The width would be 236 and the lengths would be 118
Use the equations 2L+W=472 and W*L=MAX
Change the first equation to W=472-2L and plug this into the other equation
(472-2L)(L)=MAX
472L-2L^2=M (take derivative)
472-4L=0 (set to 0 to find the max value)
4L=472
L=118
Plug into original to get W=236
Hope this helps!
Answer:
4.3
Step-by-step explanation:
6.8^2-5.3^2=x^2
AC=4.3
Notice that
13 - 9 = 4
17 - 13 = 4
so it's likely that each pair of consecutive terms in the sum differ by 4. This means the last term, 149, is equal to 9 plus some multiple of 4 :
149 = 9 + 4k
140 = 4k
k = 140/4
k = 35
This tells you there are 35 + 1 = 36 terms in the sum (since the first term is 9 plus 0 times 4, and the last term is 9 plus 35 times 4). Among the given options, only the first choice contains the same amount of terms.
Put another way, we have
![\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k=0}^{35} (9 + 4k)](https://tex.z-dn.net/?f=%5Cdisplaystyle%209%20%2B%2013%20%2B%2017%20%2B%20%5Ccdots%20%2B%20149%20%3D%20%5Csum_%7Bk%3D0%7D%5E%7B35%7D%20%289%20%2B%204k%29)
but if we make the sum start at k = 1, we need to replace every instance of k with k - 1, and accordingly adjust the upper limit in the sum.
![\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k-1=0}^{35+1} (9 + 4(k-1))](https://tex.z-dn.net/?f=%5Cdisplaystyle%209%20%2B%2013%20%2B%2017%20%2B%20%5Ccdots%20%2B%20149%20%3D%20%5Csum_%7Bk-1%3D0%7D%5E%7B35%2B1%7D%20%289%20%2B%204%28k-1%29%29)
![\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k=1}^{36} (5 + 4k)](https://tex.z-dn.net/?f=%5Cdisplaystyle%209%20%2B%2013%20%2B%2017%20%2B%20%5Ccdots%20%2B%20149%20%3D%20%5Csum_%7Bk%3D1%7D%5E%7B36%7D%20%285%20%2B%204k%29)