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Ans is 0, because when anything multiply with zero the answer is zero.

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16y ago

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What the answer too 10x-2y30?

10x-2y=30


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It is simply y0/y.


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.


How do you solve x y0 5x 2y -3?

To solve the equation ( xy^0 + 5x + 2y - 3 = 0 ), note that ( y^0 = 1 ), simplifying the equation to ( x + 5x + 2y - 3 = 0 ) or ( 6x + 2y - 3 = 0 ). Rearranging gives ( 2y = 3 - 6x ), leading to ( y = \frac{3 - 6x}{2} ). This represents a linear relationship between ( x ) and ( y ).


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 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


Is the education in Ecuador similar to the US?

y0


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in ohio y0


Is this equation linear or nonlinear y0?

linear (A+)


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The equation of a sphere with radius r, centered at (x0 ,y0 ,z0 ) is (x-x0 )+(y-y0 )+(z-z0 )=r2


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You get bad boy spankings from y0 mama


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.