-9x-5y=12
C) (2, -6) -18+30=12
An outlier<span> is an observation that lies an abnormal distance from other values in a random sample from a population. In a sense, this definition leaves it up to the analyst (or a consensus process) to decide what will be considered abnormal. Before abnormal observations can be singled out, it is necessary to characterize normal observations.
Basically the ones that are far away from the others.
Thus, the outliers for this graph are K and F
</span>
Answer:
Choice A) 
.
Step-by-step explanation:
What are the changes that would bring 
to 
?
- Translate to the left by
unit, and - Stretch vertically (by a factor greater than .)
. The choices of listed here are related to :
- Choice A)
; - Choice B)
; - Choice C)
; - Choice D)
.
The expression in the braces (for example 
as in ) is the independent variable.
To shift a function on a cartesian plane to the left by 
units, add to its independent variable. Think about how 
, which is to the left of , will yield the same function value.
Conversely, to shift a function on a cartesian plane to the right by units, subtract from its independent variable.
For example, 
is unit to the left of . Conversely, 
is unit to the right of . The new function is to the left of . Meaning that should should add to (rather than subtract from) the independent variable of . That rules out choice B) and D).
- Multiplying a function by a number that is greater than one will stretch its graph vertically.
- Multiplying a function by a number that is between zero and one will compress its graph vertically.
- Multiplying a function by a number that is between
and zero will flip its graph about the -axis. Doing so will also compress the graph vertically. - Multiplying a function by a number that is less than will flip its graph about the -axis. Doing so will also stretch the graph vertically.
The graph of is stretched vertically. However, similarly to the graph of this graph , the graph of increases as increases. In other words, the graph of isn't flipped about the -axis. should have been multiplied by a number that is greater than one. That rules out choice C) and D).
Overall, only choice A) meets the requirements.
Since the plot in the question also came with a couple of gridlines, see if the points 
's that are on the graph of fit into the expression 
.
]Eigenvectors are found by the equation
implying that
. We then can write:
![A-\lambda I = \left [ \begin{array}{cc} 4-\lambda & 2 \\ 5 & 1-\lambda \end{array}\right ]](https://tex.z-dn.net/?f=%20A-%5Clambda%20I%20%3D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%204-%5Clambda%20%26%202%20%5C%5C%205%20%26%201-%5Clambda%20%5Cend%7Barray%7D%5Cright%20%5D%20)
And:
Gives us the characteristic polynomial:
So, solving for each eigenvector subspace:
![\left [ \begin{array}{cc} 4 & 2 \\ 5 & 1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} -x \\ -y \end{array} \right ]](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%204%20%26%202%20%5C%5C%205%20%26%201%20%5Cend%7Barray%7D%20%5Cright%20%5D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bc%7D%20x%20%5C%5C%20y%20%5Cend%7Barray%7D%20%5Cright%20%5D%20%3D%20%5Cleft%20%5B%20%5Cbegin%7Barray%7D%7Bc%7D%20-x%20%5C%5C%20-y%20%5Cend%7Barray%7D%20%5Cright%20%5D%20)
Gives us the system of equations:
Producing the subspace along the line
We can see then that 3 is the answer.
Answer:
2x^2(3x^2-2x-1)
Step-by-step explanation:
12x^4 - 8x^3 - 4x^2
everything is a multiple of 4 so divide the whole thing by 4
12x^4 - 8x^3 - 4x^2/4= 3x^4-2x^3-1x^2
write it in disturbed form
2(3x^4-2x^3-1x^2)
now if u see we can see that the X will be same, so we also have to factor it out and 2 is the exponent that all can give
2x^2(3x^2-2x-1)
