Solution What Are The Solutions To The Equation Y X 3 3x 2 4x 12
1 Answer Massimiliano In this way y' = (2x 4) ⋅ √x −(x2 4x 3) ⋅ 1 2√x (√x)2 = = (2x4)√x⋅2√x−x2−4x−3 2√x x = = (2x 4) ⋅ 2x −x2 − 4x − 3 2x√x = = 4x2 8x − x2 −4x −3 2x√x = 3x2 4x − 3 2x√x Answer linkGet stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!
Subtract y from both sides x^ {2}4x1y=0 − x 2 − 4 x 1 − y = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 4 for b, and 1y for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a} This equation is in standard form a x 2 b x c = 0How to graph your problem Graph your problem using the following steps Type in your equation like y=2x1 (If you have a second equation use a semicolon like y=2x1The table below shows some values of the function #y=x^2 3# (a) Complete the table (b) Use the midordinate rule with six ordinates to estimate the area bounded by #y=x^2 3#, the yaxis, the xaxis and the line x=6 6 The figure below is a sketch of a
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Statement I The ordinate of a point describing the circle x 2 y 2 = 25 decreases at the rate of 15 cm/s The rate of change of the abscissa of the point when ordinate equals 4 cm is 2 cm/s Statement II xdx ydy = 0 An object moves around x^2 y^2 = 25 (which represents a circle whose radius is 5 meters) at a constant speed At time t = 0 seconds, the object is at (5, 0) When t = 1, it is at (4, 3) Where is the object when t = 2? If tangents pq and pr are drawn from a point on the circle x^2 y^2 = 25 to the ellipse x^2/16y^2/b^2=1, (b
The Equation Of The Tangent To The Circle X 2 Y 2 25 Passing Thr
