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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 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
What is the distance between points 1 2 and 4 -1?
Use Pythagoras to find the distance between two points (x0,.y0) and (x1, y1): distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((4 - 1)² + (-1 -2)²) → distance = √(3² + (-2)²) → distance = √(9 + 9) → distance = √18 = 3 √2
When x0 how many solutions is this?
x0 = 1 because any number raised to the power of 0 is always equal to 1
How do you integrate of e -2x for x0?
The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.
The measure of the supplement of an angle exceeds twice the measure of the supplement of the complemant of the angle by 40?
The answer is -13 1/3ohere is the detailed calculation for the problem:Let x0 be the angle, then;(180 - x0) - 2[180 - (90 - x0)] =40(180 -x0) - 2[90+x0]=40180 -x0 - 180 - 2x0=40-3x0=40hencex0= -13 1/3oAny comments are welcome
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.
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 quadrant does x0 and y0 lie?
None. The coordinate lines between the quadrants don't belong to any of the quadrants.
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
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.
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 distance between the points 5 9 and 5 13?
Pythagoras can be used to find the distance between any two points (x0, y0) and (x1, y1): Distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((5 - 5)² + (13 - 9)²) = √(0 + 4²) = √(4²) = 4.
What is the distance between the points (-7 -13) and (8 -5)?
To find the distance between two points (x0, y0) and (x1, y1) use Pythagoras: distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((8 - -7)² + (-5 - -13)²) → distance = √(15² + 8²) → distance = √289 → distance = 17 units.
What is Newton raphson's method in r programing?
It's a method used in Numerical Analysis to find increasingly more accurate solutions to the roots of an equation. x1 = x0 - f(x0)/f'(x0) where f'(x0) is the derivative of f(x0)
What causes the graph of a rational function to have a vertical asymptote?
When your input variable causes your denominator to equal zero. * * * * * A rational function of a variable, x is of the form f(x)/g(x), the ratio of two functions of x. Suppose g(x) has a zero at x = x0. That is, g(x0) = 0. If f(x0) is not also equal to 0 then at x = x0 the rational function would involve division by 0. But division by 0 is not defined. Depending on whether the signs of f(x) and g(x) are the same or different, as x approaches x0 the ratio become increasingly large, or small. These "infinitely" large or small values are the asymptotes of the rational function at x = x0. If f(x0) = 0, you may or may not have an asymptote - depending on the first derivatives of the two functions.
what is 4E8374832E374684237eX72372+399x0+273646728-1888?
0! You said x0! anything x0=0!
