Exponents can make the problem smaller and look less confusing. If you see a problem with 5*5*5*5*5*5*5*5 then you're going to eventually loose count and will have to keep repeating. But if you put 5 to the 8th power then you know right away just to multiply 5, (8) times,
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)
Width = 5, length = 9
Area = length * width
length = (2 * width) - 1
sub equation for length
45 = (2w - 1) * w
45 = 2w^2 - w
2w^2 - w - 45 = 0
(2w - 9)(w - 5)
2w - 9 = 0
w = 9/2, l = 8 no
w -5 = 0
w = 5, l = 9 yes
By adding 20+20+19+13+x and dividing by 5 you will receive the answer. Therefore by substituting numbers, 8 will give you the correct average.
20+20+19+13+8 = 80
80/5 = 16.
Therefore x = 8
If I have helped you out, help me out by making me the brainliest answer.
Answer:
The answer is A B E
Step-by-step explanation:
