Answer:
Step-by-step explanation:
There are several ways to solve this quartic equation. But since the coefficients, they repeat a=1,b=2,c=1,d=2, but e=0, and they are multiple of each other, then it is more convenient to work with factoring as the method of solving it.
As if it was a quadratic one.
![x^{4}-2x^{3}- x^{2} + 2x = 0\\x(x^{3}-2x^{2}-x+2)=0 \:Factoring \:out\\x[(x^{3}-2x^{2})+(-x+2)]=0 \:Grouping\\x[\mathbf{x^{2}}(x-2)+\mathbf{-1}(x+2)]=0 \:Rewriting\:the\:first\:factor\\x(x^{2}-1)(x-2)\:Expanding \:the \:first \:factor\\x(x-1)(x+1)(x-2)=0\\x=0,x=1,x=-1,x=2\\S=\left \{ 0,-1,1,2 \right \}](https://tex.z-dn.net/?f=x%5E%7B4%7D-2x%5E%7B3%7D-%20x%5E%7B2%7D%20%2B%202x%20%3D%200%5C%5Cx%28x%5E%7B3%7D-2x%5E%7B2%7D-x%2B2%29%3D0%20%5C%3AFactoring%20%5C%3Aout%5C%5Cx%5B%28x%5E%7B3%7D-2x%5E%7B2%7D%29%2B%28-x%2B2%29%5D%3D0%20%5C%3AGrouping%5C%5Cx%5B%5Cmathbf%7Bx%5E%7B2%7D%7D%28x-2%29%2B%5Cmathbf%7B-1%7D%28x%2B2%29%5D%3D0%20%5C%3ARewriting%5C%3Athe%5C%3Afirst%5C%3Afactor%5C%5Cx%28x%5E%7B2%7D-1%29%28x-2%29%5C%3AExpanding%20%5C%3Athe%20%5C%3Afirst%20%5C%3Afactor%5C%5Cx%28x-1%29%28x%2B1%29%28x-2%29%3D0%5C%5Cx%3D0%2Cx%3D1%2Cx%3D-1%2Cx%3D2%5C%5CS%3D%5Cleft%20%5C%7B%200%2C-1%2C1%2C2%20%5Cright%20%5C%7D)
Answer: (10.66)(19.64) = 209.3624
I see the cosine in problem 7 and I see the cotangent in problem eight. So I know that you're talking about trig functions in class. Problem 9 is trying to find out if you know what they're good for. You have a right triangle and you know the lengths of all three sides. So you can easily find the sine or the cosine or the tangent or the cotangent of the angle. Pick one and calculate the number. Then use your calculator to find the angle that has that number for the trig function that you chose.
Answer:
D
Step-by-step explanation:
Not much to do here but just convert it all to decimals
