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Phoenix [80]
3 years ago
6

Two computers working together can finish a search in 30 seconds. One of these computers can finish in 50 seconds. How long woul

d it take the second computer to finish the same search if it were working alone?
Mathematics
1 answer:
Natalka [10]3 years ago
3 0
Let x be the first computer. Let y be the first computer. We know that Work = 1 / Time.

If the work together, work can be done in 30 seconds. Thus, we have:
1/x + 1/y = 1/30

If x will finish in 50 seconds, we need to compute how many seconds y can finish the same work. Thus solving for y, we have the equation below:
1/50 + 1/y = 1/30
1/y = 1/30 - 1/50
1/y = 1/75

The second computer can finish the same work in 75 seconds.
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Answer:

The given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Step-by-step explanation:

Here, the given expression is:  (\frac{5}{6} a^{9}p^{5})  ^{3}

Now, starting from the outer most bracket.

As we know :

(abc)^{n}   = (a)^{n} \times (b)^{n}  \times (c)^{n}

and (a^m)^{n} = a^{(m \times n)}

⇒ (\frac{5}{6} a^{9}p^{5})  ^{3} = (\frac{5}{6})^{3} \times (a^{9})^{3}   \times (p^{5}) ^{3}\\

=\frac{125}{216}  \times (a)^{(9\times3) } \times (p)^{(5 \times 3)}\\= \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Hence, the given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

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