Recent content by 7yler

  1. 7

    Simplification Help: Can't Simplify Further?

    The original question was: suppose x^{6}e^{-2y}=ln(xy) Find \frac{dy}{dx} by implicit differentiation. That is the answer that I have come to, but it is said to be wrong.
  2. 7

    Simplification Help: Can't Simplify Further?

    I have worked a question down to this, but I am not sure how to simplify any further.
  3. 7

    Find Inverse of f(x)=\frac{7e^{x}-6}{e^{x}+8}: Solve x

    That was a huge help. Thank you.
  4. 7

    Find Inverse of f(x)=\frac{7e^{x}-6}{e^{x}+8}: Solve x

    Find a formula for the inverse of f. f(x)=\frac{7e^{x}-6}{e^{x}+8} I set f(x) equal to y, and operated on it until I got: ln(y)=ln(7e^{x}-6)-ln(e^{x}+8) But I'm stuck. I'm not sure how to isolate x.
  5. 7

    Find Integral of sqrt(e^(9x)) - Incorrect Solution Explained

    Find the integral. \int\sqrt{e^{9x}} dx I figured that \int\sqrt{e^{9x}} dx is equal to \int(e^{9x/2}) dx so the integral should simply be e^{9x/2}+C Why isn't this correct?
  6. 7

    Simplifying math expression help

    I'm really not sure how to approach simplifying something such as this. Could anyone offer any advice?
  7. 7

    Optimizing Work in a Parabolic Water Tank: Finding the Area of Cross-Sections

    The ends of a "parabolic" water tank are the shape of the region inside the graph of y = x^{2} for 0 ≤ y ≤ 4; the cross sections parallel to the top of the tank (and the ground) are rectangles. At its center the tank is 4 feet deep and 4 feet across. It is 5 feet long. Rain has filled the tank...
  8. 7

    Particle Launched over 13.7m Gorge - Speed & Angle Calculation

    I realized what I was doing wrong. Thanks for the tip.
  9. 7

    Particle Launched over 13.7m Gorge - Speed & Angle Calculation

    A particle is launched over a gorge that is 13.7 m straight across and 100 m deep. The particle is launched at an angle of 14° above the horizontal, and lands with 1.7 m to spare. What was the particle's launch speed? If the particle was launched at a different angle, but fell 0.5m short...
  10. 7

    Motorcycle Kinematic Equation

    v_{0}= 0 km/s a=0.0042 km/s^{2} v= 108/3600 km/s delta x= 1.1 km 108/3600 = 0 + 0.0042(t) t=108/15.12 for the time of acceleration delta x=(1/2)(0.0042(108/15.12)^{2} delta x=0.1071428571 km 1.1 km - 0.1071428571 km = 0.9928571429 km 0.9928571429 km = (108/3600)*t t=19.85714286 s...
  11. 7

    Motorcycle Kinematic Equation

    I have worked this out and gotten that it takes 27 seconds to catch up with the car, but I'm told that that is wrong. This problem is frustrating me. I really don't understand what I'm doing wrong.
  12. 7

    How to integrate the last two terms in that integral

    Wow. I'm really off of my game tonight. I finally got it worked out. Thank you everyone for your help. I truly appreciate it.
  13. 7

    How to integrate the last two terms in that integral

    So 2ab is 1/2. Then if we have the square root of (a+b)^2, that equals (a+b), and b still has x in the denominator, so how do I integrate that?
  14. 7

    How to integrate the last two terms in that integral

    I do not know how to complete the square since x is in the denominator. I can get it to (3x^{2})^{2}+(\frac{1}{12x^{2}})^{2}= -\frac{1}{2}
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