Answer:
6.66 yards of ribbon.
Step-by-step explanation:
If she has 10 yards of ribbon and divides it equally into three parts, those 3 parts would each be 3.33 yards long or 3 (1/3) yards long.
If she uses two of those parts on gifts, then she would use 3.33 x 2 or 6.66 yards of ribbon.
Answer:
Step-by-step explanation: I hope you understand this better
Answer:
C(p) = 4,96 (in thousands of dollars)
l = 2980 $ invest in labor
k = 2980 $ invest in equipment
Step-by-step explanation:
Information we have:
Monthly output P = 450*l*k ⇒ k = P/450*l
But the production need to be 4000
Then k = 4000/450*l
Cost of production = l * k (in thousands of dollars)
C(l) = l + 4000/450*l
Taking derivatives (both members of the equation)
C´(l) = 1 - 400 /45*l² ⇒ C´(l) = 0 ⇒ 1 - 400/45l² = 0
45*l² - 400 = 0 ⇒ l² = 400/45
l = 2.98 (in thousands of dollars)
l = 2980 $ And
k = 400/45*l ⇒ k 400/45*2.98
k = 2.98 (in thousands of dollars)
C(p) = l + k
C(p) = 2980 + 2980
C(p) = 5960 $
Answer:
C. 15.6
Step-by-step explanation:
Perimeter of WXYZ = WX + XY + YZ + ZW
Use the distance formula,
to calculate the length of each segment.
✔️Distance between W(-1, 1) and X(1, 2):
Let,


Plug in the values





✔️Distance between X(1, 2) and Y(2, -4)
Let,


Plug in the values





✔️Distance between Y(2, -4) and Z(-2, -1)
Let,


Plug in the values





✔️Distance between Z(-2, -1) and W(-1, 1)
Let,


Plug in the values





✅Perimeter = 2.24 + 6.08 + 5 + 2.24 = 15.56
≈ 15.6
Your question is a little ambiguous, but I am assuming that you meant to say the function 
Thus, I am solving your question based on assuming the function such as

But, it would still clear your concept, no matter what the function is.
Answer:
we conclude that

The graph is also attached.
Step-by-step explanation:
Given the function

We know that the domain of a function is the set of input or argument values for which the function is real and defined.
As the function has no undefined points nor domain constraints.
Thus, the domain is

Therefore, we conclude that

The graph is also attached.