First computer takes x min. For 1 min it does 1/x part of work.

Second computer takes (x+24) min. For one min it does 1/(x+24) part of work.

Both computers for one min do (1/x+1/(x+24)) part of work.

At the same time, both computers for one min do 1/9 part of work.

$\frac{1}{x} + \frac{1}{x+24} = \frac{1}{9} 
\\ \\ \frac{x+24+x}{x(x+24)} = \frac{1}{9} 
\\ \\9(2x+24)=x^{2} +24x
\\ \\ 18x+216=x^{2} +24x
\\ \\ x^{2}+6x-216=0
\\ \\ D=b^{2}-4ac=36+4*216 = 900
\\ \\ x= \frac{-b+/- \sqrt{D} }{2a} = \frac{-6+/-30}{2} 
\\ \\ x_{1}=-18, x_{2}=12

We can use only positive number here. So, 

for the first computer x=12 min,

 for second computer x+24=12+24= 36 min

Answer 12 min and 36 min.$

7 0

3 years ago

A disc of unit radius is tossed at random onto a large rectangular floor, which is tiled with $4 \times 6$ tiles. Find the proba

goldenfox [79]

Answer:

1/3

Step-by-step explanation:

In order for the disc to be entirely contained in a rectangular tile, its center must be at least 1 unit from the nearest edge. Which means there's a 2 by 4 region that the center can lie in.

So the probability is (2×4) / (4×6) = 8/24 = 1/3.

4 0

3 years ago