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:
whats question ?
Step-by-step explanation:
0.625 is the answer to this
If the half-life is t, then every t days, the amount of the radioactive isotope will be cut in half.
(1/2)^(number of half-lives) = 3%
number of half-lives = ln(0.03) / ln(0.5)
This gives the number of half-lives as 5.06.
Then 300 days = (5.06)(length of 1 half-life)
length of 1 half-life = 300 / 5.06 = 59.29 days
The answer is c
the rhombus above is formed by 4 triangles
area of one triangle= 9*13.5*0.5=60.75
area of 4 triangles = 4*60.75=243
