<span>y = sqrt(25-x^2) at point (3,4)
The derivative gives us the slope at 3 to be:
-2x
------------ at x=3: -3/4
2sqrt(25-x^2)
</span><span>so we have a vector that is parallel to the slope of the tangent line is: <4,-3>
</span>
<span>the mag = 5 so; unit tangent = <4/5 , -3/5>
</span>
<span>since perpendicular lines have a -1 product between slopes we get the normal to be...
<3/5,4/5>
</span>
<span>It is <4,-3> because it is rise over run. Rise is y component of vector and run is x component of vector.</span>
7 + 28n + 4
28n + (7 + 4)
28n + 11
Answer:
almost 6 servings
Step-by-step explanation:
2/3=.667
4/.667=5.99
What is the slope? We need the slope to find y