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3 years ago

The slope of the tangent to the curve y^3x + y^2x^2 = 6 at (2,1)

1 answer:

7 0

Y^3x + y^2x^2 = 6

Using implicit differentiation 

3 y^2.y'.x+y^3+2 y.y'.x^2+2 y^2.x=0

y'(3 y^2.x+2 y. x^2) = -y^3-2 y^2.x 

y' = -[y^2(y+2 x)]/[x y(3 y+2 x)] 

= -[y(y+2 x)]/[x(3 y+2 x)] 

Substituting x =2 and y = 1:

y' = -(1)(1+4)/[2(3+4)] 
 = -5/14 

The slope of the tangent of the curve at (2,1) will be:
m = -5/14

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2 years ago

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Step-by-step explanation:

Hi there!

Let x = the number of years that pass

Let y = the height of the tree (ft)

We're given that the 6-foot tree grows at a rate of 4 ft per year. This means that the height of the tree will be equal to 6 ft, the original height, plus another 4 ft every year that passes.

Height of tree = 6 feet + 4 feet × number of years that pass

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I hope this helps!

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3 years ago

PLEASE HELP WILL GIVE THANKS AND FIVE STARS IF RIGHT ANSWER

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a^2 = 16 and b^2 = 9

a = +4 and -4

b = +3 and -3

Center is (-4, -3)

Vertices is (-4 + a, -3) and (-4 - a, -3)

Vertices is (-1, -3) and (-7, -3)

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