Unknown. C is the hypotenuse, but this problem is solvable.
The answer is D. 3.3x2 and 2.75x2 then add the answers together.
The answer is a. Because when word phrases say “a number”, they are referring to the variable. For instance, x is the variable so x is “a number”.
Answer:
240 ft squared
Step-by-step explanation:
there are 3 rectangles so l*w for each of them
(12*6)+(13+6)+(5*6)
72+78+30
sa of a triangle is b*h divided by 2
5*12 divided by 2=30
there are two equal triangles so
180+30+30=240
Firstly, we'll fix the postions where the
women will be. We have
forms to do that. So, we'll obtain a row like:
![\underbrace{\underline{~~~}}_{x_2}W_2 \underbrace{\underline{~~~}}_{x_3}W_3 \underbrace{\underline{~~~}}_{x_4}... \underbrace{\underline{~~~}}_{x_n}W_n \underbrace{\underline{~~~}}_{x_{n+1}}](https://tex.z-dn.net/?f=%5Cunderbrace%7B%5Cunderline%7B~~~%7D%7D_%7Bx_2%7DW_2%20%5Cunderbrace%7B%5Cunderline%7B~~~%7D%7D_%7Bx_3%7DW_3%20%5Cunderbrace%7B%5Cunderline%7B~~~%7D%7D_%7Bx_4%7D...%20%5Cunderbrace%7B%5Cunderline%7B~~~%7D%7D_%7Bx_n%7DW_n%20%5Cunderbrace%7B%5Cunderline%7B~~~%7D%7D_%7Bx_%7Bn%2B1%7D%7D)
The n+1 spaces represented by the underline positions will receive the men of the row. Then,
![x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m~~~(i)](https://tex.z-dn.net/?f=x_1%2Bx_2%2Bx_3%2B...%2Bx_%7Bn-1%7D%2Bx_n%2Bx_%7Bn%2B1%7D%3Dm~~~%28i%29)
Since there is no women sitting together, we must write that
. It guarantees that there is at least one man between two consecutive women. We'll do some substitutions:
![\begin{cases}x_2=x_2'+1\\x_3=x_3'+1\\...\\x_{n-1}=x_{n-1}'+1\\x_n=x_n'+1\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dx_2%3Dx_2%27%2B1%5C%5Cx_3%3Dx_3%27%2B1%5C%5C...%5C%5Cx_%7Bn-1%7D%3Dx_%7Bn-1%7D%27%2B1%5C%5Cx_n%3Dx_n%27%2B1%5Cend%7Bcases%7D)
The equation (i) can be rewritten as:
![x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m\\\\ x_1+(x_2'+1)+(x_3'+1)+...+(x_{n-1}'+1)+x_n+x_{n+1}=m\\\\ x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-(n-1)\\\\ x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-n+1~~~(ii)](https://tex.z-dn.net/?f=x_1%2Bx_2%2Bx_3%2B...%2Bx_%7Bn-1%7D%2Bx_n%2Bx_%7Bn%2B1%7D%3Dm%5C%5C%5C%5C%0Ax_1%2B%28x_2%27%2B1%29%2B%28x_3%27%2B1%29%2B...%2B%28x_%7Bn-1%7D%27%2B1%29%2Bx_n%2Bx_%7Bn%2B1%7D%3Dm%5C%5C%5C%5C%0Ax_1%2Bx_2%27%2Bx_3%27%2B...%2Bx_%7Bn-1%7D%27%2Bx_n%2Bx_%7Bn%2B1%7D%3Dm-%28n-1%29%5C%5C%5C%5C%0Ax_1%2Bx_2%27%2Bx_3%27%2B...%2Bx_%7Bn-1%7D%27%2Bx_n%2Bx_%7Bn%2B1%7D%3Dm-n%2B1~~~%28ii%29)
We obtained a linear problem of non-negative integer solutions in (ii). The number of solutions to this type of problem are known: ![\dfrac{[(n)+(m-n+1)]!}{(n)!(m-n+1)!}=\dfrac{(m+1)!}{n!(m-n+1)!}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5B%28n%29%2B%28m-n%2B1%29%5D%21%7D%7B%28n%29%21%28m-n%2B1%29%21%7D%3D%5Cdfrac%7B%28m%2B1%29%21%7D%7Bn%21%28m-n%2B1%29%21%7D)
[I can write the proof if you want]
Now, we just have to calculate the number of forms to permute the men that are dispposed in the row: ![m!](https://tex.z-dn.net/?f=m%21)
Multiplying all results:
![n!\times\dfrac{(m+1)!}{n!(m-n+1)!}\times m!\\\\ \boxed{\boxed{\dfrac{m!(m+1)!}{(m-n+1)!}}}](https://tex.z-dn.net/?f=n%21%5Ctimes%5Cdfrac%7B%28m%2B1%29%21%7D%7Bn%21%28m-n%2B1%29%21%7D%5Ctimes%20m%21%5C%5C%5C%5C%0A%5Cboxed%7B%5Cboxed%7B%5Cdfrac%7Bm%21%28m%2B1%29%21%7D%7B%28m-n%2B1%29%21%7D%7D%7D)