Well, let's see . . .
You said that (a) men can dig (c) holes in (b) hours.
So . . . It takes (a) men (b / c) hours to dig one hole.
And . . . It takes One man (a b / c) hours to dig one hole.
Now . . . There are (b) holes to be dug.
It would take one man (a b / c)·(b) = (a b² / c) hours to dig them all.
But if you had 'x' men, it would only take them (c) hours to do it.
So c = (a b² / c) / x
x c = (a b² / c)
x = a b² / c² men .
Now, I lost the big overview while I was doing that,
and just started following my nose through the fog.
So I have to admit that I'm not that confident in the answer.
But gosh durn it. That's the answer I got, and I'm stickin to it.
Rrarrup !
11% = 11/100
2/9 + 11/100 + 1/10 = 389/900
389/900 x 350 = 151.3
350 - 151.3 = 198.7
Let 
. Then
lies in the second quadrant, so
So we have
and the fourth roots of are
where 
. In particular, they are
Answer:
(2, 1)
Step-by-step explanation:
The best way to do this to avoid tedious fractions is to use the addition method (sometimes called the elimination method). We will work to eliminate one of the variables. Since the y values are smaller, let's work to get rid of those. That means we have to have a positive and a negative of the same number so they cancel each other out. We have a 2y and a 3y. The LCM of those numbers is 6, so we will multiply the first equation by a 3 and the second one by a 2. BUT they have to cancel out, so one of those multipliers will have to be negative. I made the 2 negative. Multiplying in the 3 and the -2:
3(-9x + 2y = -16)--> -27x + 6y = -48
-2(19x + 3y = 41)--> -38x - 6y = -82
Now you can see that the 6y and the -6y cancel each other out, leaving us to do the addition of what's left:
-65x = -130 so
x = 2
Now we will go back to either one of the original equations and sub in a 2 for x to solve for y:
19(2) + 3y = 41 so
38 + 3y = 41 and
3y = 3. Therefore,
y = 1
The solution set then is (2, 1)
