Asked by Jesse Stone
what is the slope of the line tangent to the curve (/x)^2 at point (4,16)
* (/x)= the squareroot of x
* (/x)= the squareroot of x
Answers
Answered by
Anonymous
(√x)<sup>2</sup> is just x.
The point (4,16) lies on the curve for x<sup>2</sup>
Wanna take another stab at it?
The point (4,16) lies on the curve for x<sup>2</sup>
Wanna take another stab at it?
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