Given:
The function is
![f(x)=(x+3)^2(x-5)^6](https://tex.z-dn.net/?f=f%28x%29%3D%28x%2B3%29%5E2%28x-5%29%5E6)
To find:
The zeros of the given function.
Solution:
The general form of polynomial is
...(i)
where, a is a constant,
are zeros of respective multiplicities
.
We have,
![f(x)=(x+3)^2(x-5)^6](https://tex.z-dn.net/?f=f%28x%29%3D%28x%2B3%29%5E2%28x-5%29%5E6)
On comparing this with (i), we get
![c_1=-3,m_1=2](https://tex.z-dn.net/?f=c_1%3D-3%2Cm_1%3D2)
![c_2=5,m_2=6](https://tex.z-dn.net/?f=c_2%3D5%2Cm_2%3D6)
It means, -3 is a zero with multiplicity 2 and 5 is a zero with multiplicity 6.
Therefore, the correct option is B.
Answer:
2077.42
Step-by-step explanation:
There are 52 weeks in a year, so there are 26 "biweekly" weeks in a year.
Cory receieves $54013 per year, so he gets 54013 / 26 dollars biweekly.
We can compute that number, getting 2077.42 dollars.
You will save $172.65 if you buy an item listed as 575.60
Answer:
It would be the third answer
Step-by-step explanation:
Answer:
1.38
Step-by-step explanation:
1.7-0.4=1.8
1.8/1.3=1.38
On graph place point at (1,8) and draw a triangle