algebra helper

A:Yes you can, but it is a sort of a tricky process. You have to go into the hard drive and copy the folder. If you need any other info or directions, you can e-mail me.

A:let us group b4 and b3 together and b and 3(b^0) together 5b^4-15b^3-b+3 = 5b^3(b-3)-1(b-3) (take 5b^3 common in 1st 2 and -1 in second 2 =(b-3)(5b^3-1) now 5b^3-1 cannot be factored as rational roots

A:Problem 1: (1/3) + (1/4) = 1/T where T is the time it will take for Fay and Carlo to prepare surgical equipment together. Problem 2: (1/6) + (1/8) = 1/T where T = time it will take the mason and helper to put up a tile wall if they work together. Problem 3: (1/5) + (1/S) = 1/3 where S = time it will take to mow the lawn if Stanley would do it alone. I trust that given the above equations, you can solve for the respective unknowns.

A:There are two distinct ways of proving that equality: (1) Directly from the expression C(n,k) = n!/(k!(n−k)!) C(n,r)C(r,s) = (n!/(r!(n−r)!))(r!/(s!(r−s)!)) = = (n!/(n−r)!)(1/s!(r−s)!) = = (n!/(n−s)!(n−r)!)((n−s)!/s!(r−s)!) = = (n!/s!(n−s)!)((n−s)!/(n−r)!(r−s)!) = = C(n,s)C(n−s,r−s) Where in the last equality we used n−r = n−s−(r−s). (2) Combinatorially; this is a more abstract proof but, mathematically speaking, it's preferred: Suppose that, from among a set of n people, you want to select a committee of r persons and, from this, a subcommittee of s. The number of ways to do this is given by: Choosing the committee first: you have C(n,r) ways of doing this; then selecting the subcommittee from the r persons already selected on the first step: you have C(r,s) ways of doing this. Therefore: Number of ways of selecting committee plus subcommittee = C(n,r)C(r,s) But you have another ways of doing this: suppose that you select the subcommittee first: you have C(n,s) ways of doing this; then completing the committee by selecting r−s persons from the remaining n−s. With this approach, you have: Number of ways of selecting committee plus subcommittee = C(n,s)C(n−s,r−s) If you think about it a little, you'll notice that there is a bijection from the first selection method to the second, so the number of distinct ways of doing it must be the same, so: C(n,r)C(r,s) = C(n,s)C(n−s,r−s)

A:(n C r) * (r C s) = [n!/r!(n-r)!] * [r!/s!(r-s)!] = [n!/s!(n-s)!] * [(n-s)!/(n-r)!(r-s)!] = [n!/s!(n-s)!] * [(n-s)!/((n-s)-(r-s))!(r-s)!] = (n C s) * (n-s C r-s)