Find the remainder when (x3 – 2) is divided by (x – 1). what is the remainder?
2 answers:
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Answer:
Part A
The width of the garden is approximately 3.987 meters
Part B
Please find attached the drawing of the solution
Step-by-step explanation:
Part A
The given parameters are;
The dimensions of the backyard of the home = 25 m by 30 m
The width of the garden to be built in the yard = Uniform width
The shape of the grass left inside = Rectangular
The area of the grass at the center = The area of the garden
Let 'x' represent the width of the garden, we have;
The length of the rectangular grass area at the center, l = 30 - 2·x
The width of the rectangular area of grass at the center, w = 25 - 2·x
The area of the rectangular backyard, A = 30 m × 25 m = 750 m²
The area of the rectangular backyard, A = (The area of the garden) + (The area of the rectangle of grass inside)
The area of the rectangle of grass, GA = (30 - 2·x)·(25 - 2·x) = The area of the garden
The area of the rectangular backyard, A = 750 = (30 - 2·x)·(25 - 2·x) + (30 - 2·x)·(25 - 2·x) = 2 × (30 - 2·x)·(25 - 2·x) = 8·x² - 220·x + 1,500
∴ 750 = 8·x² - 220·x + 1,500
8·x² - 220·x + 1,500 - 750 = 0
8·x² - 220·x + 750 = 0
x = (220 ± √(220² - 4 × 8 × 750))/(2 × 8)
x ≈ 23.513, or x = 3.987
When x = 25.513, the width of the rectangle of grass inside, w = 25 - 2 × 23.513 = -22.026, which is not a natural (physically possible)
Therefore, the possible width of the garden, x ≈ 3.987
Part B
The drawing of the solution created with MS Visio is attached
Answer:
Step-by-step explanation:
We are given:
Since ⇒
Riemann sum is area under the function given. And it is asked to find Riemann sum for the left endpoint.
Note:
If it will be asked to find right endpoint too,
The average of left and right endpoint Riemann sums will give approximate result of the area under and it can be compared with the result of integral of the same function in the interval given.
So,
Result are close but not same, since one is approximate and one is exact; however, by increasing sample rates (subintervals), closer result to the exact value can be found.