Answer:
the last one D
Step-by-step explanation:
Slope is (y2-y1)/(x2-x1)
(2-2)/(1-1)
slope is 0 which means it is a horizontal line.
Answer:
the expected area of the resulting circular region is 616.38 m²
Step-by-step explanation:
Given that:
otherwise
The expected area of the resulting circular region is:
= ![E(\pi r^2)](https://tex.z-dn.net/?f=E%28%5Cpi%20r%5E2%29)
= ![\pi E (r^2)](https://tex.z-dn.net/?f=%5Cpi%20E%20%28r%5E2%29)
To calculate ![E(r^2)](https://tex.z-dn.net/?f=E%28r%5E2%29)
![E(r^2) = \int\limits^{15}_{13} {r^2} \ f(r) \ dr](https://tex.z-dn.net/?f=E%28r%5E2%29%20%3D%20%5Cint%5Climits%5E%7B15%7D_%7B13%7D%20%7Br%5E2%7D%20%20%5C%20f%28r%29%20%5C%20dr)
![E(r^2) = \int\limits^{15}_{13} \ \dfrac{3r^2}{4}(1-(14-r)^2)dr](https://tex.z-dn.net/?f=E%28r%5E2%29%20%3D%20%5Cint%5Climits%5E%7B15%7D_%7B13%7D%20%5C%20%5Cdfrac%7B3r%5E2%7D%7B4%7D%281-%2814-r%29%5E2%29dr)
![E(r^2) = \dfrac{3}{4} \int\limits^{15}_{13} \ r^2 (1-196-r^2+28r) dr](https://tex.z-dn.net/?f=E%28r%5E2%29%20%3D%20%5Cdfrac%7B3%7D%7B4%7D%20%5Cint%5Climits%5E%7B15%7D_%7B13%7D%20%5C%20r%5E2%20%281-196-r%5E2%2B28r%29%20dr)
![E(r^2) = \dfrac{3}{4} \int\limits^{15}_{13} \ r^2 (28r^3-r^4-195r^2)dr](https://tex.z-dn.net/?f=E%28r%5E2%29%20%3D%20%5Cdfrac%7B3%7D%7B4%7D%20%5Cint%5Climits%5E%7B15%7D_%7B13%7D%20%5C%20r%5E2%20%2828r%5E3-r%5E4-195r%5E2%29dr)
![E(r^2) = \dfrac{3}{4}[\dfrac{28 r^4}{4}-\dfrac{r^5}{5}-\dfrac{195r^3}{3}]^{^{15}}}__{13}](https://tex.z-dn.net/?f=E%28r%5E2%29%20%3D%20%5Cdfrac%7B3%7D%7B4%7D%5B%5Cdfrac%7B28%20r%5E4%7D%7B4%7D-%5Cdfrac%7Br%5E5%7D%7B5%7D-%5Cdfrac%7B195r%5E3%7D%7B3%7D%5D%5E%7B%5E%7B15%7D%7D%7D__%7B13%7D)
![E(r^2) = \dfrac{3}{4} [ \dfrac{28 \times 50625}{4} - \dfrac{759375}{5} - \dfrac{195 \times 3375}{3} ]-[ \dfrac{28 \times 28561}{4} - \dfrac{371293}{5} - \dfrac{195 \times 2197}{3} ]](https://tex.z-dn.net/?f=E%28r%5E2%29%20%3D%20%5Cdfrac%7B3%7D%7B4%7D%20%5B%20%5Cdfrac%7B28%20%5Ctimes%2050625%7D%7B4%7D%20-%20%20%5Cdfrac%7B759375%7D%7B5%7D%20-%20%5Cdfrac%7B195%20%5Ctimes%203375%7D%7B3%7D%20%5D-%5B%20%5Cdfrac%7B28%20%5Ctimes%2028561%7D%7B4%7D%20-%20%5Cdfrac%7B371293%7D%7B5%7D%20-%20%5Cdfrac%7B195%20%5Ctimes%202197%7D%7B3%7D%20%5D)
![E(r^2) = \dfrac{3}{4} [ 354375-151875-219375-199927+74258.6+142805]](https://tex.z-dn.net/?f=E%28r%5E2%29%20%3D%20%5Cdfrac%7B3%7D%7B4%7D%20%5B%20354375-151875-219375-199927%2B74258.6%2B142805%5D)
![E(r^2) = \dfrac{3}{4} [261.6]](https://tex.z-dn.net/?f=E%28r%5E2%29%20%3D%20%5Cdfrac%7B3%7D%7B4%7D%20%5B261.6%5D)
![E(r^2) = 196.2](https://tex.z-dn.net/?f=E%28r%5E2%29%20%3D%20196.2)
Recall:
The expected area of the resulting circular region is:
= ![E(\pi r^2)](https://tex.z-dn.net/?f=E%28%5Cpi%20r%5E2%29)
= ![\pi E (r^2)](https://tex.z-dn.net/?f=%5Cpi%20E%20%28r%5E2%29)
where;
![E(r^2) = 196.2](https://tex.z-dn.net/?f=E%28r%5E2%29%20%3D%20196.2)
Then
The expected area of the resulting circular region is:
= ![\pi \times 196.2](https://tex.z-dn.net/?f=%5Cpi%20%5Ctimes%20196.2)
= 616.38 m²
If the sequence is quadratic, then the <em>n</em>-th term (<em>n</em> ≥ 1) is
![p_n = an^2 + bn + c](https://tex.z-dn.net/?f=p_n%20%3D%20an%5E2%20%2B%20bn%20%2B%20c)
We're given the first four terms,
![\begin{cases}p_1=-4\\p_2=1\\p_3=12\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dp_1%3D-4%5C%5Cp_2%3D1%5C%5Cp_3%3D12%5Cend%7Bcases%7D)
Using the formula for the <em>n</em>-th term, this turns into a system of equations,
![\begin{cases}a+b+c=-4\\4a+2b+c=1\\9a+3b+c=12\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Da%2Bb%2Bc%3D-4%5C%5C4a%2B2b%2Bc%3D1%5C%5C9a%2B3b%2Bc%3D12%5Cend%7Bcases%7D)
Solve the system:
• Eliminate <em>c</em> :
(4<em>a</em> + 2<em>b</em> + <em>c</em>) - (<em>a</em> + <em>b</em> + <em>c</em>) = 1 - (-4) ===> 3<em>a</em> + <em>b</em> = 5
(9<em>a</em> + 3<em>b</em> + <em>c</em>) - (<em>a</em> + <em>b</em> + <em>c</em>) = 12 - (-4) ===> 8<em>a</em> + 2<em>b</em> = 16
• Multiply the second equation by 1/2 to get 4<em>a</em> + <em>b</em> = 8, then eliminate <em>b</em> and solve for <em>a</em> :
(4<em>a</em> + <em>b</em>) - (3<em>a</em> + <em>b</em>) = 8 - 5 ===> <em>a</em> = 3
• Solve for <em>b</em> and <em>c</em> :
3<em>a</em> + <em>b</em> = 9 + <em>b</em> = 5 ===> <em>b</em> = -4
<em>a</em> + <em>b</em> + <em>c</em> = 3 - 4 + <em>c</em> = -4 ===> <em>c</em> = -3
Then the rule for the <em>n</em>-th term is
![p_n = \boxed{3n^2 - 4n - 3}](https://tex.z-dn.net/?f=p_n%20%3D%20%5Cboxed%7B3n%5E2%20-%204n%20-%203%7D)