Oh goodness! Let x=1. We have (6+a)(14+b)=260. So 84+14a+6b+ab=260. I will leave the rest to you. Just plug in 2 for x and solve.
By the binomial theorem, \(\left(x-\frac{1}{x}\right)^5=x^5-\frac{1}{x^5}-5x^3+\frac{5}{x^3}+10x-\frac{10}{x}=x^5-\frac{1}{x^5}-5(x^3-\frac{1}{x^3})+10(x-1/x)\). Plugging, our expression is equivalent to \(x^5-\frac{1}{x^5}-5(x^3-\frac{1}{x^3})+20=32\rightarrow x^5-\frac{1}{x^5}-5(x^3-\frac{1}{x^3})=12\). Now we just need to find x^3+1/x^3. By the binomial theorem, \(x^3-\frac{1}{x^3}-3(x+1/x)=8\rightarrow x^3-\frac{1}{x^3}=14\). Plugging, \(x^5-\frac{1}{x^5}-70=12\rightarrow \boxed{x^5-\frac{1}{x^5}=82}\)
We use complementary counting: what we don't want. How many options for the first digit? 1,3,4,6,7,8,9=7. Second? 0,1,3,4,6,7,8,9=8. Third? 0,1,3,4,6,7,8,9=8. So 7*8*8=448 in total. There are 900 total 3 digit numbers, so 900-448=452. Ans.
WLOG let f(x) be linear. So f(x)=ax+b. So \(f(x+1)=a(x-1)+b=ax-a+b\rightarrow ax-a+b-ax-b=-a\). Note that this is constant, so f(x) must be quadratic. Let f(x)=ax^2+bx+c. So \(f(x+1)=a(x+1)^2+b(x+1)+c=ax^2+2ax+bx+a+b+c\rightarrow \)\(f(x+1)-f(x)=2ax+c\implies 2a=6\rightarrow \boxed{a=3} \)
(a) \(\frac{d}{dx}7x^{-1}=\boxed{-7x^{-2}}\) by the power rule
I don't understand the other problems
Let there be a liters of full acid, b liters of 10% acid. We have \(a+b=39, \frac{a+0.1b}{a+b}=0.4\rightarrow a+0.1b=15.6\rightarrow \boxed{a=13, b=26}\)
\(\left(\frac{2}{a}+\frac{1}{b}\right)^2=\frac{4}{a^2}+\frac{1}{b^2}+\frac{4}{ab}=81\rightarrow 57+\frac{4}{ab}=81\rightarrow \frac{4}{ab}=24\rightarrow \frac{1}{6}=ab\rightarrow\boxed{3ab=\frac{1}{2}}\)
do you mean 222?
