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P varies directly as q and inversely as r?
P=q/r* * * * *The correct answer is P = k*q/r where k is the constant of proportionality.
Why is the sum of the reciprocals of all of the divisors of a perfect number equal to 2?
Suppose N is a perfect number. Then N cannot be a square number and so N has an even number of factors.Suppose the factors are f(1) =1, f(2), f(3), ... , f(k-1), f(k)=N.Furthermore f(r) * f(k+1-r) = N for r = 1, 2, ... k so that f(r) = N/f(k+1-r)which implies that 1/f(r) = f(k+1-r)/NThen 1/f(1) + 1/(f(2) + ... + 1/f(k)= f(k)/N + f(k-1)/N + ... + f(1)/N= [f(k) + f(k-1) + ... + f(1)] / N= 2N/N since, by definition, [f(k) + f(k-1) + ... + f(1)] = 2N
The square of r increased by a quantity that is fifty times the cube of k-?
The square of r increased by a quantity that is fifty times the cube of k can be written as r squared + 50 (k cubed). It cannot be solved any further.
R varies directly as s find r when s equals 7 and k equals 5?
Since r varies directly with s, we have r = ks (replace k with 5 and s with 7) r = (5)(7) r = 35
If given the 10th term in the above geometric sequence which is 1536 what would you need to do to find the 11th term?
To find any term of a geometric sequence from another one you need the common ration between terms: t{n} = t{n-1} × r = t{1} × r^(n-1) where t{1} is the first term and n is the required term. It depends what was given in the geometric sequence ABOVE which you have not provided us. I suspect that along with the 10th term, some other term (t{k}) was given; in this case the common difference can be found: t{10} = 1536 = t{1} × r^9 t{k} = t{1} × r^(k-2) → t{10} ÷ t{k} = (t{1} × r^9) ÷ (t{1} × r^(k-1)) → t{10} ÷ t{k} = r^(10-k) → r = (t{10} ÷ t{k})^(1/(10-k)) Plugging in the values of t{10} (=1536), t{k} and {k} (the other given term (t{k}) and its term number (k) will give you the common ratio, from which you can then calculate the 11th term: t{11} = t(1) × r^9 = t{10} × r
