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March 31, 2015

March 31, 2015

Posted by **Henry** on Monday, June 18, 2012 at 5:48pm.

a) y = -3x^2 + 5x - 4

b) f(x) = 6/x - 3/(x^2)

c) f(x) = (3x^2 - 4x)(x^3 + 1)

the answers i got for each was:

a) y'= -6x + 5

b) f'(x)= -6/x^2 + 6/x^3

c) f'(x)= 15x^4 + 6x - 16x^3 - 4

please double check and let me know if these are correct, thank you.

- Calculus - please check my work -
**MathMate**, Monday, June 18, 2012 at 5:56pmAll correct, well done!

- Calculus - please check my work -
**Henry**, Monday, June 18, 2012 at 6:13pmthank you! can u help me on this question please

determine the coordinates of two points on the plane with equation 5x + 4y - 3z = 6

- Calculus - please check my work -
**MathMate**, Monday, June 18, 2012 at 7:50pmAny point on the plane will satisfy the equation.

Since there are three variables x,y and z, you can fix any two variables and solve for the third.

Example:

Assuming arbitrarily y=6, z=6

=>

5x+4(6)-3(6)=6

5x+24-18=6

5x=0

x=0

So (0,6,6) is a point on the plane.

- Calculus - please check my work -
**Henry**, Tuesday, June 19, 2012 at 12:29amthank you

- Calculus - :) -
**MathMate**, Tuesday, June 19, 2012 at 7:36amYou're welcome!

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