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vampirchik [111]
3 years ago
9

You have a 10 gallon jug and a 3 gallon jug. Both jugs are unmarked. You need exactly 5 gallons of water. How can you get it if

a water faucet is available?
Mathematics
1 answer:
Lady_Fox [76]3 years ago
8 0

Answer:

Step-by-step explanation:

You have a 10 gallon jug and a 3 gallon jug and need exactly 5 gallons of water.

This what you do:

- Fill the 3 gallons jug full and pour into the 10 gallons keg

- Repeat the above a second time making a total of 6 gallons of water in the 10 gallons jug.

- Fill the 3 gallons jug a third time and pour into the 10 gallons jug making 9 gallons

- Fill the 3 gallons jug the fourth time and pour into the 10 gallons jug till full leaving 2 gallons of water remaining in the 3 gallons jug.

- Then empty out the 10 gallons jug and pour the 2 gallons of water in the 3 gallons jug into the 10 gallons jug.

- Then, fill the 3 gallons jug again and pour into the 10 gallons jug holding the 2 gallons of water making a total of 5 gallons of water in the 10 gallons jug.

Hope, that helps!!!

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Using the quotient property we would have

\begin{gathered} \sqrt{\frac{75r^9}{8^8}}=\frac{\sqrt{75r^9}}{\sqrt{8^8}} \\  \end{gathered}

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From the simplified expression as shown above

\frac{\sqrt{75r^9}}{\sqrt{8^8}}=\frac{\sqrt{3\times25\times r^9}}{\sqrt{(2^3)^8}}

Thus;

\frac{\sqrt{3\times25r^9}}{\sqrt{(2^3)^8}}=\frac{\sqrt{25}\times\sqrt{3}\times\sqrt{r^9}}{\sqrt{2^{^3\times8}}}=\frac{5\sqrt{3r^9}}{\sqrt{2^{24}}}

Therefore

\begin{gathered} Using\text{ fraction index law we could simplify the denominator} \\ \frac{5\sqrt{3r^9}}{2^{\frac{24}{2}}}=\frac{5\sqrt{3r^9}}{2^{12}} \end{gathered}

We can not simplify the 3 and the r raised to power of 9 as their power is not even, hence the final answer is given below

\frac{5\sqrt{3r^9}}{2^{12}}=\frac{5\sqrt{3r^9}}{4096}

The final answer is :

\frac{5\sqrt{3r^9}}{4096}

8 0
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