First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.
Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.
![\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20cot%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B6%7D%7D%7B%5Cstackrel%7Bopposite%7D%7B-7%7D%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bhypotenuse%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B6%5E2%2B%28-7%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B36%2B49%7D%5Cimplies%20c%3D%5Csqrt%7B85%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)


Answer:
48 minutes
Step-by-step explanation:
first divide 6 miles by 2.5 miles, that will be 2.4 miles, then multiply by 20
Dy/dx = dy/dt * dt/dx
xy = 4
y + x(dy/dx) = 0 by implicit differentiation.
x(dy/dx) = -y
dy/dx = -y/x
<span>dy/dx = dy/dt * dt/dx dy/dt = -2
</span>
<span>-y/x = -2 * dt/dx
</span>
y/(2x) = dt/dx
dt/dx = y/(2x)
dx/dt = 2x/y
When x = -3, xy = 4, y = 4/x = 4/-3 = -4/3
dx/dt = 2*-3/(-4/3) = -6 *-3/4 = 18/4 = 9/2 = 4.5
dx/dt = 4.5
Answer:
x + 27 = - 12
x = - 39
Step-by-step explanation:
Answer:

Step-by-step explanation:
we know that
In the right triangle ABC
The function sine of angle 68 degrees is equal to divide the opposite side SB by the hypotenuse AC
so

substitute the values and solve for AC


