Answer:
show the graph .
Step-by-step explanation:
The described picture frame can be visualized into two separate parts. The first area is equal to the area using the outermost dimensions for the length and width.
Area = Length x Height
Area = (20 in) x (14 in)
Area = 280 in²
We are given that the area is only equal to 192 in². We subtract this value from the computed area.
Difference = 280 in² - 192 in²
difference = 88 in²
This area is equal to the area of the hollow space inside the frame. That is equal to,
height = 20 in - 2x
length = 14 in - 2x
The area,
88 = (20 - 2x)(14 - 2x)
Simplify the right hand side of the equation.
88 = 280 - 68x + 4x²
Divide the equation by 4,
22 = 70 - 17x + x²
Transposing,
x² - 17x - 48 = 0
The factors of the equation is 58.2.
Thus, the thickenss is equal to 58.2 in
Answer:
28 units²
Step-by-step explanation:
Assume that your net has the dimensions shown below.
Its area is
2 yellow rectangles = 2(4 × 2) = 2 × 8 = 16 units²
2 grey rectangles = 2(4 × 1) = 2 × 4 = 8 units²
2 green rectangles = 2(2 × 1) = 2 × 2 = <u> 4 units²</u>
TOTAL = 28 units²
The prism's surface area is 28 units².
Answer:
1:3
Step-by-step explanation:
I think i am right but I might not be hope this helps
a. Answer: D: (∞, ∞)
R: (-∞, ∞)
<u>Step-by-step explanation:</u>
Theoretical domain is the domain of the equation (without an understanding of what the x-variable represents).
Theoretical range is the range of the equation given the domain.
c(p) = 25p
There are no restrictions on the p so the theoretical domain is All Real Numbers.
Multiplying 25 by All Real Numbers results in the range being All Real Numbers.
a) D: (∞, ∞)
R: (-∞, ∞)
*********************************************************************************
b. Answer: D: (0, 200)
R: (0, 5000)
<u>Step-by-step explanation:</u>
Practical domain is the domain of the equation WITH an understanding of what the x-variable represents.
Practical range is the range of the equation given the practical values of the domain.
The problem states that p represents the number of cups. Since we can't have a negative amount of cups, p ≥ 0. The problem also states that Bonnie will purchase a maximum of 200 cups. So, 0 ≤ p ≤ 200
The range is 25p → (25)0 ≤ (25)p ≤ (25)200
→ 0 ≤ 25p ≤ 5000
b) D: (0, 200)
R: (0, 5000)
