The answer is 35 i think but i am not sure
Answer:
Step-by-step explanation:
we are given
(A)
(f×g)(x)=f(x)*g(x)
now, we can plug it
we can simplify it
(B)
Domain:
Firstly, we will find domain of f(x) , g(x) and (fxg)(x)
and then we can find common domain
Domain of f(x):
we know that f(x) is undefined at x=0
so, domain will be
∪
Domain of g(x):
Since, it is polynomial
so, it is defined for all real values of x
now, we can find common domain
so, domain will be
∪..............Answer
Range:
Firstly, we will find range of f(x) , g(x) and (fxg)(x)
and then we can find common range
Range of f(x):
we know that range is all possible values of y for which x is defined
since, horizontal asymptote will be at y=0
so, range is
∪
Range of g(x):
Since, it is quadratic equation
so, its range will be
now, we can find common range
so, range will be
∪.............Answer
Answer:
= 3.134 × 10^9
(scientific notation)
= 3.134e9
(scientific e notation)
= 3.134 × 10^9
(engineering notation)
(billion; prefix giga- (G))
= 3134000000
(real number)
Step-by-step explanation:
Answer: C
Step-by-step explanation:
For this problem, we need to know the standard form of a sine function and the meaning of each part.
Standard form: ![y=asin[b(x-h)]+k](https://tex.z-dn.net/?f=y%3Dasin%5Bb%28x-h%29%5D%2Bk)
a=amplitude
b=period
h=phase shift
k=vertical replacement/shifting
Now that we know the standard form and the components, we know that we can forget about k and plug in 0 for h. This would leave us with ![y=asin[b(x)]](https://tex.z-dn.net/?f=y%3Dasin%5Bb%28x%29%5D)
. We know that the amplitude is 2, therefore, a=2. To find the period, you divide 2π by the given period. 
, therefore, b=1/2.
[plug in a=2]
![y=2sin[b(x)]](https://tex.z-dn.net/?f=y%3D2sin%5Bb%28x%29%5D)
[plug in b=1/2]
Therefore, C is the correct answer.
