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Mathematics 20 Online
OpenStudy (anonymous):

pls give me the steps to find the slope of tangent for 5x^2-2 x=5

OpenStudy (anonymous):

we hav the equation, y=5x^2-2x-5; dy/dx,will giv u the slope @ (x,y) and so we hav , dy/dx=10x-2

OpenStudy (anonymous):

so the slope = 10x-2 @ the point (x,y).

OpenStudy (anonymous):

the answer isnt right bracket

OpenStudy (anonymous):

your equation is 5x^2- 2 , differetian it gives the slop d/dx [5x^2-2] = 10x >>> at x = 5 slop = 50

OpenStudy (anonymous):

slope**

OpenStudy (anonymous):

yoyo, sorry i was misled by u not putting a comma after the equation...please be carefull , cheers :)

OpenStudy (anonymous):

that is fine thank u soo much for ur help

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