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)


Slope of AB=(2-3)/(1+1)=-1/2; AC=(-1-3)/(-3+1)=-4/-2=2; BC=(-1-2)/(-3-1)=-3/-4=3/4.
The product of two sides that are perpendicular is -1. Slopes of AB and AC are perpendicular so angle A is a right angle. ABC is a right-angled triangle.
Answer:
1.0010353034
Step-by-step explanation:
The difference of two squares factoring pattern states that a difference of two squares can be factored as follows:

So, whenever you recognize the two terms of a subtraction to be two squares, you can factor it as the sum of the roots multiplied by the difference of the roots.
In this case, the squares are obvious:
is the square of
, and
is the square of 
So, we can factor the expression as
![(x+2)^2 - (y+2)^2 = [(x+2)+(y+2)] - [(x+2)+(y+2)]](https://tex.z-dn.net/?f=%20%28x%2B2%29%5E2%20-%20%28y%2B2%29%5E2%20%3D%20%5B%28x%2B2%29%2B%28y%2B2%29%5D%20-%20%5B%28x%2B2%29%2B%28y%2B2%29%5D%20)
(the round parenthesis aren't necessary, I used them only to make clear the two terms)
We can simplify the expression summing like terms:
![(x+2)^2 - (y+2)^2 = [(x+2)+(y+2)][(x+2)-(y+2)] = (x+y+4)(x-y)](https://tex.z-dn.net/?f=%28x%2B2%29%5E2%20-%20%28y%2B2%29%5E2%20%3D%20%5B%28x%2B2%29%2B%28y%2B2%29%5D%5B%28x%2B2%29-%28y%2B2%29%5D%20%3D%20%28x%2By%2B4%29%28x-y%29%20)
Answer:
The value of x is -9, 3/4 and 3.
Step-by-step explanation:
In order to find the value of x, tou have to let f(x) equals to 0 :
Let f(x) = 0,
(x-3)(x+9)(4x-3) = 0
x - 3 = 0
x = 3
x + 9 = 0
x = -9
4x - 3 = 0
4x = 3
x = 3/4