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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.
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.
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
Why is 10 power 0 equal to 1?
Any number to the power zero is equal to one. That can be derived from the following index law: xa*xb = xa+b (x not zero) Now let b = 0 so that the above becomes xa*x0 = xa+0 so xa*x0 = xa (since a+0 = a) That is, any number multiplied by x0 is the number itself. That can be true only if x0 is the multiplicative identity, that is, only if x0 = 1.
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 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)
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
what is 4E8374832E374684237eX72372+399x0+273646728-1888?
0! You said x0! anything x0=0!
What hooks up to X0?
On a transformer connection H1 and H2 are the primary connections. X1 and X2 are the secondary connections. If your transformer has a split secondary that is grounded, that terminal is X0. The sequence is X1 - X0 - X2. The X0 usually indicates that there is a connection to a neutral wire along with the ground wire.
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.
When x0 how many solutions is this?
x0 = 1 because any number raised to the power of 0 is always equal to 1
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.
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 can be learned from the first derivative?
let f be a function and f' the first derivative.If f'>0 the function is genuinely ascending.If f'
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.
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
