Calculus (Normal Line, please check my work)
The slope of the line normal to the graph of 4 sin x + 9 cos y = 9 at the point (pi, 0) is: Derivative: 4cosx - 9siny(dy/dx) = 0 (dy/dx) = (-4cosx) / (-9siny) (dy/dx) = (4) / 0 Normal line = -1 / (4/0) Does this mean that the slope of the normal line is undefined, or did I do ...
Find the point on the graph of y = x^2 + 1 thats closest to the point 8, 1.5. Hint: Remember the distance formula.
Nevermind, that 4.42 was a mistake and my very original answer of 1.105940354 was absolutely correct!!! This is the right answer, I know it!
Okay, so does this change my original answer of approximately 1.64 to 4.42??? The 4.42 came from putting your new values in the quadratic equation.
Now I'm lost, I don't get why you changed the signs.
I rechecked and found that 3x^2-22x+28 has the correct signs. Knowing this equation and the values I found from the quadratic equation, would you say that the 1.639079157 term is correct? (The 2.69 square inches came from squaring the 1.639079157).
I think that you might have gotten the equation wrong, I think that it should be: 3x^2 - 22x + 28. When I put this equation into the quadratic equation, I got 5.694254177 and 1.639079157. So the squares that need to be cut out should have an area of approximately 2.69 square i...
Calculus (Optimization, Still Need Help)
I just wanted to correct something for my equation, it should be: V = (14 - 2x)(8 - 3x)(x), which simplifies to V = 112x - 44x^2 - 4x^3. Take the derivative: V' = 112 - 88x - 12x^2 Now all I need are the roots, any help? I think I found one around 1.10594, possibly?
A rectangular piece of cardboard, 8 inches by 14 inches, is used to make an open top box by cutting out a small square from each corner and bending up the sides. What size square should be cut from each corner for the box to have the maximum volume? So far I have: V = (14 - 2x...
Pretty sure I figured it out, 4/27. I found this by simplifying: ((1/3pi (h - 2/3 h))(4/9 r^2)) / (1/3 pi r^2 h)
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