A walker has travelled 9 km along a trail. If he has completed 80% of the trail, how much further does he still have to go?
1 answer:
You might be interested in
At the start, the tank contains
(0.25 lb/gal) * (100 gal) = 25 lb
of sugar. Let be the amount of sugar in the tank at time
. Then
.
Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of
and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,
Separate variables, integrate, and solve for <em>S</em>.
Use the initial value to solve for <em>C</em> :
The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time
has passed. At this time, we want the tank to contain
(0.5 lb/gal) * (5 gal) = 2.5 lb
of sugar, so we pick <em>P</em> such that
Notice the picture,
recall your SOH, CAH, TOA
you have,
opposite side, 165
adjacent side, 50
and the angle
that means, we'll need Mrs. tangent
thus
![\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(\theta)=\cfrac{165}{50} \\ \quad \\ tan^{-1}\left[ tan(\theta) \right]=tan^{-1}\left[ \cfrac{165}{50} \right] \\ \quad \\ \theta=tan^{-1}\left[ \cfrac{165}{50}\right] \\ \uparrow \\ \textit{angle of elevation}\iff\textit{angle of depression}](https://tex.z-dn.net/?f=%5Cbf%20tan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%5Cimplies%20tan%28%5Ctheta%29%3D%5Ccfrac%7B165%7D%7B50%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%20tan%5E%7B-1%7D%5Cleft%5B%20tan%28%5Ctheta%29%20%5Cright%5D%3Dtan%5E%7B-1%7D%5Cleft%5B%20%5Ccfrac%7B165%7D%7B50%7D%20%5Cright%5D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%5Ctheta%3Dtan%5E%7B-1%7D%5Cleft%5B%20%5Ccfrac%7B165%7D%7B50%7D%5Cright%5D%0A%5C%5C%20%5Cuparrow%20%20%5C%5C%0A%20%5Ctextit%7Bangle%20of%20elevation%7D%5Ciff%5Ctextit%7Bangle%20of%20depression%7D)
recall your SOH, CAH, TOA
you have,
opposite side, 165
adjacent side, 50
and the angle
that means, we'll need Mrs. tangent
thus