0.33333... is 1/3
0.72727272... /(1/3) = 0.727272727.....*3
100*0.72727272... = 72.72727272....
1* 0.72727272... = 0.7272727....
subtract
99*0.727272727 = 72
0.72727272.... = 72/99 = 8/11
8/11 *3 = 24/11 = 2 and 3/11
Thanks Qube73 for acknowledging Asinus's solution.
I have done it independantly of asinus and got the same answer.
So maybe you would like to explain why you believe our answer is incorrect.
asinus answered one of your questions and you gave him no feed back at all.
You are expected to respond to answerers, especially if they have obviously put time into the answer.
\(f(x)=ax^3+bx^2+cx+d\\ a=1\\ f(x)=x^3+bx^2+cx+d\\ -b=r+s+t\\ c=rs+rt+st\\ -d=rst\)
So I am asked to find c
\(f(x)=x^3+bx^2+cx+d\\ f(6)=6^3+b*6^2+c*6+d\\ f(6)=216+36b+6c+d\\~\\ f(-6)=-216+36b-6c+d\\~\\ f(6)-f(-6)=432+12c=7-2\\ 12c=-427\\ c=35\frac{7}{12}\)
You need to check for careless errors.
Cube the monomial (2b6m4)3
(2b^6m^4)^3
\((2b^6m^4)^3\\ =2^3(b^6)^3(m^4)^3\\ =8b^{18}m^{12}\)
Thanks Alan,
That was good revision for me. :)
Thanks guest.
14 is correct. I think guest did it with a calculator.
This is fine but I will give a non-calculator answer
Find N such that 27*N (mod29) = 1
\(27\equiv-2 \mod 29\\ 30\equiv 1 \mod 29\\ -2*-15 = 30 \equiv 1 \mod29\\ -15\equiv 29-15 \mod29 \equiv 14 \mod29\\ so\\ 27*14=1 \mod29\\ \text{So 14 is the modular inverse of 27 (mod 29)}\)
You question makes no sense to me.
For instance what is b.0. ??
number of games = 3n
number of wins = (n-4)*2 n is geater or equal to 4
percent win = [ 2(n-4)/3n ] *100
\(\frac{2n-8}{3n}*100\ge63\\ (2n-8)*100\ge189n\\ 200n-800\ge189n\\ 11n\ge800\\ n\ge72.7\)
So I get 73 weeks
I have no idea if anbody has actually answered this yet but here is a 'proper' answer.
We can see that 1000343 = 1000000 + 343 = \(100^3+7^3\)
I know that \(a^3+b^3=(a+b)(a^2-ab+b^2)\)
a=100 and b=7
1000343=(100+7)(10000-700+49)
1000343=(107)(9349)
+118690 
