[コンプリート!] consider the parabola y=x^2 the shaded area is 263326-Consider the parabola y=x^2 the shaded area is

Solve This 10 Consider The Parabola Y X2 The Shaded Area Is 1 232 533 734 Physics Motion In A Straight Line Meritnation Com

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 AnswerX=hpThe directrix is x= 15 kendraBada kendraBada Mathematics Middle School answered Consider the parabola (y 1)2 = 2(x 1) Which equation is used to find the directrix? between y = 4x − x2 and y = x then subtract from the integral of the first (between a and b) the integral of the second (again, between a and b) Part 1 Points of intersection occurs when 4x −x2 = x This occurs when either x = 0 or x = 3 (we could, but don't actually need to calculate ya and yb)

Consider the parabola y=x^2 the shaded area is

Consider the parabola y=x^2 the shaded area is- A_("Solid")=2 y=x^2 is a parabola (shown in purple) opening up with its vertex on the origin y=1 is a horizontal line (shown in red) The region bound by these two curves is shaded in yellow The solid described in the problem is based on this yellow region with thin square plates perpendicular to the yellow base and parallel to the xaxisShare It On Facebook Twitter Email 1 Answer 1 vote answered by Jay Chaubey (k points) selected by Akanksha01 Best answer = 7/6 square units

Answered In The Figure The Equation Of The Bartleby

Answered In The Figure The Equation Of The Bartleby

Area y=x^21, (0, 1) \square!Answer (1 of 2) This is a classic integration problem since we are looking for area under a curve * ybounds Note that you aren't told what the bounds for y are One assumption you should always automatically make (unless otherwise specified) is that the lower bound is located at y = 0 and,Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!

Calculus Find the Area Between the Curves y=x^2 , y=4xx^2 y = x2 y = x 2 , y = 4x − x2 y = 4 x x 2 Solve by substitution to find the intersection between the curves Tap for more steps Substitute x 2 x 2 for y y into y = 4 x − x 2 y = 4 x x 2 then solve for x x3Find the area of the region bounded by the curves y= x^2 2 , y=x , x =0 and x = 3applications integration,application of integration area,133/6 Consider the following The x y coordinate plane is given A line y = x − 6, a curve y = x 2 − 6x, and a shaded region are graphed The line enters the window at y = −6 on the negative yaxis, goes up and right, passes through the point (1, −5) crossing the curve, crosses the xaxis at x = 6crossing the curve, and exits the window in the first quadrant

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