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:
2n-4
Step-by-step explanation:
You're subtracting 4 from the product of n times 2
Answer:
1
Step-by-step explanation:
Probability is given by number of possible outcomes ÷ number of total outcomes
Assuming we stop rolling the six-sided die once our sum is 290 ( exceeds 285)
Number of possible outcomes = 75, number of total outcomes = 290
Probability (75 rolls are needed to get this sum) = 75/290 = 0.259
Probability (more than 75 rolls are needed to get this sum) = 1 - 0.259 = 0.741
Probability (at least 75 rolls are needed to get this sum) means that either 75 rolls or more than 75 rolls are needed to get this sum = 0.259 + 0.741 = 1
A) The area of the rectangle ABCD is base times height. 20in x 12in = 240 in sq.
B) The area of the triangle AED is base times height divided by two. (14in x 12in)/2 = 84in sq.
C) Figure EBCD is a right trapezoid (having one pair of parallel sides, while the others are slanted and forming a right angle).
D) Area of EBCD is the area of the rectangle minus the area of the triangle. 240in sq - 84in sq = 156in sq.
9514 1404 393
Answer:
2πr cm
Step-by-step explanation:
The circumference of a circle of radius r is ...
C = 2πr
For a radius of (r cm), the circumference is ...
A = 2π(r cm) = 2πr cm
For a radius of (2r cm), the circumference is ...
B = 2π(2r cm) = 4πr cm
Then the difference between the circumferences is ...
B -A = (4πr cm) -(2πr cm) = 2πr cm . . . . difference in circumference
