Answer:
-2x-10
Step-by-step explanation:
just distribute the -2 :)
please mark me as brainlist
Answer:
C(p) = 4,96 (in thousands of dollars)
l = 2980 $ invest in labor
k = 2980 $ invest in equipment
Step-by-step explanation:
Information we have:
Monthly output P = 450*l*k ⇒ k = P/450*l
But the production need to be 4000
Then k = 4000/450*l
Cost of production = l * k (in thousands of dollars)
C(l) = l + 4000/450*l
Taking derivatives (both members of the equation)
C´(l) = 1 - 400 /45*l² ⇒ C´(l) = 0 ⇒ 1 - 400/45l² = 0
45*l² - 400 = 0 ⇒ l² = 400/45
l = 2.98 (in thousands of dollars)
l = 2980 $ And
k = 400/45*l ⇒ k 400/45*2.98
k = 2.98 (in thousands of dollars)
C(p) = l + k
C(p) = 2980 + 2980
C(p) = 5960 $
a = amount invested at 5%
b = amount invested at 2%
now, we know the total invested by Mr Wilson was 20000, so whatever "a" and "b" might be, we know that a + b = 20000.
![a + b = 20000\implies b = 20000-a \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{5\% of a}}{\left( \cfrac{5}{100} \right)a}\implies 0.05a~\hfill \stackrel{\textit{2\% of b}}{\left( \cfrac{2}{100} \right)a}\implies \begin{array}{llll} 0.02b\\\\ \stackrel{substituting}{0.02(20000-a)} \end{array}](https://tex.z-dn.net/?f=a%20%2B%20b%20%3D%2020000%5Cimplies%20b%20%3D%2020000-a%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B5%5C%25%20of%20a%7D%7D%7B%5Cleft%28%20%5Ccfrac%7B5%7D%7B100%7D%20%5Cright%29a%7D%5Cimplies%200.05a~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B2%5C%25%20of%20b%7D%7D%7B%5Cleft%28%20%5Ccfrac%7B2%7D%7B100%7D%20%5Cright%29a%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%200.02b%5C%5C%5C%5C%20%5Cstackrel%7Bsubstituting%7D%7B0.02%2820000-a%29%7D%20%5Cend%7Barray%7D)
now, we know the total of earned interest is 550 bucks, so then
