The function in variable w giving the cost C (in dollars) of constructing the box is C(w) = 30w² + 270/w. The result is obtained by using the formula of volume and area of the box.
<h3>How to determine the function?</h3>
We have a rectangular storage container without a lid.
- Volume, V = 10 m³
- Length, l = 2w
- Width, w = w
- Base costs $15/m²
- Sides costs $9/m²
The formula of volume of the box is
V = l × w × h
Where
- l = length
- w = width
- h = height
So, the height is
10 = 2w × w × h
10 = 2w² × h
h = 10/2w²
h = 5/w²
To find the total cost, calculate the area of base and sides of the box!
See the picture in the attachment!
The base area is
A₁ = 2w × w = 2w² m²
The sides area is
A₂ = 2(2wh + wh)
A₂ = 2(3wh)
A₂ = 6wh
A₂ = 6w(5/w²)
A₂ = 30/w m²
The total cost is
C = $15(2w²) + $9(30/w)
C = $30w² + $270/w
The function of the total cost is
C(w) = 30w² + 270/w
Hence, the function of constructing the box is C(w) = 30w² + 270/w.
Learn more about function of area here:
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Answer:
Step-by-step explanation:
The form of the equation you are given is called "point-slope" form. The "slope" in this case is the per-hour fee. The point is (9 h, $195). Point-slope form generally looks like this:
y -k = m(x -h) . . . . . line with slope m through point (h, k)
Here, you have m=15, (h, k) = (9, 195), so the equation looks like ...
y -195 = 15(x -9)
__
The "one-time fee" is the cost when hours are zero.
y -195 = 15(0 -9)
y = 195 -9(15) = 60 . . . . add 195 to both sides, and evaluate
The one-time fee is $60.
Answer:
area of the two faces plus sides
Answer:
D. The graph of G(x) is the graph of F(x) flipped over the y-axis and
stretched vertically.
Step-by-step explanation:
since there is a negative, there must be a reflection. the reflection is over the y-axis because it is outside the x-value (-F(x) not F(-x))
the same goes for the vertical stretch. we know it is a vertical stretch because the 3 is outside the x-value (3F(x) not F(3x))
Answer:
13 and 16
Step-by-step explanation:
let the 2 parts be x and y, then
x + y = 29 → (1) and
x² + y² = 425 → (2)
From (1) → x = 29 - y → (3)
substitute x = 29 - y into (2)
(29 - y)² + y² = 425 ( expand factor )
841 - 58y + y² + y² = 425 ( rearrange into standard form )
2y² - 58y + 416 = 0 ← in standard quadratic form
divide all terms by 2
y² - 29y + 208 = 0
Consider the factors of 208 which sum to - 29
These are - 13 and - 16, hence
(y - 13)(y - 16) = 0
equate each factor to zero and solve for y
y - 13 = 0 ⇒ y = 13
y - 16 = 0 ⇒ y = 16
substitute these values into (3)
x = 29 - 13 = 16 and x = 29 - 16 = 13
The 2 parts are 13 and 16
