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:
$1400
Step-by-step explanation:
x= profit of 6th month
(200+300+400+500+500+x)
-------------------------------------- = 550
6
(1900+x)=3300
x=3300-1900
x=1400
To avoid confusion, remember PEMDAS. This is the order of operation.
Parenthesis, Exponents, Multiply, Divide, Add, Subtract
21 ÷ 7 + 20 * 102 * 2 - 3
No parenthesis and no exponents, proceed to multiplication.
20 * 102 * 2 = 4,080
Then, perform division
21÷7 = 3
Then, do addition
3 + 4080 = 4083
Lastly, do subtraction
4,083 - 3 = 4,080
So, 21 ÷ 7 + 20 * 102 * 2 - 3 = 4,080
THEREFORE THE ANSWER IS 4,080
2
subtract 2x on both sides u get 3x=6 , divide both sides by 3 , x=2
2/3= 6/9
So, it wuld be 1/9+1/9+1/9+1/9+1/9+1/9=6/9
Hope this helps.
