has degree 3, so it takes the general form
![P(x)=ax^3+bx^2+cx+d](https://tex.z-dn.net/?f=P%28x%29%3Dax%5E3%2Bbx%5E2%2Bcx%2Bd)
It has a root
of multiplicity 2, which means
divides
exactly, and it has a root of
of multiplicity 1 so that
also is a factor. So
![P(x)=a(x-1)^2(x+3)](https://tex.z-dn.net/?f=P%28x%29%3Da%28x-1%29%5E2%28x%2B3%29)
Expanding this gives
![P(x)=a(x^3+x^2-5x+3)=ax^3+ax^2-5ax+3a](https://tex.z-dn.net/?f=P%28x%29%3Da%28x%5E3%2Bx%5E2-5x%2B3%29%3Dax%5E3%2Bax%5E2-5ax%2B3a)
![\implies\begin{cases}a=b\\-5a=c\\3a=d\end{cases}](https://tex.z-dn.net/?f=%5Cimplies%5Cbegin%7Bcases%7Da%3Db%5C%5C-5a%3Dc%5C%5C3a%3Dd%5Cend%7Bcases%7D)
The
-intercept occurs for
, for which we have
![P(0)=d=2.1](https://tex.z-dn.net/?f=P%280%29%3Dd%3D2.1)
Then
![3a=d\implies a=0.7](https://tex.z-dn.net/?f=3a%3Dd%5Cimplies%20a%3D0.7)
![a=b\implies b=0.7](https://tex.z-dn.net/?f=a%3Db%5Cimplies%20b%3D0.7)
![-5a=c\implies c=-0.14](https://tex.z-dn.net/?f=-5a%3Dc%5Cimplies%20c%3D-0.14)
So we have
![\boxed{P(x)=0.7x^3+0.7x^2-0.14x+2.1}](https://tex.z-dn.net/?f=%5Cboxed%7BP%28x%29%3D0.7x%5E3%2B0.7x%5E2-0.14x%2B2.1%7D)
<h3>
Answer: Choice A</h3>
- Domain: x > 4
- Range: y > 0
========================================================
Explanation:
We want to avoid having a negative number under the square root. Solving
leads to ![x \ge 4](https://tex.z-dn.net/?f=x%20%5Cge%204)
So it appears the domain could involve x = 4 itself; however, if we tried that x value, then we'd get a division by zero error.
So in reality, the domain is x > 4.
-------------
The range of y = sqrt(x) is the set of positive real numbers. So y > 0 is the range for this equation. Shifting left and right does not affect the range, so the range of y = sqrt(x-4) is also y > 0.
We are dividing a positive number (3) over some positive number in the denominator. Overall, the expression
is positive because positive/positive = positive.
Therefore, the range of the given equation is y > 0
-------------
The graph is shown below. We have a vertical asymptote at x = 4 and a horizontal asymptote at y = 0. The green curve is fenced in the upper right corner (northeast corner).
The most appropriate statement is the interquartile range for the Wolverines, 30 is less than the IQR for the panthers, 40.
<h3>What is the correct statement?</h3>
The box plot is used to show the distribution of data. The box plot can be used to determine the range, interquartile range and median of the data set.
The range is the difference between the two ends of the whiskers.
Range for the Wolverines = 96 - 35 = 61
Range for the Panthers = 107 - 33 = 74
The interquartile range is the difference between the first and third lines on the box
IQR for the Wolverines = 85 - 55 = 30
Range for the Panthers = 90 - 50 = 40
To learn more about box plots, please check: brainly.com/question/1523909
#SPJ1
Answer:
D = .44P
Step-by-step explanation:
We need to find the slope of the line
m = (y2-y1)/ (x2-x1)
Using two points
m = (22-4.4) /(50-10)
= 17.6/40
= .44 lb/ in^2 ft
We can use the point slope form of the equation
y-y1 = m(x-x1) where y=D and x=P
D-4.4 = .44 (P-10)
Distribute
D-4.4 = .44P - 4.4
Add 4.4 to each side
D -4.4+4.4 = .44P -4.4+4.4
D = .44P
9514 1404 393
Answer:
r = 14.5
Step-by-step explanation:
AD is the perpendicular bisector of chord EC, so is the diameter of the circle. The radius is half the diameter.
r = AD/2 = (AB +BD)/2 = (25 +4)/2 = 29/2
r = 14.5
__
Segments BE and BC are 10 = √(25·4)