If you visualize the problem, there are two concentric circles, the pool, and the pool plus the walkway. So, we have to subtract the area of these two concentric circles to find the walkway.
Bigger circle: Pool plus walkway
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diameter = 18 + 4(2) -- this is because there is 2 ft of walkway at each far end
diameter = 26 ft
Area = pi*(26/2)^2
Area = 530.93 ft2
Smaller circle:pool
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diameter = 18 ft
Area = pi*(8/2)^2
Area = 50.27 ft2
Area of walkway
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A = 530.93 - 50.27
A = 480.66 ft2
Then the cost would be
Cost = $4.25 * 480.66
Cost = $2,042.81
Answer:
135°
Step-by-step explanation:
seg AB=seg OB
tan ∠AOB = 1
∠AOB = 45°
∠AOC = 135°
The quotient is x^3 + 4x^2 -x + 1.
Solution:
By polynomial grid division, we start by the divisor 3x + 10 placed on the column headings.
3x 10
x^3 3x^4
We know that 3x^4 must be in the top left which means that the first row entry must be x^3. So the row and column multiply to 3x^4. We use this to fill in all of the first row, multiplying x^3 by the terms of the column entries.
3x 10
x^3 3x^4 10x^3
4x^2
We now got 10x^3 though we want 22x^3. The next cubic entry must then be 12x^3 so that the overall sum is 22x^3.
3x 10
x^3 3x^4 10x^3
4x^2 12x^3
Now we have 40x^2, so the next quadratic entry must be -3x^2 so that the overall sum is 37x^2.
3x 10
x^3 3x^4 10x^3
4x^2 12x^3 40x^2
-x -3x^2 -10x
This time we have -10x, so the next linear entry must be 3x so that the overall sum is 7x.
3x 10
x^3 3x^4 10x^3
4x^2 12x^3 40x^2
-x -3x^2 -10x
1 3x 10
The bottom and final term is 10, which is our desired answer. Therefore, we can now read the quotient off the first column:
3x^4+22x^3+37x^2-7x+10 / 3x + 10 = x^3 + 4x^2 -x + 1