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.
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.
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?
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...
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...
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...
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.