras: For a line joining two points (x0, y0) to (x1, y1) there is a right angle triangle: One side (leg) joins (x0, y0) to (x1, y0) with length (x1 - x0) One side (leg) joins (x1, y0) to (x1, y1) with length (y1 - y0) The hypotenuse joins (x0, y0) to (x1, y1) → length hypotenuse = √((x1 - x0)² + (y1 - y0)²) → length between (0, 0) and (-15, 8) is given by: distance = √((-15 - 0)² + (8 - 0)²) = √((-15)² + (8)²) = √(225 + 64) = √289 = 17 units
How can you find an equation line between two pair of points?
Assuming you want the equation of the straight line between the two points (x0, y0) and (x1, y1), the equation is: y - y0 = m(x - x0) where m is the gradient between the two points: m = (y1 - y0) ÷ (x1 - x0) Note: if the two x coordinates are equal, that is x0 = x1, then the equation of the line is x = x0.
Given that the points 16 -5 and -40 16 lie on a line what is the equation of the line?
The gradient m between two points (x0, y0) and (x1, y1) is given by m = change_in_y/change_in_x = (y1 - y0)/(x1 - x0) Equation of a line through a point (x0, y0) with gradient m is given by: y - y0 = m(x - x0) Thus for the points (16, -5) and (-40, 16): m = (16 - -5) / (-40 - 16) = 21/-56 = -3/8 y - -5 = -3/8(x - 16) → 8y + 40 = -3x +48 → 8y + 3x = 8
Explain how to find the distance between the points (28 -17) and (-15-17) on a coordinate plane?
To find the distance between any two points on the Cartesian plane use Pythagoras: The distance between (x0, y0) and (x1, y1) is given by: distance = √((x1 - x0)² + (y1 - y0)²) → distance between (28, -17) and (-15, -17) is: distance = √((x1 - x0)² + (y1 - y0)²) = √((-15 - 28)² + (-17 - -17)²) = √((-43)² + (0)) = √1849 = 43 ------------------------ In this case, the y-coordinates are the same (y0 = y1 = -17), so this becomes: distance = √((x1 - x0)² + (y0 - y0)²) = √((x1 - x0)² + 0²) = √((x1 - x0)²) = |x1 - x0| The vertical bars around the expression mean the absolute value of the expression, which is the numerical value of the expression ignoring the sign. distance = |x1 - x0| = |-15 - 28| = |-43| = 43.
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.
Implement full subtractor with two half subtractors and OR gate?
You use the first HS with entries of x0 and y0 . then you get D0=X0 XOR Y0 and B0=X0 NOT*Y0.(*=AND) Then you use the sacond HS with entries of D0(from the first HS) and B-1(borrow from the previous level). then you get D1=X0 XOR Y0 XOR B-1(exactly like D in FS) and B1=D0 NOT*B-1. Then you use the OR gate with entries B0(from the first HS) and B1(from the second HS) and you get B=B0 OR B-1 AND D0 NOT. If you check the options you can get their similar to the borrow of FS
What is the slope of the line joining 2 0 and 1 3?
between points (x0, y0) and (x1, y1): slope = change_in_y/change_in_x → slope = (y1 - y0)/(x1 - x0) → slope = (3 - 0)/(1 - 2) = 3/(-1) = -3
What is bresanham algorithm?
If you're wondering about the "Bresenham line algorithm", it is an algorithm (a process used for making a desired result) that plots a geometrical line (consiting of infinate points, as if you draw a straight line on a piece of paper) and translates it to a computer screen (composed of pixels, or video "points" that have a specific amount of sized points) The algorithm is as follows: function line(x0, x1, y0, y1)boolean steep := abs(y1 - y0) > abs(x1 - x0)if steep thenswap(x0, y0)swap(x1, y1)if x0 > x1 thenswap(x0, x1)swap(y0, y1)int deltax := x1 - x0int deltay := abs(y1 - y0)int error := -(deltax + 1) / 2int ystepint y := y0if y0 < y1 then ystep := 1 else ystep := -1for x from x0 to x1if steep then plot(y,x) else plot(x,y)error := error + deltayif error ≥ 0 theny := y + ysteperror := error - deltax This algorithm was taken from the site http://en.wikipedia.org/wiki/Bresenham's_line_algorithm
