donselma

thanks, not sure how I did that one . . . guess you're right . . .

yep . . . the negative (technically inverse signs) cancel out . . . then just remember the exponent rule that you multiple one exponent by y when it is to exponent y . . .

it's c . . .

you can't just solve by inspection (save your time!) . . . 36 is way too long, but it has to be less than 120 b/c the new machine can do that on its own . . . hence, (B) . . .

problem with 10/2 and 9/2 is that x/y isn't greater than 1, it's simply 10/9 . . .

yeah, what do they mean by roots, numbers that solve the equation?

C too

I think for (1) you can solve with just (B). You can set up a single equation with one variable . . . You have x +400 = x/.75

yeah, Mike, since we're just finding a percent, a ratio, the actual number doesn't it matter . . . it doesn't matter if it was 100 miles or 2 million miles, the percent will still be the same so we don't need (A). B tells us what that percent is by giving us y, the proportion of the time traveled at each rate . . .

okay, since x/y > 1 (B) we know that /x/>/y/ . . . Thus, the only way 2x - 2y could be 1, is for both to be positive b/c if both we negative the answer could be no greater -1 if both were negative because /X/>/y/ . . . try an example if it's 2(-3) - 2y = 1, y would have to be -3/2, but then there's no one (B) x/y > 1 could hold . . .

(A) gives us the ration of m:w:c, but we don't know out of how many (B) tells us the number of women . . . together with the ratio we found out in (A) we can determine the number of men . . .

first, foil it out . . . (n^2 - 1)/24 . . . so could be any number since we don't know anything about n yet, but it's easier to work with . . . (A) Tells us a lot, not everything we need to know but we can get rid of half of our 24 choices (0-23, since a remainder of 24 would actually just be no remainder) . . . so only odds are possible choices, since n must be odd as it is not divisible by two . . . (B) Doesn't help b/c we can no longer assume (A) which means it could be some even numbers . . . © Put them together . . . can eliminate anything with a three or a two, so all even's plus 3,9,15,21 . . . leaving only numbers than only have factors of 5, 7, 11, 13, 17, 19 . . . and other prime numbers greater than 3 . . . any prime number greater than 3 divided by 24 has a remainder of . . . okay, I'm officially lost . . . but looking back, we know the number has to be even because it's going to be a prime > 3 squared, which is an odd, and then have one subtracted, making it an even . . .

okay, definetly © . . . (A) sounds good b/c it means it can't be something to the zero power, which eliminates every possible choice except 1 . . . but here's the trick, also -1, since -1^2 will give us 1 . . . that's why (A)'s no good . . . add in (B) that z has to be greater than zero and 1 is the only possible number it could be . . . n could still be anything since 1 to any power is still 1, but we are only asked to solve for z . . .

sorry, Mike, I almost had it but it got erased . . . pretty much you know it can be a=8 or 7 or 6 because they are all much greater than 8!, and they said 8! has to be a multiple! 8!=2**3*2*2(4)*5*3*2(6)*7*2*2*2(8), or 2^6*3^2*5*7 . . . 8^6 would be way too high as would 7 or 6 . . . 4^6 is still going to get you something too high because 3^2*5*7 is less than 350 and 4^6 (which is 2^12 so pull out 2^6) . . . ahh! okay, I don't know!!! But I can tell you it's definetly not (D) b/c based on (A) it could be 2^6 or 8^2 or even 4^4 (unless assuming distinctness) . . .

hey, Mike . . . tricky one . . . go ahead and do some algebra . . . You'll get m+x > m + (mx)/n Now subtract the m's . . . Left with x > (mx)/n Following so far? Now you can divide by (1/x) . . . That will give you 1 > m/n I honestly don't know why we need (2) since the x's will divide out, I'm not really an expert, but a root can't be getting lost could it . . . how does x tell us anything, because if x is negative the other side will be number also (-x) on the one side versus -xm/(n) on the other . . . the only thing I can think is that the x could equal zero and that's why we need to know because in that case the two sides could be equal to each other, but then if x is non-negative the left hand side will be greater since m/n is a fraction . . .