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At the given point, find the slope of the curve or the line that is tangent...
At the given point, find the slope of the curve or the line that is tangent to the curve, as requested. y® + x3 = y2 + 11x, tangent at (0,1) 11 O A. y=- 8 11 OB. y=- EX-1 11 O C. y= 6*+1 11 OD. y= *+1
Find the slope of a line tangent to the curve of the given equation at the given point. Sketch the curve and the tangent line. y=x? -5; (4,11) The slope is (Simplify your answer.) Enter your answer in the answer box and then click Check Answer. 1 part remaining Clear All
Find the slope of the tangent line to the given polar curve at the point specified by the value of θ. r = 6sin(θ) θ = π/3 Find the slope of the tangent line to the given polar curve at the point specified by the value of θ. r = 4 - sin(θ) θ = π/4 Find the slope of the tangent line to the given polar curve at the point specified by the value of θ. r = 9/θ...
(1 point) Use implicit differentiation to find the slope of the tangent line to the curve defined by 5xy + 7xy = 36 at the point (3,1). The slope of the tangent line to the curve at the given point is
Find the slope of the line tangent to the polar curve at the given point. r= 5 sine (25) r=5 sin 0;
Find the equation of the tangent line to the curve at the given point using implicit differentiation. Remember: equation of a line can be found by y-y1=m(x-x1) where m is the slope of the line and (x1,y1) is any point on the line. Curve: at (1,1)
(1 point) Find the slope of the tangent line to the polar curve ?=cos(4?)r=cos(4θ) at the point corresponding to ?=?/3θ=π/3. The tangent line has slope (1 point) Find the slope of the tangent line to the polar curve r = cos(40) at the point corresponding to 0 = a/3. The tangent line has slope
(1 point) Find the slope of the tangent line to the curve 4x + 3y + 4xy = 47+160 at the point (8,5). The slope is
The slope of the tangent line to a curve is given by f'(x) = 4x + 3x - 2. If the point (0,8) is on the curve, find an equation of the curve.
2 The slope of the tangent line to the curve y = at the point (8, is: 1 32 The equation of this tangent line can be written in the form y = mx + b where: m is: bis: